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Zamach na pałac prezydencki w Kabulu (2013)
 Zamach w Kabulu (2016)
 Zamachy w Afganistanie (10 stycznia 2017) (m.in. w Kabulu)

redpajama_set_name: RedPajamaWikipedia


The Young Leaders Council (YLC) is the peer-elected student leadership body for the YouthBuild field. YLC members launch their three-year terms as current students voted into office at YouthBuild USA's Annual AmeriCorps Conference of Young Leaders. YLC members represent their states at various national trainings and YouthBuild policy-making forums. Both the YLC and the National Alumni Council (NAC) ensure that youth voice plays a central role in decision making and strategic planning for the YouthBuild field. YouthBuild students who serve on The Young Leaders Council (YLC) engage in leadership roles such as policy input to YouthBuild USA, advocacy, support for local programs, activism and action, and self-development. Representatives of the YLC and the NAC serve on the Policy Council of the YouthBuild USA Affiliated Network in equal numbers with directors. The Policy Council sets program design and performance standards for affiliation and accreditation.


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ACCREDITATIONS AND MEMBERSHIPS
British Safety Council Member
The British Safety Council has campaigned tirelessly to protect workers from accidents, hazards and unsafe conditions, and played a decisive role in the political process that has led to adoption of landmark safety legislation in the UK. Its members in more than 60 countries are committed to protecting and improving the well-being of workers believing that a healthy and safe work environment is also good for business.
UK Fire Association Member
The UK-FA exists to support small and medium sized fire trade enterprises, creating a network of experts who provide services and solutions to each other and UK consumers.
The East Anglian Festival Network (EAFN)
exists to bring together all the amazing local people (organisers, workers, venues, suppliers, contractors, traders, performers & Festival goers) that form the community that put on and enjoy over Seven Hundred (700!) wonderful East Anglian Festivals, Fairs, Carnival and Fun Days held each year across East Anglia (Cambs, Essex, Norfolk, Suffolk).
STAFF MEMBERSHIPS AND ACCREDITATION'S
Institute of Occupational Safety and Health (IOSH)
IOSH are a world-leading developer of certificated safety and health training courses. The courses provide insights and practical tools so that individuals and organisations can create safer and healthier working environments. By promoting a safety and health culture in which everyone plays their part, the courses help to reduce sickness absence, improve productivity and bring about other organisational benefits.
International Institute of Risk and Safety Management (IIRSM)
IIRSM is an educational charity and professional body for those managing risk in all its forms. They are independent and speak freely, championing risk management to drive positive change for both employee and organisational well-being. They do this through setting standards, connecting people and organisations, and providing practical education, training, advice and resources.
Institute of Fire Safety Managers (IFSM)
The Institute of Fire Safety Managers is a professional body of highly respected individuals and companies who all share the same objective.
"To raise the awareness of fire safety at a local, national and international level, promoting fire prevention, fire protection and reducing the risk from fire as far as reasonably practicable"
Occupational Safety and Health Consultants Register Ltd (OSHCR)
A network of professional bodies and stakeholders worked together in partnership to develop this register, which has established a benchmark standard for safety consultants and helped to raise the quality of advice being given to businesses.
This minimum standard has been set at a degree level qualification, with at least two years experience and active engagement in a continuing professional development scheme.
Businesses can have confidence in choosing a health and safety consultant from the register, as they are bound by their professional body's code of conduct and are committed to providing sensible and proportionate health and safety advice.
https://www.oshcr.org/member/philip-adams/
Association of First Aiders (AoFA)
The Association of First Aiders – AoFA is an association formed out of the membership of the First Aid Cafe website, UK's largest independent community of over 14,000 first aiders and first aid trainers.
TRAINING APPROVALS
UK Asbestos Training Association (UKATA)
UKATA is a leading authority on asbestos training, a not for profit association established in 2008 with a set purpose in mind: to be recognised as the asbestos industry's most eminent training association.
UKATA is instantly recognised and respected for the exacting standards in asbestos training delivered by its Members in accordance with current HSE Legislation.
Qualifications Network UK (QNUK)
QNUK has been involved with regulated qualifications since 2005. An Awarding Organisation (AO) regulated by Ofqual, Qualifications Wales and CCEA Regulation in Northern Ireland.
Qualifications Network is known as a leading First aid Awarding body / organisation in the UK, however, they also offer a wide range of qualifications in food safety, health and safety, security, alcohol licencing, teaching, assessing and verification.
The Construction Industry Training Board Approved Training
A CITB Approved Training Organisation - or ATO - is an organisation which provides construction training courses and qualifications to a defined and industry-agreed training standard.

redpajama_set_name: RedPajamaCommonCrawl


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Smacking your kid could make them more likely to be an abuser – or a victim
May 4, 2017 Written by Admin
Hitting your children could increase their risk of becoming abusers — or of being victims of abusers, according to new research.
Gender activists say the new study is all the more reason to outlaw corporal punishment in the home to help to break South Africa's cycles of abuse.
As part of a study, 10 178 men and 3 106 women from six low- and middle-income countries in Asia and the Pacific were interviewed about their experiences of violence in the home, whether as children, partners or parents. Researchers argue that the findings may also be applicable to other countries.
The research, published this month in The Lancet Global Health journal, showed that on average at least 52% of participants said they had experienced some form of childhood trauma, such as sexual, physical or emotional abuse. Among women, this was strongly linked to being subjected to intimate partner violence as adults.
The study also found that women who were ill-treated as children, who were in abusive relationships, or who had witnessed their mothers being abused were more likely to resort to harsh parenting practices, such as smacking or beating.
About 75% of women, and slightly fewer men, said they or their partners smacked their children at least sometimes.
Men who witnessed their mothers being abused and who were mistreated as children were also more likely to grow up to be abusive partners. Men's use of corporal punishment in the home was also associated with childhood trauma, researchers found.
But the study also revealed that female partners often have a greater influence on whether men raise their hands to children. The research showed that men are more likely to beat their children if they see their female partners do it first.
Study co-author and director of the gender and health unit of the Medical Research Council (MRC), Rachel Jewkes, says the data illustrate that women aren't only victims of violence. "Violence against children is not only a man's problem. Women are part of what's driving the acceptability of violence in society," she says.
In South Africa, one in five children will experience sexual abuse by the age of 17, according to a 2016 University of Cape Town study published by the Centre for Justice and Crime Prevention.
Meanwhile, a 2009 MRC report found that one in three men admitted to having raped a woman. More than half of these men said they had been abused as children.
Should the government tell you how to raise your kids?
Dean Peacock, the co-executive director of the gender equality advocacy organisationSonke Gender Justice, warns that being raised in an abusive home doesn't necessarily mean children will grow up to be abusers or victims of abuse, and that many factors such as alcohol and inequality contribute to violence.
He says the connection between childhood abuse and violent adults does mean that banning corporal punishment such as beating and smacking in the home could be key to protecting women and children and to ending the vicious cycle of abuse.
"All too often what starts off as discipline ends up as abuse," he says.
Although physical discipline in the home is not against the law, corporal punishment in schools was outlawed in South Africa in 1997. A 2013 study published by the Centre for Justice and Crime Prevention found that nearly 74% of schools in some provinces still reported physically disciplining pupils.
Peacock says allowing corporal punishment in the home sends a clear message that violence is an appropriate response to difficulty.
"Why are we against hitting women but willing to hit children, who are the most vulnerable in society?" he asks.
Although some parents may oppose a government ban on physically disciplining children, evidence and not personal beliefs should inform policy, he says.
Early interventions can help turn the tide
Both Jewkes and Peacock say programmes aimed at young people, whether as new parents or students, could help to change parenting and gender norms that breed violence.
Peacock adds that resources for survivors must also be made available in communities. "Young men who experience violence in the home and don't get help are more likely to perpetrate violence," he says.
Lisa Vetten, a research associate at the Wits Institute for Social and Economic Research, says there might not be the budget for the kind of programmes that are needed. Social welfare budgets have largely stagnated in the past decade and international donors are reluctant to fund programmes in a middle-income country, she says.
Peacock says the Lancet study should also bolster calls for a national plan to tackle violence in South Africa. "This study is one more bit of evidence that should be prompting our government's response to children's experience and exposure to violence. It's a travesty that we have no national strategic plan to address domestic and gender-based violence in South Africa."
child abusecorporal punishmentgender-based violence
[From our archives] Diepsloot: Where men think it's their right to rape
How corporal punishment's legality aggravates child abuse
Cosatu to hold special meeting to address Zuma booing
Greening the Future 2017: How you can save the world


In addition to economic development, Colorado communities are looking to the future by planning for students and tomorrow's workforce. Ballot questions in a number locations asked voters to allow school districts to have the option of investing in telecommunications if necessary. They don't have faith that incumbents will keep up with their growing needs.
Colorado Mountain College, also unsure of the future, asked voters in six different communities for permission to provide their own Internet, if necessary. Voters in all locations said "yes."
Out From Under The "Dark Cloud"
"We didn't know exactly what we'd do," Turner said. "But we no longer are under this dark cloud of not being able to be innovative."
SB 152 first passed through the state legislature in 2005 after heavy lobbying from Comcast and CenturyLink. Legislators and lobbyists backing the law argued its intent was taxpayer protection but the past 10 years have proved otherwise. The real motivation behind the bill was to protect incumbent de facto monopolies and prevent potential competition by municipal networks.
The law hurts taxpayers by discouraging private investment. It prevents local governments from working with private sector ISP partners who may want to use publicly owned fiber infrastructure. It stalls economic development because employers can't get the connectivity they need. It stifles growth in the small communities that need growth the most.
These communities have waited patiently for incumbents to invest in better infrastructure but communities will no longer wait and watch while places like Longmont, Rio Blanco, and Estes Park leave them behind.
Until the State Legislature decides to strike SB 152 and the expensive hoops communities must jump through to opt out of it, places like Fort Collins, Steamboat Springs, and Pitkin County will be forced to spend precious public dollars on this type of referenda.
It's not that we want to compete with the private sector — it's that the private sector isn't providing the level of service the community needs.
Now that these communities have recovered the right to determine their broadband destiny, they have a choice. They can rest in the comfort of knowing they comply with the law or explore endless possibilities now open to them. They can stop pounding drums and start innovating.
Update: We learned about the community of Ophir after publishing our article, which originally reported 43 communities. The correct number, as listed below, is 44 communities.
November 18th update: Virgil Turner from Montrose has been helping us keep tabs on the statewide elections and he sends word that Gunnison County, Ouray County, and Delta County also passed measures to reclaim local authority. Their results were 73 percent, 83 percent, and 76 percent respectively. We originally reported on the cities of Gunnison, Ouray, and Delta, but were not aware of the county referenda. With the addition of these three, the grand total of communities that chose to reclaim local authority this election cycle comes to 47.


text:
export function autoCorrelate(buf, sampleRate) {
    var MIN_SAMPLES = 4;	// corresponds to an 11kHz signal
    var MAX_SAMPLES = 1000; // corresponds to a 44Hz signal
    var SIZE = 1000;
    var best_offset = -1;
    var best_correlation = 0;
    var rms = 0;

    if (buf.length < (SIZE + MAX_SAMPLES - MIN_SAMPLES))
        return -1;  // Not enough data

    for (var i = 0; i < SIZE; i++) {
        var val = (buf[i] - 128) / 128;
        rms += val * val;
    }
    rms = Math.sqrt(rms / SIZE);

    for (var offset = MIN_SAMPLES; offset <= MAX_SAMPLES; offset++) {
        var correlation = 0;

        for (var i = 0; i < SIZE; i++) {
            correlation += Math.abs(((buf[i] - 128) / 128) - ((buf[i + offset] - 128) / 128));
        }
        correlation = 1 - (correlation / SIZE);
        if (correlation > best_correlation) {
            best_correlation = correlation;
            best_offset = offset;
        }
    }
    if ((rms > 0.01) && (best_correlation > 0.01)) {
        return sampleRate / best_offset;
    }
    return -1;
}

export function noteFromPitch(frequency) {
    var noteNum = 12 * (Math.log(frequency / 440) / Math.log(2));
    return Math.round(noteNum) + 69;
}

export function centsOffFromPitch(frequency, note) {
    return Math.floor(1200 * Math.log(frequency / frequencyFromNoteNumber(note)) / Math.log(2));
}

function frequencyFromNoteNumber(note) {
    return 440 * Math.pow(2, (note - 69) / 12);
}

redpajama_set_name: RedPajamaGithub


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Summer Glau
Terminator DVD details, commentary clips
Posted September 15, 2009 by admin
Blu-ray.com posted a review of Terminator: The Sarah Connor Chronicles – The Complete Second Season [Blu-ray] last week, along with a list of the special features (scroll all the way down).
The region 1 DVD comes out next week (Sept. 22). No details on the bonus material for region 2 yet. (The DVD doesn't come out until November.)
Television Without Pity has revealed the winners of the Tubey Awards. The show won in 16 categories, including Best Drama, Best Returning Show, Most Painful Cancellation, Best Single Episode (Allison from Palmdale), Best Season Finale, Best Smackdown (Cameron vs. police station), Best Onscreen Death (Derek) and Favourite Actress (Summer Glau).
Next stop, the Scream Awards – if you haven't voted yet, head over to Spike.com.
In the meantime, some preview clips from the DVD showed up on YouTube yesterday. Embedded below are the ones from Adam Raised a Cain, with commentary from Josh Friedman, Summer Glau, Thomas Dekker, John Wirth and James Middleton. Big thanks to Summer-Glau.net for posting them. (Make sure you visit for more clips.)
Finally, the Blu-ray extras:
Disc One
Disc one features a pair of commentary tracks, the first for the episode Samson and Delilah and featuring Executive Producer Josh Friedman and Actors Lena Heady, Thomas Dekker, Summer Glau, and Shirley Manson. Commentary track two, available on Allison from Palmdale, features Executive Producers Josh Friedman, James Middleton, and John Wirth, and Actors Thomas Dekker and Summer Glau. Also included is The Storyboard Process: Cameron Goes Bad (1080i, 2:55), a brief piece that looks at the importance of storyboards to the filmmaking process, including a side-by-side storyboard vs. final product comparison.
Disc Two
While none of the five episodes found on disc two sport a commentary track, the disc does offer two "Terminated Scenes," one each for the episodes The Tower is Tall but the Fall is Short (1080p, 1:04) and Complications (1080p, 0:26). Also included here under the "Behind the Story" heading is Cameron vs. Rosie: Fight Rehearsal (1080i, 5:27), an informative piece that looks at the process of bringing this fight sequence to life, featuring raw behind-the-scenes footage, interview clips with the crew, and storyboard analysis.
Disc Three
Only a pair of "Terminated Scenes" for Strange Things Happen at the One-Two Point (1080p, 0:50) and Earthlings Welcome Here (1080p, 1:04) are included on disc three.
Disc Four
Like disc three, disc four features only a collection of "Terminated Scenes," this time one from Today is the Day — Part 1 (1080p, 1:19) and four from Today is the Day — Part 2 (1080p, 2:50 combined).
Disc Five
Disc five contains the bulk of the supplementary material, headlined by another pair of commentary tracks. Track one, accompanying Adam Raised a Cain, features Executive Producers Josh Friedman, James Middleton, and John Wirth, and Actors Thomas Dekker and Summer Glau. The second track with the same participants is available for the series finale episode Born to Run. Collision With the Future: Deconstructing the Hunter Killer Attack (1080i) is an interactive piece that takes viewers on an in-depth tour of the many layers of work that came together to create the season's climactic sequence. Viewers may choose from one of four boxes — one each for "Production," "Direction," "Special FX," and "Visual FX" — from a main menu that features all four pieces playing simultaneously with an overlaying commentary with Creator/Executive Producer/Writer Josh Friedman. The Continuing Chronicles: Terminator is a collection of eight featurettes presented in 1080i high definition: Write the Future (12:39), Conceptualization (8:18), Blood and Metal (7:39), Designing Deconstruction (7:32), Choreographing Chaos (7:21), War Stories (9:15), Setting the Tempo (13:15), and Motivations (9:51). Rounding out disc five are three "Terminated Scenes" — two for To the Lighthouse (1080p, 0:43 & 1:32) and one for Born to Run (1080p, 0:27) — and a gag reel (1080p, 6:04). [Blu-ray.com]
Posted in: DVD News, Television News, Terminator: The Sarah Connor Chronicles | Tagged: Adam Raised a Cain, James Middleton, John Wirth, Josh Friedman, Summer Glau, Terminator: The Sarah Connor Chronicles, Thomas Dekker
The Sarah Connor Chronicles nominated for four Saturns
Posted March 10, 2009 by admin
The Saturn nominations were announced earlier today and The Sarah Connor Chronicles is up for Best Network Television Series. Big congratulations to Josh Friedman and his entire team!
The leads – Lena Headey, Thomas Dekker and Summer Glau – all scored nods the acting categories.
Here is the list of TV nominees:
NETWORK TELEVISION SERIES
SYNDICATED/CABLE TELEVISION SERIES
PRESENTATION ON TELEVISION
"24: Redemption"
"The Last Templar"
"The Librarian: The Curse of the Judas Chalice"
ACTOR IN TELEVISION
Bryan Cranston, "Breaking Bad"
Matthew Fox, "Lost"
Michael C. Hall, "Dexter"
Timothy Hutton, "Leverage"
Edward James Olmos, "Battlestar Galactica"
Noah Wiley, "The Librarian: The Curse of the Judas Chalice"
ACTRESS IN TELEVISION
Lena Headey, "Terminator: The Sarah Connor Chronicles"
Jennifer Love Hewitt, "The Ghost Whisperer"
Evangeline Lilly, "Lost"
Mary McDonnell, "Battlestar Galactica"
Anna Paquin, "True Blood"
Kyra Sedgwick, "The Closer"
Anna Torv, "Fringe"
SUPPORTING ACTOR IN TELEVISION
Henry Ian Cusick, "Lost"
Thomas Dekker, "Terminator: The Sarah Connor Chronicles"
Michael Emerson, "Lost"
Josh Holloway, "Lost"
Adrian Pasdar, "Heroes"
Milo Ventimiglia, "Heroes"
SUPPORTING ACTRESS IN TELEVISION
Jennifer Carpenter, "Dexter"
Summer Glau, "Terminator: The Sarah Connor Chronicles"
Yunjin Kim, "Lost"
Elizabeth Mitchell, "Lost"
Hayden Panettiere, "Heroes"
Katee Sackhoff, "Battlestar Galactica"
GUEST STARRING ROLE IN A TELEVISION SERIES
Kristen Bell, "Heroes"
Alan Dale, "Lost"
Kevin Durand, "Lost"
Robert Forster, "Heroes"
Jimmy Smits, "Dexter"
Sonya Walger, "Lost"
Posted in: Television News, Terminator: The Sarah Connor Chronicles | Tagged: Lena Headey, Saturn Awards, Summer Glau, Terminator: The Sarah Connor Chronicles, Thomas Dekker
Terminator – from the Wondercon panel
Posted March 2, 2009 by admin
Josh Friedman, Summer Glau and Shirley Manson were at Wondercon over the weekend to talk about The Sarah Connor Chronicles and io9's Charlie Jane Anders (best T:SCC reviewer ever) has a report on what was shown in the clip reel from the last six episodes as well as some general stuff about the show, e.g. the reason why "Desert Cantos" sucked and something about the last three episodes being among the top four this season.
For the full report and all the spoilers, head over to io9.com.
Here is the part that involves John Henry (highlight to read):
We saw Savannah, Catherine Weaver's daughter skipping and playing hopscotch in ominous gray tunnels, until she comes upon the mysterious room where the artificial intelligence John Henry is. "Would you like to play hide and seek?" John Henry says. Later on, Ellison is demanding to know where Savannah is, and John Henry won't answer. John Henry asks Catherine Weaver what would happen if people knew she wasn't Savannah's real mother, and Catherine asks if she's being threatened.
Ellison is telling Catherine he's sorry, and she says he should be. And then John Connor is decking Ellison and yelling that he'll kill him. And Sarah meets Jesse, but doesn't seem to know who she is. And Sarah meets Catherine Weaver(!).
Posted in: Television News, Terminator: The Sarah Connor Chronicles | Tagged: Josh Friedman, Summer Glau, Terminator: The Sarah Connor Chronicles, Wondercon
Sarah Connor episodes in March/April – from Fox & the Terminator blog
Fox has issued a release with all the March highlights, including info about episodes 2×17 – 2×20 of The Sarah Connor Chronicles.
Here is the combined info about all the remaining episodes from The Futon Critic and the Terminator blog:
TERMINATOR: THE SARAH CONNOR CHRONICLES (8:00-9:01 PM ET/PT) – "Ourselves Alone"
Written by Toni Graphia and Daniel T. Thomsen
Directed by Jeff Woolnough
The return of Cameron's (Summer Glau) glitch results in regrettable consequences for everyone, and Riley (Leven Rambin) fears Cameron has discovered her secret.
TERMINATOR: THE SARAH CONNOR CHRONICLES (8:00-9:01 PM ET/PT) – "Today is the Day, Part 1 of 2"
Written by Ashley E. Miller & Zack Stentz
Directed by Guy Bee
In the first part of a game-changing two-part installment, Jesse (guest star Stephanie Jacobsen) flashes back to a life-altering mission aboard her submarine, the Jimmy Carter. Meanwhile, Sarah (Lena Headey) and John (Thomas Dekker) decide they can't continue living in their house and make plans to leave.
TERMINATOR: THE SARAH CONNOR CHRONICLES (8:00-9:01 PM ET/PT) – "Last Voyage of the Jimmy Carter/Today Is the Day Part 2 of 2"
The exciting two-parter continues as Jesse's submarine mission in the future takes a fatal turn that has unforeseen consequences for John, Sarah and Derek (Brian Austin Green) in the present. Also, Sarah deals with Cameron, who has become a liability, and John steps up and makes a life-or-death choice.
TERMINATOR: THE SARAH CONNOR CHRONICLES (8:00-9:01 PM ET/PT) – "To The Lighthouse"
Written by Natalie Chaidez
Directed by Guy Ferland
Fearing for her life, Sarah stashes John in a safe house with the only person she believes she can rely on, Charley Dixon (guest star Dean Winters). Cameron and Derek struggle to work together after she reveals a big secret, and Weaver's (Shirley Manson) entire effort is at risk when John Henry (Garret Dillahunt) is compromised – but by whom?
TERMINATOR: THE SARAH CONNOR CHRONICLES (8:00-9:01 PM ET/PT) – "Adam Raised a Cain"
Written by Toni Graphia
Directed by Charles Beeson
TERMINATOR: THE SARAH CONNOR CHRONICLES (8:00-9:01 PM ET/PT) – "Born to Run"
Written by Josh Friedman
Directed by Jeffrey Hunt
Posted in: Television News, Terminator: The Sarah Connor Chronicles | Tagged: Ashley Miller, Brian Austin Green, Daniel Thomsen, Dean Winters, Garret Dillahunt, Josh Friedman, Lena Headey, Leven Rambin, Natalie Chaidez, Stephanie Jacobsen, Summer Glau, Terminator: The Sarah Connor Chronicles, Thomas Dekker, Toni Graphia, Zack Stentz


Vanessa Gray, 23, and her newly formed non-profit corporation has been given control of Lincoln Cemetery, a place where she has been caretaking and leading the way for other volunteers for the past 14 months.
On Monday, February 27, a Gulfport woman's preservation dream for a local cemetery became publically true. Her newly formed non-profit corporation has been given control of Lincoln Cemetery, a place where she has been caretaking and leading the way for other volunteers for the past 14 months.
Most of the approximately 6,000 people buried there are black and over 1,000 are American armed forces veterans, according to an August 2, 2016 City of Gulfport resolution that gave Vanessa Gray's efforts a vote of continuing historical significance.
"I came for a special moment for the city – the Lincoln Cemetery resolution – to express my support," said U.S. Congressman David Jolly, R-District 13, at the August 2016 council meeting.
According to a Monday, February 27 Tampa Bay Times news report, a black congregation, Greater Mount Zion AME Church in St. Petersburg, "wanted to take over the neglected property." And, Pinellas County was planning to give the church $90,000 from the BP oil spill settlement to maintain it.
Since December 2015, "I've made it my mission to restore Lincoln Cemetery back to what it was intended to be: A peaceful, serene memorial garden where people can walk through and pay their respects to their loved ones and our fallen heroes buried there," says Gray on the Facebook page dedicated to the Lincoln Cemetery Society.
The Times reported the church learned on February 27, 2017 that property owner Richard S. Alford of St. Petersburg had transferred ownership to Gray's nonprofit named the Lincoln Cemetery Society, Inc. on February 8, 2017.
Gray, 23, incorporated the nonprofit on June 20, 2016 and serves as its president. Her mother, Sharon Butler, 49, is vice president and is also a Gulfport resident.
And, "now that the ownership of the cemetery is clear, the focus cannot just be on maintenance and preservation. [It must also include] developing the funding and infrastructure that will sustain it [into] perpetuity," said Gray in the press release.
Currently, the cemetery faces $31,788.26 in code enforcement liens because the city maintained the property up until May 2016 and the amount keeps accruing interest, said Gulfport City Manager Jim O'Reilly on Wednesday, March 1. In the past, the city assessed the property $878 per month to mow the grass and periodically charged additional fees for special pick ups to take care of larger items like tree limbs.
O'Reilly, Gray and Butler met on Monday, February 27 to address the lien.
"We are treating this like any other private property," said O'Reilly. The city's due diligence is to collect the lien.
Established in 1926, the cemetery is considered an historical treasure covering nine acres at 58th St. S.
For more information, visit lincolncemeterysociety.org.


The RSPCA is appealing for information after a swan was shot with an air rifle in Cheshire.
A concerned member of the public found the bird with a large wound above his leg in a field in Well Lane, Weaverham, on Sunday 5 March. They contacted the RSPCA, and Animal Collection Officer Lauren Bradshaw collected the bird.
Nursery school children in Pembroke got an unexpected education in wildlife – after a swan crash landed into their playground.
RSPCA Cymru rescued the swan, which was found at Golden Manor Nursery, on Golden Lane, on Tuesday 10 January.
The bird was found unable to take-off, due to a lack of space.
She was safely rescued by an RSPCA inspector, and released at the nearby Mill Pond, where she flew away strongly, and showed no signs of distress.
A swan was rescued by the RSPCA after he crash landed in a building site – and got himself wedged behind a brick wall.
The swan was completely stuck when RSPCA Animal Collection Officer (ACO) Cara Gibbon arrived at the site in Dovedale Road, Birmingham, on Wednesday afternoon (4 January).
A swan rescued by RSPCA Cymru – having been found exhausted outside a Morfa shopping centre – has now been returned safely to the wild.
The bird shocked Christmas shoppers at the Morfa Bay precinct on Thursday 1 December, with a sizeable crowd gathering to watch as the exhausted bird sat struggling outside shops.
A cygnet who had got in a bit of a flap on top of a cinema in Tamworth had to be rescued by the RSPCA.
Staff at Odeon cinema, in Bolebridge Street, called the RSPCA after spotting the swan on top of the 50ft-high roof.


Prozone® Ozone Generator cleans and rejuvenates the air we breathe, refreshing the atmosphere and bringing outdoor freshness indoors.
In nature ozone is produced by the action of high energy discharges in lightning on oxygen in the atmosphere. Prozone® recreates this natural process. The ozone produced destroys harmful micro-organisms with which it comes into contact and breaks down malodour-causing compounds - therefore neutralizing the unpleasant smells they produce.


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Bishop urges lawmakers to end 'stranglehold' on immigration to U.S.
'We need a habitat on the moon,' says former NASA flight director
Crookston Diocese reaches $5 million settlement with abuse survivors
Mercy sisters honored for efforts to help New Hampshire end death penalty
'What's more important than life?' deacon asks Wisconsin crowd at rally
A woman holds a sign in Spanish that translates to "My heart hurts from having an abortion," during the Jan. 20 Arizona for Life Rally in Phoenix. The event was one of many held across the country to mark the anniversary of the U.S. Supreme Court's 1973 decision legalizing abortion. (CNS photo/Jesus Valencia, Catholic Sun)
By Kevin Wondrash • Catholic News Service • Posted February 2, 2018
MADISON, Wis. (CNS) — Rain or shine, life is precious and a gift from God.
More than 500 people were probably reflecting on that as they marched from St. Patrick Church in Madison to the Wisconsin Capitol Jan. 22 — the 45th anniversary of the U.S. Supreme Court's Roe v. Wade decision to legalize abortion at all stages of pregnancy in all 50 states.
Sponsored by the Diocese of Madison, the event was the first of its kind in about 20 years in Madison and was one of many held around the country, including in Washington, to mark the 1973 decision.
"It's important (to march for life) because it's exactly what God calls us to do," said Deacon Jim Hoegemeier, associate director of Madison's diocesan Apostolate for Persons with Disabilities.
"What we're called to do is be that model that other people can follow," he said. "What more important is there than life? That's why Christ came, to give us life. … We're marching to basically fulfill what God calls each one of us to do and who to be."
Before the march to the Capitol, schoolchildren, members of area parishes and staff from diocesan offices gathered for a Rosary for Life and Mass at St. Patrick Church, which is part of downtown's Cathedral Parish.
Several priests from around the diocese came to pray the rosary and concelebrate the Mass.
Msgr. Kevin Holmes, rector of Cathedral Parish, welcomed all who were there to thank God "for the gift of life and to beg his grace for the legal protection of life in our country."
Later in his homily, he said the pro-life movement of past four decades has been a "peaceful and serene witness to the dignity and sanctity of human life."
He called it "the least self-interested social movement in the history of our country. Every single person involved in this movement has already been born. We are not trying to look out for ourselves."
"One abortion is one too many," he added.
After Mass, as people exited St. Patrick into the rain, umbrellas were hurriedly opened, hoods were quickly raised over people's heads, signs were held and a line formed ready to head to the Capitol.
As everyone gathered at the top of the steps, Msgr. Holmes led a prayer of thanksgiving to God for "your countless blessings" and for "the gift of life, which you have made the basis of every other gift that you give us."
He added in part: "We give you thanks for whatever opposition, scorn, or hatred that we encounter in giving witness to the sanctity of life in accord with the words of your son in the beatitudes that we are blessed because our reward will be in heaven."
Prayer was followed by 60 seconds of silence for those who have lost their lives to abortion in the past 45 years. More than 60 million lives have been lost to abortion since the high court's 1973 decision.
The participants went inside the Capitol and took with them one of dozens of roses and the name of a state lawmaker to whom to give it.
Across the country in Arizona, a chilly overcast day did not deter an estimated 4,500 people from gathering Jan. 20 to celebrate life and raise awareness for a growing pro-life movement at the Arizona for Life March and Rally in downtown Phoenix.
Mary Fowler, a 15-year-old who attends St. Timothy Parish in Mesa, Arizona, was excited to carry one of the banners at the start of the march, adding that she is proud to show what she believes in by fighting to end abortion.
"These babies are people too," Fowler told The Catholic Sun, newspaper of the Phoenix Diocese. "I just wanted to fight for those who couldn't fight for themselves."
That was the message of the march and rally, organized by the Arizona Life Coalition — to be a voice for the voiceless.
Phoenix Bishop Thomas J. Olmsted, one of the speakers at the rally, talked of Catholics' mission to be that voice and to lift up the dignity of human life.
"By God's providence, you and I are living in 2018," he said. "It's God's plan that we be his sons and daughters, that we be his voice on behalf of the voiceless in this time in history."
Jim Ward of Blessed Sacrament Parish in Scottsdale, Arizona, said that pro-lifers have been fighting an abortion war ever since Roe v. Wade and feels rallies like the one in Phoenix are necessary to boost the morale of pro-lifers who may feel weary from the battle.
"I think we should show that we are celebrating life, and it's joyful and beautiful and a blessed thing," Ward said. "I think that is great as a witness to the whole community, and it's great for the people here. Sometimes you get tired of working, but it's nice to be inspired."
The event began at Cesar Chavez Plaza in downtown Phoenix. Thousands of people then made the roughly one-mile march to hear the speakers.
Marchers carried pink and blue balloons that read, "I am the voice." Parishes from around the diocese carried banners, and individuals decorated signs with Bible verses, J.R.R. Tolkien quotes and even Dr. Seuss verses.
Jamie Jeffries, who runs the nondenominational group Let Life Live, said she was most looking forward to the actual march because it was a way for her to show Arizona her goal of "having an abortion-free AZ."
"There's so many people here, and I'm excited to see how long we can stretch out on the actual march and see how many people here are moving as one," Jeffries said. "That's cool to me, moving as one, as a whole community, as a whole unit."
Among the teenagers in attendance was 15-year-old Pierce Vittone of St. Thomas the Apostle in Phoenix.
"I'm looking forward to meeting new people who are with the organization, so I can know how to stand up for life and those who are defenseless," Vittone said. "These babies don't have anyone to protect them, and they need people to protect them."
Wondrash is a reporter at the Catholic Herald, newspaper of the Diocese of Madison. Contributing to this story was Margaret Naczek of The Catholic Sun, newspaper of the Diocese of Phoenix.
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text:
\section{Introduction}
\noindent Dynamical systems are studied by different branches of mathematics in many different forms.
In the simplest setting, a dynamical system $(X, T)$ is comprised of a set $X$ and a transformation $T: X\rightarrow X$. By placing different requirements on $X$ and $T$,  structure can be added to the system that will influence its behavior.

Central to the analysis of a dynamical system is the analysis of the orbits of points in the system. Given a point $x \in X$, the orbit of $x$ is the sequence $x, T(x), T^2(x), \ldots.$ If our dynamical system has certain global properties, then this  guarantees the existence of points with certain orbits.

\begin{theorem}[Birkhoff's recurrence theorem]
Let $X$ be a compact topological space and $T:X\rightarrow X$ a continuous transformation. Then there exists $x \in X$ and a sequence $n_1, n_2, \ldots$, such that
\[\lim_i T^{n_i}(x)\rightarrow x.\]
\end{theorem}
Such an $x$ is called a \textit{recurrent point} of the system $(X,T)$. Comparable results hold  if we place a probability measure on the space $X$ and require $T$ be a measure-preserving transformation.

The standard proof of Birkhoff's recurrence theorem, shows the existence of a point $x$ with the following stronger property (see for example \cite[Theorem 2.3.4]{Tao_2009}). For every neighborhood $N$ of $x$, there is a bound $b$, such that for all $n$, there is a $k <b$ with $T^{n+k}(X) \in N$. Such a point $x$ is called an \textit{almost periodic point} of the system $(X,T)$.

The objective of this paper is to analyze the reverse mathematical strength of the existence of recurrent points and almost periodic points. The motivation for this work  lies not just in the intrinsic interest of Birkhoff's recurrence theorem but in the fact that this is the simplest of a family of recurrence theorems, that have widespread applications.
In this respect,  Birkhoff's recurrence theorem is similar to Ramsey's theorem and the reverse mathematical study of Ramsey's theorem has been remarkable fruitful.
Two examples will illustrate the importance of recurrence theorems. Firstly,  Furstenberg's multiple recurrence theorem, a theorem of measure-preserving systems  can be used to prove Szemeredi's theorem.
A second example, which has been studied from a reverse mathematical perspective, is the Auslander-Ellis theorem. This theorem states that if $X$ is compact metric space, with metric $d$, and $T:X\rightarrow X$ is a continuous transformation, then for any point $x$, there exists a point $y$ such that:
\begin{enumerate}
\item $y$ is an almost periodic point of the system.
\item $(\forall \epsilon)(\exists n)(d(T^n(x), T^n(y)) < \epsilon)$.
\end{enumerate}
Blass, Hirst and Simpson have shown that $\msf{ACA}_0^{+}$ proves the Auslander-Ellis theorem \cite{Blass_Hirst_Simpson_1987}. It is an open question as to whether or not it follows from $\msf{ACA}_0$ \cite{Montalban_2011}.
The Auslander-Ellis theorem can be used to prove Hindman's theorem. Hindman's theorem states that if the integers are colored with finitely many colors, then there exists an infinite set $S$ such that $\{n \colon n$ is a finite sum of elements of $S\}$ is homogenous.
 Blass, Hirst and Simpson also showed that the strength of Hindman's Theorem lies between $\msf{ACA}_0^{+}$ and $\msf{ACA}_0$ \cite{Blass_Hirst_Simpson_1987}. Recent work, particular of Towsner, has shed further light on the difficult question of  calibrating the  strength of Hindman's theorem \cite{Beig_Town_2012, Towsner_2011b,Towsner_2011a,Towsner_2012}.

In this paper we will investigate topological dynamical systems where $X$ is a closed subset of Cantor space and $T$ is a continuous transformation. This is a very important class of topological dynamical systems because subsets of natural numbers can be coded as elements of Cantor space. The proof of Hindman's theorem via  the Auslander-Ellis theorem uses such systems. In the next section we will develop and formalize two principles.
\begin{enumerate}
\item $\mathsf{RP}$: Every topological dynamical system on Cantor space contains a recurrent point.
\item $\mathsf{AP}$: Every topological dynamical system on Cantor space contains an almost periodic point.
\end{enumerate}
In Section~\ref{sect: RP}, we will show that over $\msf{RCA}_0$, $\mathsf{RP}$ is equivalent to $\msf{WKL}$. This is perhaps a little surprising because the set of recurrent points of a system is not closed.
The principal $\mathsf{AP}$ is more unusual.
In Section~\ref{section min}, we analyze the standard proof of the existence of an almost periodic point and show this requires $\msf{ACA}_0$ to carry out.
However, from the perspective of reverse mathematics,  this proof is not optimal. In fact, over $\msf{RCA}_0$, the principle $\mathsf{AP}$ lies \textit{strictly} between $\msf{WKL}$ and $\msf{ACA}$. This is the first natural example of a principle with this property. The separation between $\mathsf{AP}$ and $\msf{WKL}$ is established in Section~\ref{sect: AP WKL sep}, and the separation between $\msf{ACA}$ and $\mathsf{AP}$ is established in Section~\ref{sect: ACA AP}.

 Harrington proved that $\msf{WKL}_0$ is conservative for $\Pi^1_1$ sentences over $\msf{RCA}_0$. In Section~\ref{sect: conservation}, we show that $\msf{RCA}_0 +\mathsf{AP}$ also has this property.

 The $\mathrm{PA}$ degrees are those Turing degrees that contain a complete extension of Peano Arithmetic. This is a very well-studied class of upwards-closed Turing degrees.
 From a computability-theoretic perspective, the proof that the principle $\mathsf{AP}$ lies strictly between $\msf{WKL}$ and $\msf{ACA}$  establishes the existence of an interesting  upwards-closed strict subclass of the $\mathrm{PA}$ degrees. In Section~\ref{sect: PA degrees}, we conclude with a number of questions about these  $\mathrm{PA}$ degrees.

\section{Topological Dynamics in $\msf{RCA}_0$}

\noindent A standard definition of a topological dynamical system on Cantor space is the following.

\begin{definition}
\label{def: system}
A pair $(C,F)$ is a \textit{topological dynamical system} on $2^{\omega}$ if $C$ is a non-empty closed  subset of $2^{\omega}$, and $F$ is a continuous transformation of $2^{\omega}$ such that for all $X \in C$, $F(X) \in C$.
\end{definition}

Note that this definition requires $F$ to be defined on all elements of $2^{\omega}$ rather than just those elements of $C$. This is not a limitation because of the Tietze extension theorem which can be proved in $\msf{RCA}_0$ \cite[Theorem II.7.5]{Simpson_2009}.
  Sometimes $F\!\upharpoonright{C}$ is required to be a homeomorphism of $C$ but we will not consider that possibility here.
From now on, we will often refer to a topological dynamical system as simply a system.

We  need to consider how we encode a system inside a model of 2\textsuperscript{nd} order arithmetic. We need to be careful with our choice of encoding to ensure that it behaves sensibly for models of $\msf{RCA}_0$. We will follow the standard approach of  encoding a closed set as the set of paths through a tree. The approach we take for encoding a continuous transformation is also standard.

\begin{definition}
\label{def:endo}
A function $f:2^{<\omega} \rightarrow 2^{<\omega}$ \emph{encodes a continuous transformation of $2^{\omega}$} if
\begin{enumerate}
\item $f$ is total.
\item $f$ is order preserving.
\item $(\forall l)(\exists m)(\forall \sigma \in \{0,1\}^m)|f(\sigma)|\ge l$.
\end{enumerate}
\end{definition}
Given such an $f$, we will  denote by $F$ the transformation of Cantor space encoded.  In particular,  if $X \in 2^{\omega}$, we will denote by $F(X)$,
\[ \lim_{l \in \omega} f(X\upharpoonright{l}).\]
It could be objected that the third condition should really be stated as for all $X$ and $l$, there exists an $m$ such that $|f(X\upharpoonright{m})| >l$, and the equivalence of this statement with the third condition is a consequence of the compactness of Cantor space. However, because we are working over $\msf{RCA}_0$  we will use the stronger definition above.
Let $\lambda$ denote the empty string. For convenience, we will assume that any such $f$ has the additional property that for any string $\sigma \ne \lambda$, $|f(\sigma)| < |\sigma|$. This does not result in any loss of generality because given any $f$ encoding a continuous transformation of $2^{\omega}$, we can uniformly find a $\hat{f}$ with this property that encodes the same continuous transformation of $2^{\omega}$ as $f$.

\begin{definition}[$\msf{RCA}_0$]
\label{def: system2}
A pair $(C, f)$ encodes a system  if
\begin{enumerate}[(i)]
\item The set $C$ is a tree.
\item The set $f$ encodes a continuous transformation of $2^{\omega}$.
\item $[C] \ne \emptyset$.
\item For all $\sigma \in C$ we have that $f(\sigma) \in C$.
\end{enumerate}
\end{definition}

The final condition ensures that if  $(C, f)$ encodes a system inside a model $\mathcal{M}$ then if we extend $\mathcal{M}$ by adding additional reals, then $(C, f)$ still encodes a system inside the extended model. This rules out pathological cases. For example, $f$ could map paths in $[C]$ that are not in the model outside of $[C]$. Then if we extend our model by adding such a path $(C,f)$ would no longer be a system.

It is impossible to discuss recurrence points and almost periodic points without discussing orbits.
Given a system $(C,f)$ and $X \in C$  we need to show that even with the limited induction available in $\msf{RCA}_0$, the orbit of $X$ is well-defined.

Given a function $f$ encoding a continuous transformation of $2^{\omega}$, define
$f:2^{<\omega} \times \omega \rightarrow 2^{<\omega}$ by
\[f(\sigma, k) =\begin{cases}
\sigma & k=0 \\
f(f(\sigma, k-1)) & k>0.
\end{cases}
\]
When convenient, we will write $f^k(\sigma)$ for $f(\sigma, k)$. Observe that $f^1 =f$. We define $F^k(X) = \lim_{l\in \omega} f^k(X\upharpoonright{l})$. The following lemma shows that given Definitions \ref{def:endo} and \ref{def: system2}, we can talk sensibly about orbits of a point in $\msf{RCA}_0$.

\begin{lemma}[$\msf{RCA}_0$]
\label{lemma:basic}
If $(C,f)$ is a system then for all $k$
\begin{enumerate}
 \item \label{i:system} $f^k$ encodes a continuous transformation of Cantor space.
\item \label{iii:system} $(C, f^k)$ is a system.
\item \label{ii:system} For all $X \in 2^{\omega}$, $F(F^k(X))=F^{k+1}(X)$.
 \end{enumerate}
\end{lemma}
\begin{proof}
\eqref{i:system}. The function $f^k$ is total because it has been defined by primitive recursion from $f$. It is order preserving by $\Pi_1$ induction on $k$ for the formula
\[(\forall k)(\forall \sigma, \tau, \rho, \pi)((\sigma \preceq \tau \wedge \rho = f^k(\sigma)  \wedge \pi = f^k(\tau) )\rightarrow \rho \preceq \pi).\]
The third condition is established by fixing $l$, then inducting on $k$ for the formula
\[ (\forall k)(\exists m)(\forall \sigma \in \{0,1\}^m)(\forall \tau \in \{0,1\}^{<l})(f^k(\sigma) \ne \tau).\]
This formula holds trivially for the case $k=0$ (take $m >l$). For the case $k=n+1$, let $m_n$ witness the above formula for $k=n$.
Now let $m$ be such that for all $\sigma \in \{0,1\}^m$, $|f(\sigma)| \ge m_n$. The existence of $m$ comes from our assumption on $f$.  Note that if $\sigma \in \{0,1\}^m$, then making use of the associative law (provable in $\msf{RCA}_0$) we have
\[f^{k+1}(\sigma)  = f^k(f(\sigma)) = f^k(\rho)\]
where $|\rho| \ge m_n$ and so $|f^k(\rho)| \ge l$ and in particular $f^k(\rho)$ cannot equal any string of length strictly less that $l$.

\eqref{iii:system}. The first three conditions for $(C, f^k)$ to be are system are met trivially. Fix $\sigma \in C$. We will show that for all $k$, $f^k(\sigma) \in C$.
The set $\{n : f^n(\sigma) \not \in C\}$ is computable. Hence, if it is non-empty, it has a least element $k$. As $\sigma \in C$, $k$ cannot be $0$. Let $\tau = f^{k-1}(\sigma)$ so $\tau \in C$. But $f^k(\sigma) = f(f^{k-1}(\sigma)) = f(\tau) \in C$, a contradiction.

\eqref{ii:system}. Fix a $k$ and $X \in C$. For any $l$, there exists some $m_l$ such that
$F^k(X)\upharpoonright{l} \preceq f^k(X\upharpoonright{m_l})$. Hence
\begin{multline*}
F(F^k(X)) = \lim_l f(F^k(X)\upharpoonright{l}) \preceq\lim_l f(f^k(X\upharpoonright{m_l})) \\   = \lim_l f^{k+1}(X\upharpoonright{m_l})
=F^{k+1}(X). \qedhere
\end{multline*}
\end{proof}

The orbit of $X$ under $F$ is the sequence $\seq{F^k(X) \colon k \in \omega}$. Note that this is uniform and hence
$\oplus_{k \in \omega} F^{k}(X)$ exists by recursive comprehension in any model of $\msf{RCA}_0$ that includes $X$ and $f$.

\section{Recurrent Points}
\label{sect: RP}
\noindent We call $X$ a \emph{recurrent point} of a  topological dynamical system $(C, f)$, if
$X \in [C]$ and
\[(\forall n, c)(\exists k)(F^{n+k}(X) \succeq X\upharpoonright{c}).\]

We call $X$ an \emph{almost periodic point} of a  topological dynamical system  $(C, f)$, if
$X \in [C]$ and
\[(\forall c)(\exists b)(\forall n)(\exists k <b)(F^{n+k}(X) \succeq X\upharpoonright{c}).\]

 This leads to two principles. First $\mathsf{RP}$ is the principle that every topological dynamical system on
 $2^{\omega}$ contains a recurrent point. The second $\mathsf{AP}$ is the principle that every topological dynamical system on $2^{\omega}$ contains an almost periodic point.  Over $\mathsf{RCA_0}$ we have the obvious implication that $\mathsf{AP}$ implies $\mathsf{RP}$ because an almost periodic point is a recurrent point.

The following theorem would be trivial if condition (iii)  in Definition~\ref{def: system2}, was replaced by requiring the tree $C$ to be infinite. Given such a $C$, we could simply take $(C,f)$ be our system where $f$ is the identity map. Any recurrent point of this system would have to be an element of $[C]$ hence proving $\msf{WKL}$.

\begin{theorem}
Over $\msf{RCA_0}$, $\mathsf{RP}$ implies $\msf{WKL}$.
\end{theorem}
\begin{proof}
Let $T$ be an infinite computable tree on $2^{<\omega}$. We will regard $T$ as computable tree in $3^{<\omega}$ (i.e.\ a computable subtree of $3^{<\omega}$ such that no node of $T$ contains a $2$).
We will define a  system on $3^\omega$ such that any recurrent point of the system is a path on $T$. As  $3^\omega$ is computably homeomorphic to $2^\omega$, this is sufficient to prove the theorem. The idea behind the following definition of $f$ is that if $X \in 3^\omega$ is a path on $T$, then $F(X)=X$. If $X$ is not a path on $T$, then the orbit of $X$ moves in increasing lexicographical order searching for a path on $T$, looping around if it extends $2$.  The extra branching of  $3^\omega$ allows us to move the orbit of $F(X)$ if $X$ is not a path on~$T$.

Define the following function $f:3^{<\omega} \rightarrow 3^{<\omega}$. First $f(\lambda) = \lambda$. Second if $|\sigma| >0$, let $n = |\sigma| -1$. If $\sigma \in T$, let $f(\sigma) = \sigma \upharpoonright{n}$. If $\sigma \not \in T$, then let $\pi$ be the shortest initial segment of $\sigma$ such that $\pi \not \in T$. Because $T$ is a subtree of $2^{<\omega}$, if $\pi$ contains a $2$, then $\pi$ must end with $2$. Define
\[f(\sigma) =
\begin{cases}
\rho 1 0^\omega \upharpoonright{n} & \pi = \rho 0 \vee \pi = \rho 02 \\
\rho 2 0^\omega \upharpoonright{n} & \pi = \rho 1 \vee \pi = \rho 12 \\
0^n & \pi =2.
\end{cases}\]

It is not difficult to verify that $(3^\omega, f)$ is a system. Let $\le_{lex}$ be the lexicographical ordering on finite strings.  (Recall that under this ordering $\sigma \le_{lex} \tau$ if $\sigma \preceq \tau$ or $\sigma(i) < \tau(i)$ for the least $i$ where these strings differ.)

\begin{claim} Let $n >0$. Let $\sigma_0, \sigma_1, \ldots, \sigma_n$ be a finite sequence of strings such that $\sigma_0 \succ \sigma_n$, for all $i <n$, $f(\sigma_i)=\sigma_{i+1}$ and $\sigma_n \not \in T$. Then for some $k < n$, $2 \preceq \sigma_k$.
\end{claim}
\begin{proof}
Consider $S = \{ i \le n \colon \sigma_i \le_{lex} \sigma_n \wedge \sigma_i \not \in T\}$. The set $S$ is not empty as it contains $n$. As $S$ is computable, it has a least element $l$.
Now $l \ne 0$ as $\sigma_0$ is a strict extension of $\sigma_n$. Let $k=l+1$. First $\sigma_k \not \in T$ as otherwise $f(\sigma_k) = \sigma_l \in T$. By minimality of $l$ we have that $\sigma_k \not \le_{lex} \sigma_n$ and in particular $\sigma_k \not \le_{lex} \sigma_l$. Now because $\sigma_k \not \in T$, the definition of $f$ implies  that $\sigma_k \succeq 2$.
\end{proof}

\begin{claim}If $|\tau|> |\sigma|$, $\sigma <_{lex} \tau$, and  $\tau \in T$ then
$f(\sigma)<_{lex} \tau$.
\end{claim}
\begin{proof}
If $\sigma \prec \tau$, then $\sigma \in T$ and so $f(\sigma)$ is an initial segment of $\sigma$ and  the result holds.
Otherwise let $\xi$ be the least common initial segment of $\sigma$ and $\tau$. So $\xi0 \preceq \sigma$. Because $\xi$ is on the tree, either $f(\sigma)$ extends $\xi0$ or $f(\sigma) = \xi10^j$ for some $j$. However as $|\tau|> |\sigma| \ge |f(\sigma)|$ this implies that in either case $f(\sigma) <_{lex} \tau$.
\end{proof}

Let $R$ be a recurrent point for this system. Assume  $R  \not \in [T]$. Take $\sigma  \prec R$ such that $\sigma \not \in T$. As $R$ is a recurrent point, there exists a sequence
$\sigma_0, \ldots, \sigma_n$ such that $\sigma_n = \sigma \prec \sigma_0 \prec R$ and
$f(\sigma_i) =\sigma_{i+1}$. By the  1\textsuperscript{st} claim for some $k<n$, $2 \preceq \sigma_k$.

Let $\tau \in T$ such that for all $i \le n$, $|\tau| > |\sigma_i|$. Now $\sigma_{k+1}$ is a string of all $0$'s,  and  $|\sigma_{k+1}| <|\tau|$. Hence  $\sigma_{k+1} <_{lex} \tau$.  It follows that $\sigma_n <_{lex} \tau$ by inducting over the 2\textsuperscript{nd} claim. Now as $\sigma_0 \succeq \sigma_n$ and $\sigma_n \not \preceq \tau$ because $\sigma_n \not \in T$, this implies that
$\sigma_0 <_{lex} \tau$.

But this is impossible. If $\sigma_0 <_{lex} \tau$ then again by inducting over the 2\textsuperscript{nd} claim, for all $i$, $\sigma_i <_{lex} \tau$ and so $\sigma_i \not \succeq 2$. This contracts  the fact that $\sigma_k \succeq 2$.
Hence $R \in [T]$.
\end{proof}

\begin{theorem}Over $\mathsf{RCA_0}$, $\mathsf{WKL}$ implies $\mathsf{RP}$.
\end{theorem}
\begin{proof}
Let $(C, f)$ be a system.
We can define the set of recurrent points of this system, $\mathcal{R}$, as follows.
\[\mathcal{R} = \{ X \in [C] \colon (\forall c)( \exists n, l > c )(f^n(X\upharpoonright{l}) \succeq X\upharpoonright{c})\}.\]
This shows that $\mathcal{R}$ is  $\Pi^0_2$ in  $C\oplus f$.
In order to prove that $\mathsf{WKL}$ implies $\mathsf{RP}$, it is sufficient to  show  there is a non-empty $\Pi^0_1(C\oplus f)$ class contained in $\mathcal{R}$.
We will construct a computable sequence of finite sets of strings
$\seq{ U_i \colon i\in \omega}$ and let
$\bigcap_i [U_i] \cap [C]$ be our $\Pi^0_1(C\oplus f)$ class. We will ensure that if $X \in [U_i] \cap C$, then for some  $n, l>i$, $f^n(X\upharpoonright{l}) \succeq X\upharpoonright{i}$, hence if $X \in  \bigcap_i [U_i] \cap [C]$, then $X\in \mathcal{R}$ and so $X$ is a recurrent point of $(C,f)$.

The difficulty with defining $U_i$, is that it is possible that $[U_i] \cap [C]$ might be empty.  To avoid this occuring, we will ensure that for all $i$, there is an $s_i$ such that, $\bigcup_{n<s_i} F^{-n}([U_i]) \supseteq [C]$. This means that no $[U_i]$ can be removed entirely from $[C]$ because otherwise  $[C]$ would  either be empty or, for some $X\in [C]$, there would be some $n$ such that $F^{n}(X) \not \in [C]$. In either case $(C,f)$ would not be a system.

Let $U_0 =V_0 =\{\lambda\}$ and $s_i=0$. We will assume that we are given $U_i$ and $V_i$,  both finite sets of strings and $s_i$ a number such that:
\begin{enumerate}
\item $(\forall \tau \in U_i) (\exists n)
(i \le n \le s_i \wedge f^n(\tau) \succeq \tau \upharpoonright{i})$.
\item $(\forall \sigma \in V_i)(\exists n \le s_i)
(\exists \tau \in U_i) (f^n(\sigma) \succeq  \tau)$.
\item $V_i$ is an open cover of $C$.
\end{enumerate}
These conditions hold trivially for the case $i=0$. We inductively define $U_{i+1}[s]$ and $V_{i+1}[s]$ as follows.
\begin{align*}
U_{i+1}[s] = \{ \sigma \in 2^{<\omega}  &\colon ((\exists \tau \in U_i) (\sigma \succ \tau))  \wedge \\
& (\exists n)((i \le n \le s )\wedge(f^n(\sigma) \succeq \sigma \upharpoonright{i})) \}.\\
V_{i+1}[s] = \{  \sigma \in 2^{<\omega} &\colon (\exists n \le s)(\exists \tau \in U_{i+1}[s])( f^n(\sigma) \succeq \tau)\}.
\end{align*}
\noindent These definitions imply that:
\begin{enumerate}
\item $[U_{i+1}[s]] \subseteq [U_i]$.
\item $[U_{i+1}[s]] \subseteq [U_{i+1}[s+1]]$.
\item $[V_{i+1}[s]] \subseteq [V_{i+1}[s+1]]$.
\end{enumerate}

\begin{claim}
 $\bigcup_s [V_{i+1}[s]] \supseteq C$.
\end{claim}
\begin{proof}
By applying bounding, there is some  $h > \max\{|\tau| \colon \tau \in V_i\}$ such that
\[(\forall \sigma \in \{0,1\}^h)(\forall m \le s_i)(|f^m(\sigma)| \ge i+1).\]
 Take $X \in C$. By the pigeon-hole principle there is some $\sigma \in \{0,1\}^h$ and $j, k \in \omega$ such that $F^j(X) \in [\sigma]$
and $F^{j+k}(X) \in [\sigma]$.
We can also ensure that $k \ge i+1$.
 Now applying $\msf{WKL}$, we know that $\sigma$ extends some element of $V_i$. From the definition of $V_i$, this means that for some $m \le s_i$,  $f^m(\sigma)$ extends some $\tau \in U_i$. Let $Y= F^{j+m}(X)$ so $Y \succ f^m(\sigma) \succeq \tau$. Further
$F^k(Y) = F^{j+k+m}(X) \succ f^m(\sigma) \succeq \tau$ as well. Finally,
$Y\upharpoonright{i+1} = F^k(Y)\upharpoonright{i+1}$ because $|f^m(\sigma)| \ge i+1$. Take $l$ such that $f^k(Y\upharpoonright{l}) \succeq f^m(\sigma)$.
Thus $Y\upharpoonright{l}  \in U_{i+1}[\max\{k,l\}]$ and $X \in [V_{i+1}[s]]$ where $s >\max\{j+m,k,l\}$ is large enough such that $f^{j+m}(X\upharpoonright s) \succeq Y\upharpoonright{l}$.
\end{proof}

Hence by compactness, there is some  least $s_{i+1}$ such any string of length $s_{i+1}$ in $C$ extends some element of  $V_{i+1}[s_{i+1}]$. We define $U_{i+1} =  U_{i+1}[s_{i+1}]$ and $V_{i+1} =  V_{i+1}[s_{i+1}]$. Note that $[U_{i+1}]$ is a closed set in Cantor Space.
Fix $i$. Take any $X \in [C]$. We know that $X \in [V_i]$ and hence there exists some $n \le s_i$ such that $F^n(X) \in [U_i]$. By Lemma~\ref{lemma:basic}, $F^n(X) \in [C]$. Hence $[U_i] \cap [C] \ne \emptyset$.
Thus $\seq{[U_i] \cap [C] \colon i \in \omega}$ is a nested sequence of non-empty closed sets and so by $\msf{WKL}$ contains an element $R$, which is a recurrent point of $(C,f)$.
\end{proof}

Note that the proof given uses the fact that the sets $[U_i]$ are clopen. The proof can be extended to certain spaces which do not have a basis of clopen sets, such as the unit interval, by adding the condition that for any open set $E  \in U_{i+1}$, we have that the closure of $E$ is contained in  $[U_i]$. To find our recurrent point, we take a point  $R \in \bigcap_i (\mathrm{cl}[U_i] \cap [C])$. The same argument shows that this set is not empty (instead of adding $Y\upharpoonright{l}$ to $U_{i+1}$ find an $\epsilon$ such that $\mathrm{cl}[B(Y; \epsilon)] \subset [U_i]$ and add this ball to  $U_{i+1}$). However, $R$ must be an interior point of each set $\mathrm{cl}[U_i]$ and  so as $R\in \bigcap_i [U_i]$, $R$ is a recurrent point.

\section{Minimal Systems}
\label{section min}

\noindent We  now investigate the principle $\mathsf{AP}$.
The standard proof that every topological dynamical system has an almost periodic point uses the existence of minimal subsystems. We call a system $(C,f)$ \emph{minimal}, if for any system $(D,f)$ such that $[D] \subseteq [C]$, we have that $[D]=[C]$.

Let $(C, f)$ be a system. By Zorn's lemma, $(C, f)$ contains a minimal subsystem $(D,f)$. Now by the following standard lemma, which we can formalize in $\mathsf{WKL_0}$, every point of $D$ is an almost periodic point of $(D,f)$ and hence an almost periodic point of  $(C, f)$.

\begin{lemma}
$\mathsf{WKL_0}$ proves that any path in a minimal system is almost periodic.
\end{lemma}
\begin{proof}
Let $(D, f)$ be a minimal system and take any $X \in [D]$. Assume $X$ is not an almost periodic point. If so  there exists some $\tau \prec X$ such that
\begin{equation}
\label{e: notap}
(\forall b )(\exists n)(\forall k \le b)F^{n+k}(X) \not \succ \tau.
\end{equation}

Define $E = \{\sigma \in D \colon (\forall n \le |\sigma|)(f^n(\sigma) \not \succeq \tau)\}$.
As $X \not \in [E]$, we have that  $[E] \subsetneq [D]$. Our assumption that $|f(\sigma)| <|\sigma|$, implies that for all $\sigma \in E$, $f(\sigma) \in E$. Now $(E,f)$ cannot be a system because this would contradict the minimality of $(D,f)$.
This means that $[E] = \emptyset$. Applying $\msf{WKL}$, there exists a $b$, such that $E$ contains no string of length $b$.  Hence for all $Z \in [D]$ there is some $k <b$ with $F^k(Z) \succ \tau$. This contradicts \eqref{e: notap} and  hence our assumption that $X$ is not an almost periodic point is incorrect.
\end{proof}

 While we appealed to Zorn's lemma to construct a minimal subsystem, this is not necessary for systems in Cantor space. The reason is Cantor space contains a computable basis of open sets. This allows us to show that $\msf{ACA}_0$ implies that any system contains a minimal subsystem.
 To find a minimal subsystem simply enumerate the basis and ask in order can any element be removed. In particular, let $\{\sigma_i\}_{i \in \omega}$ enumerate the finite strings. Given a system $(C,f)$ let $C_0 =C$. If $(C_i, f)$ has been defined,  let $C_{i+1}$ be equal to $\{\tau \in C_i \colon (\forall n \le |\tau|)(f^n(\tau) \not \succeq \sigma_i)\}$ if the later set is not finite.  Otherwise let $C_{i+1} = C_i$. It is not difficult to verify that $( \bigcap_i C_i, f)$ is a minimal subsystem of $(C,f)$. This gives us the following result.

\begin{proposition}
\label{prop: ACA}
$\msf{ACA}_0$ proves that any system contains an almost periodic point.
\end{proposition}

\begin{theorem}
\label{thm: ACA minimal}
Over $\mathsf{WKL_0}$, $\msf{ACA}$ is equivalent to statement that every system contains a minimal subsystem.
\end{theorem}
\begin{proof}
The argument proceeding Proposition~\ref{prop: ACA} shows that $\msf{ACA}_0$ proves that every system contains a minimal subsystem. To show the other direction we will work over $\msf{WKL}_0$ as our base system. In order to simplify the exposition of this proof, we will work with $\Pi^0_1$ classes of reals in Cantor space as opposed to trees in $2^{<\omega}$.

First we will show how to encode a single bit of $\emptyset'$ into a system.
Let $f$ be the left-shift. Fix $n$, we will define a $\Pi^0_1$ class $C$ such that given any minimal subsystem $(D,f)$ of $(C,f)$ the set $[01] \cap D$ is empty if and only if $n \not \in \emptyset'$.
In particular, if $n \not \in \emptyset'$, then
\[C = \{0^i1^\omega \colon i \in \omega\} \cup \{1^i0^\omega \colon i \in \omega\}.\]
Observe that in this case, the only minimal subsystems of $(C,f)$ will be $(\{0^\omega\}, f)$ and $(\{1^\omega\}, f)$.

Let $S_i = \{F^n((0^i1^i)^\omega) : n \in \omega\}$.  Each $S_i$ is a minimal system with $2\cdot i$ elements. For example,
\[S_2 =\{ (0011)^\omega, 011(0011)^\omega, 11(0011)^\omega, 1(0011)^\omega\}.\]
If $n \in \emptyset'$, then we will define $C$ to be equal to $S_i$ for some $i$ compatible with our definition of $C$ at the stage $n$ enters $\emptyset'$.
Formally, let $t$ be $\infty$ if $n\not \in \emptyset'$ and let $t$ be the  least $s$ such that  that $n \in \emptyset'[s]$ otherwise.
Let $E_s =\{X  \in 2^\omega \colon (\exists l \le s)(0^l1^{s-l} \prec X \vee  1^l0^{s-l} \prec X\}$.
 Define
\[C = \begin{cases} \bigcap_s E_s & t=\infty \\
S_i & t =i <\infty.
\end{cases}\]
Observe that $S_i \subseteq \bigcap_{s< i} E_s$. Hence $C$ is a $\Pi^0_1$ class.
Let $(D,f)$ be a minimal subsystem of $(C,f)$. To determine  if $n$ is in $\emptyset'$ wait until a stage $s$ such that either $n  \in \emptyset'$ or $[01] \cap D[s] = \emptyset$ (the existence of such an $s$ when $n \not \in \emptyset'$ requires $\msf{WKL}$).

In order  to code all elements of $\emptyset'$, we use the uniformity in the definition above to build a product system as follows.
For all $n$, let $C_n$ be the set defined by the above construction.
Let $C = \Pi_n C_n$ (i.e.\ $X \in C$ if and only if for all $n$, $X^{[n]} \in C_n$ where  $X^{[n]}$ denotes the $n$th column of $X$). Let $f$ be the mapping produced by applying the left-shift to each column.  Now  if $(D, f)$ is a minimal subsystem of $(C,f)$  then we have that $n \not \in \emptyset'$ if and only if the set
$\{X \in D | X^{[n]} \succ [01]\}$ is empty. Using $\msf{WKL}_0$,  this set is empty if and only if  the associated tree is finite and we have provided a $\Sigma^0_1$ definition of the complement of $\emptyset'$.
\end{proof}

\section{Separating $\mathsf{AP}$ from $\mathsf{WKL}$}
\label{sect: AP WKL sep}

\noindent We have seen that $\msf{ACA}_0$ proves $\mathsf{AP}$. Further $\msf{RCA}_0 +\mathsf{AP}$ proves $\msf{WKL}$ because any almost periodic point is a recurrent point.
In this section we will separate  $\mathsf{AP}$ from $\msf{WKL}$.
We will show that there is a model of $\msf{WKL}$ that is not a model $\msf{RCA}_0 +\mathsf{AP}$.  The natural numbers in this model will be the true natural numbers and so  we will work with full induction. We will also regard our closed sets  as  $\Pi^0_1$ classes, as this simplifies the exposition.

The key to the separation is the following technical lemma. Let $(C,f)$ be a system. A point $X\in C$ is called a  \emph{periodic point} of $(C,f)$ if for some $n$, $F^n(X) =X$. Let $Orb(X)$ be the orbit of $X$. Note that if $X$ is a periodic point then $Orb(X)$ is a finite set.
For a finite string $\sigma$ we denote by $\sigma^n$, the string obtained by repeating $\sigma$ $n$ times and we denote by $\sigma^\omega$, the infinite sequence $\bigcup_n \sigma^n$.
\begin{lemma}
\label{lemma: no closed set}
Let $f$ be the left-shift on Cantor space. Let $P \subseteq 2^{\omega}$ be a $\Pi^0_1$ class. There is a $\Pi^0_1$ class $C$, computable uniformly in an index for $P$ such that $(C,f)$ is a system and either:
\begin{enumerate}
\item $C \cap P = \emptyset$; or \label{propc1}
\item There is a non-empty $\Pi^0_1$ class $\widehat{P} \subseteq P$ with the property that no element of $\widehat{P}$ is an almost periodic point of $(C,f)$. \label{propc2}
\end{enumerate}
\end{lemma}
\begin{proof}
The definition of $C$ is simple. Let $\{X_i\}_{i\in \omega}$ be an enumeration of the periodic points in $(2^\omega, f)$. Such an enumeration exists because any periodic point is of the form $\sigma^\omega$ for some finite string $\sigma$. We let $C=2^{\omega}$ unless at some least stage $s$ we have $Orb(X_i) \cap P = \emptyset$ for some $i <s$. If so we let $C = Orb(X_i)$ for the least $i$ for which this holds at stage $s$. The definition of $C$ is uniform because we can refine $C$ to $Orb(X_i)$ at any point.

If $C= Orb(X_i)$ for some periodic point $X_i$, then $C \cap P = \emptyset$ and condition \eqref{propc1} is meet. Hence we will consider the case that $C =2^\omega$.
If there is some computable point $X \in P$ such that $X$ is not almost periodic, then condition \eqref{propc2} holds by defining $\widehat{P} = \{X\}$. Hence we will assume that any computable point in $P$ is almost periodic.

We  inductively define a sequence of finite strings $\sigma_1, \sigma_2, \ldots.$  The strings will have the following properties.
If $i<j$ then $\sigma_i \preceq \sigma_j$. For all $i$, $1^i$ is a substring of $\sigma_{i+1}$ but $1^{i+1}$ is not. The string $\sigma_1= 0^n$ for some $n >0$.

\begin{enumerate}
\item The sequence $10^\omega$ is computable and not almost periodic. Hence $10^\omega \not \in P$ and so  there exists some $n_1 >0$ such that $[1 0^{n_1}] \cap P= \emptyset$. Let $\sigma_1 = 0^{n_1}$.
\item The sequence $11(\sigma_1 1)^\omega$ is computable and not almost periodic (the subsequence $11$ only occurs once). Hence $11(\sigma_1 1)^\omega\not \in P$, and so there exists some $n_2>0$ such that $[11(\sigma_1 1)^{n_2} ] \cap P= \emptyset$.
Let $\sigma_2 = (\sigma_1 1)^{n_2}$.
 \item Similarly  $111(\sigma_2 \sigma_1 11)^\omega \not \in P$, and so there exists some $n_3$ such that $[111(\sigma_2 \sigma_1 11)^{n_3} ] \cap P= \emptyset$.
Let $\sigma_3 = (\sigma_2\sigma_1 11)^{n_3}$.
\item In general we define $\sigma_{i+1} = (\sigma_i\sigma_{i-1}\ldots \sigma_1 1^i)^{n_{i}}$ such that
\[[1^{i+1}(\sigma_i\sigma_{i-1}\ldots \sigma_1 1^i)^{n_{i}}] \cap P= \emptyset.\]
\end{enumerate}
Consider the periodic systems generated by
$(\sigma_i)^\omega$. Because $C = 2^\omega$,  for all $i$, there is some $X_i \in Orb((\sigma_i)^\omega) \cap P$.

\begin{claim}For all $i$, $X_i(0)=0$.
\end{claim}
\begin{proof}
Take any $X_i$.
 Let $k \in \omega$ be the largest number such that $1^k$ is an initial segment of $X_i$. First $k < i$ because any substring of $1$'s in $(\sigma_i)^\omega$ has length less than $i$. Further, by construction if $1^k0$ forms an initial sequence of $X_i$ then $1^k\sigma_k$ forms an initial sequence of $X_i$, but $\sigma_k$ was choosen so that $[1^k\sigma_k] \cap P= \emptyset$. Note here we are using the fact that if $i<j$ then $\sigma_i \preceq \sigma_j$. Hence $X_i(0) =0$.
\end{proof}
\begin{claim} Fix $k$. Let $c_k =2k + \sum_{s=1}^k|\sigma_s|$. Then for all $i > k$, $X_i\upharpoonright{c_k}$  contains  $1^k$ as a substring.
\end{claim}
\begin{proof}
 Let $\tau = \sigma_k \ldots \sigma_1$. We will show by induction that for $i > k$, $\sigma_i$ is a string of the form  $\tau1^{n_1}\tau1^{n_2}\tau1^{n_3} \ldots \tau1^{n_l}$ where each $n_j \ge k$ for $j \in \{1, \ldots, l\}$.
First $\sigma_{k+1} = (\sigma_{k}\sigma_{k-1}\ldots \sigma_1 1^k)^{n_{k}}
= (\tau1^k)^{n_{k}}$ and  is clearly of this form. Fix $i \ge k+1$ and assume this holds for all $j \in \{k+1, k+2, \ldots, i\}$. Then
\[\sigma_{i+1} = (\sigma_i\sigma_{i-1}\ldots \sigma_1 1^i)^{n_{i}}
=(\sigma_i\sigma_{i-1}\ldots \sigma_{k+1}\tau 1^i)^{n_{i}}\]
and so has the desired property by induction. As $X_i$ is a left-shift of $(\sigma_i)^\omega$ and $c_k = |\tau| +2k$, the claim holds.
\end{proof}

Let $X$ be an accumulation point of $\{X_i \colon i \in \omega\}$. Hence $X$ is an element of $P$ as $P$ is closed.
The sequence $X$ has the property that
$X(0)=0$ and for all $k$, the initial segment $X\upharpoonright{c_k}$ contains a subsequence of $1^k$. Observe that the sequence $\{c_i\}$ is computable. Now define $\widehat{P} \subseteq P$ to be the following $\Pi^0_1$ class
\[\{ X \in P \colon  X(0)=0 \wedge (\forall k) (1^k\mbox{ is a substring of }X\upharpoonright{c_k})\}.\]
If all the assumptions are meet until this point, $\widehat{P}$ is non-empty and  no element of $\widehat{P}$ is an almost periodic point. Hence condition \eqref{propc2} is met.
\end{proof}

In the proof of the following theorem we will make use of the fact that if $P\subseteq 2^{\omega}$ is a $\Pi^0_1$ class and $f:2^{\omega} \rightarrow 2^{\omega}$ is a total computable function, then
both $f(P)$ and $f^{-1}(P)$ are $\Pi^0_1$ classes.

\begin{theorem}
\label{thm: WKL does not imply AP}
$\msf{WKL}_0$ does not prove $\mathsf{AP}$.
\end{theorem}
\begin{proof}
Let $f$ be the left-shift. Let $\{Q_i\}_{i\in \omega}$ be a enumeration of all $\Pi^0_1$ classes. It follows from the uniformity of Lemma~\ref{lemma: no closed set}, that we can build a system
\[ (C,g) = \prod_{e\in \omega}(C_e, f)\]
such that if $Q$ is the $e$\textsuperscript{th}  $\Pi^0_1$ class  then either
 \begin{enumerate}
 \item $\pi_e(Q) \cap C_e = \emptyset$ or
 \item There is a non-empty $\Pi^0_1$ class $\widehat{Q} \subseteq Q$ such that no element of $\pi_e(\widehat{Q})$ is almost periodic,
 \end{enumerate}
   where $\pi_e$ is the projection on the $e$\textsuperscript{th} coordinate (see Theorem~\ref{thm: ACA minimal} for an example of how to encode such a product system). While Lemma~\ref{lemma: no closed set} guarantees the existence of a non-empty  $\Pi^0_1$ subset of $\pi_e({Q})$, no element of which is almost periodic, this can be pulled-back along $\pi_e$ to obtain $\hat{Q}$.
 We will show that there is a set of $\mathrm{PA}$ degree that does not compute an almost periodic point of $(C,f)$.

\subsection*{Construction} At stage $0$, let $P_0$ be  a non-empty $\Pi^0_1$ class of sets of $\mathrm{PA}$ degree. At stage $s+1$, let
 $\Phi_s$ be the $s$\textsuperscript{th}  Turing functional. If for some $n$ the set
 $\{ X\in P_s \colon \Phi_s^X(n) \uparrow\}$ is not empty, then let $P_{s+1}$ be this set for the least such $n$.

 Otherwise, we have that  $\Phi_s$ is total on all elements of $P_s$.
 Let $Q = \Phi_s(P_s)$. Now $Q$ is a $\Pi^0_1$ class because there is a total functional that agrees with $\Phi_s$ on the elements of $P_s$. Let $e$ be an index for $Q$ as a $\Pi^0_1$ class.
 There are two possible outcomes. First  $\pi_e(Q) \cap C_e = \emptyset$ in which case let $P_{s+1}= P_s$ and note that no element of $P_{s+1}$ computes an element of $C$ via $\Phi_s$ let alone an almost periodic element.
The other possible outcome is that  there is some non-empty $\hat{Q} \subseteq Q$ such that no element of $\pi_e(\hat{Q})$ is almost periodic in $C_e$ (and hence no element of $\hat{Q}$ is almost periodic in $C$). For this outcome let
$P_{s+1} = \{X \in P_s \colon \Phi_s^X \in \hat{Q}\}$. In this case,  $P_{s+1}$ is a non-empty $\Pi^0_1$ class, no element of which computes an almost periodic point in $C$ via $\Phi_s$.

By compactness there is some  $X \in \bigcap_i P_i$. This set $X$ is of $\mathrm{PA}$ degree and $X$ does not compute an almost periodic point of $(C,g)$.  Now it is standard result that there is a model of $\msf{WKL}_0$, such that all sets in this model are Turing below $X$. This model does not contain an almost periodic point for the system $(C,g)$ and shows that $\msf{WKL}_0$ does not imply $\mathsf{AP}$.
\end{proof}

\section{Separating $\mathsf{ACA}$ from $\mathsf{AP}$}
\label{sect: ACA AP}

\noindent In this section, we will show that there exists a model of $\mathsf{RCA_0} +\mathsf{AP}$ that is not a model of $\mathsf{ACA}$. To achieve this, we will prove that every topological system on Cantor space has an almost periodic point that is low relative to the system. Because the main theorem of this section is a separation result,  we could make use  of full induction. However, we will restrict ourselves to $I\Sigma_1$ induction so that we can make use of these results in Section~\ref{sect: conservation}.

The objective is to construct an almost periodic point of a system while forcing the jump. Let $(C,f)$ be a system and let $U$ be a c.e.\ set of strings. If there is a subsystem  $(D,f)$ of $(C,f)$ such that
$[D] \cap [U]= \emptyset$, then we can replace our original system with $(D,f)$. Any almost periodic point in $(D,f)$ is an almost periodic point of $(C,f)$ and we know that such a point cannot meet $U$.

If we cannot find such a subsystem, then we will show that for some $b$ for all $X \in [C]$ there exists some
$k<b$ with $F^k(X) \in [U]$. We will use this fact to build a new system $(D,g)$ such that
$[D] \subseteq [C] \cap [U]$ and for all $X \in [C]$, $G(X) = F^k(X)$ for some $k <b$. We will show that this gives us a certain recurrence  property that allows us to build an almost period point that meets $U$.

\begin{definition}\
\begin{enumerate}
\item Let $f,g:2^{<\omega} \rightarrow 2^{<\omega}$ encode continuous transformations of $2^{\omega}$.   Call  $g$  a \textit{piece-wise combination of iterates of }$f$ if for some $l, b$ there is a function $j:\{0,1\}^l \rightarrow \{1, \ldots, b\}$ such that for all $\sigma$ with $|\sigma| \ge l$,
$g(\sigma) = f(\sigma, j(\sigma\upharpoonright{l}))$.
\item Let $(C,f), (D,g)$ be systems. We say that
 $(D,g)$ \emph{refines} $(C,f)$, written  $(D,g) \le (C,f)$ if:
 \begin{enumerate}
 \item $D \subseteq C$.
 \item  $g$  is a {piece-wise combination of iterates of }$f$.
 \end{enumerate}
\end{enumerate}
\end{definition}

Clearly if $(D,f)$ is a subsystem of $(C,f)$ then $(D,f) \le (C,f)$.

\begin{lemma}$(\msf{WKL}_0)$
\label{lemma: nbh ap}
Let $(C,f), (D,g)$ be systems such that $(D, g) \le (C,f)$. If $X \in [D]$, then
$(\exists b)(\forall n)(\exists k \le b) F^{n+k}(X) \in [D]$.
\end{lemma}
\begin{proof}
Assume this fails for some $X \in [D]$. Let $b$ witness that $(D, g) \le (C,f)$ i.e.\ $\{1, \ldots, b\}$ is range of the function $j$. Consider the set of $n$ such that
\[\{F^{n+1}(X), F^{n+2}(X), \ldots, F^{n+b}(X)\} \cap [D]= \emptyset.\]
This set is computable in $X$ and non-empty by assumption. Hence it contains a least element $l$.
By minimality and the fact that $F^0(X) = X \in [D]$ we must have that $F^l(X) \in [D]$ hence for some $k \in \{1, \ldots, b\}$,
$G(F^l(X))=F^k(F^l(X))$ and so
$F^{l+k}(X) = G(F^l(X))  \in [D]$ contradicting our assumption.
\end{proof}

\begin{lemma}$(\msf{WKL}_0)$
\label{lemma: transitive}
The refinement relation is transitive.
\end{lemma}
\begin{proof}
Let $(E,h) \le (D,g) \le (C,f)$. Clearly $[E] \subseteq [C]$. Let $j_1:\{0,1\}^{l_1} \rightarrow \{1, \ldots, b_1\}$ and
$j_1:\{0,1\}^{l_2} \rightarrow \{1, \ldots, b_2\}$ be such that for all $\sigma$, if $|\sigma| \ge \max\{l_1, l_2\}$ then
\begin{enumerate}
\item $g(\sigma) = f^{j_1(\sigma\upharpoonright{l_1})}(\sigma)$.
\item  $h(\sigma) = g^{j_2(\sigma\upharpoonright{l_2})}(\sigma)$.
\end{enumerate}
Let $b_3 = b_1 \cdot b_2$. Let $l_3 >l_2$ be sufficiently large such that for all $\sigma \in \{0,1\}^{l_3}$, for all $n < b_2$, $|g^n(\sigma)| > l_1$.

Take any string  $\sigma$ such that $|\sigma| \ge {l_3}$. Let $m = j_2(\sigma \upharpoonright{l_2})$. Then
 $h(\sigma) = g^m(\sigma)$. Further
\begin{align*}
g^m(\sigma) =& g\circ g^{m-1}(\sigma) \\
=& f^{j_1(g^{m-1}(\sigma)\upharpoonright{l_1})} \circ g^{m-1}(\sigma)\\
=&  f^{j_1(g^{m-1}(\sigma)\upharpoonright{l_1})} \circ f^{j_1(g^{m-2}(\sigma)\upharpoonright{l_1})} \circ \ldots
\circ f^{j_1(g^{0}(\sigma)\upharpoonright{l_1})}(\sigma)
\end{align*}
Hence $h(\sigma)= f^n(\sigma)$, where $n = \sum_{i=0}^{m-1} j_1(g^{i}(\sigma)\upharpoonright{l_1})$. Because $m \le b_2$,  we have that $n$ only depends on $\sigma \upharpoonright{l_3}$. Further $n \le b_3$. Hence $h$ is a piece-wise combination of iterates of $f$ and so
$(E, h) \le (C, f)$.
\end{proof}

\begin{lemma} $(\msf{WKL}_0)$
\label{lemma: meet or avoid}
Let $(C,f)$ be a system and $U$ a c.e.\ set. There is a system $(D,g)$ refining $(C,f)$ such that
either:
\begin{enumerate}
\item \label{out} $[D] \cap [U]= \emptyset$; or
\item \label{in} $[D] \subseteq [U]$.
\end{enumerate}
\end{lemma}
\begin{proof}
Define
\[D_0 = \{ \sigma \in C \colon (\forall n \le |\sigma|)(\forall \tau \in U[|\sigma|])
(f(\sigma,n) \not \succeq \tau)\}.\]
To establish that $D_0$ is a tree, let $\sigma$ and $\sigma'$ be any two strings such that $\sigma \preceq \sigma'$. Assume $\sigma \not \in D_0$. If $\sigma \not \in C$ then because $C$ is a tree $\sigma' \not \in D_0$. If $\sigma \in C$ then for some $n \le |\sigma|$ and $\tau \in U[|\sigma|]$, $f(\sigma, n) \succeq \tau$. Hence $f(\sigma', n) \succeq \tau$ and $\tau \in U[|\sigma'|]$. Thus $\sigma' \not \in D_0$.

\begin{claim}
For all $\sigma \in D_0$, $f(\sigma) \in D_0$.
\end{claim}
\begin{proof}[Proof of claim]
If $f(\sigma) \not \in D_0$, then let $\sigma' =f(\sigma)$. There is some $n \le |\sigma'|$ and $\tau \in U[|\sigma'|]$ such that
$f(\sigma', n) \succeq \tau$. Hence $f(\sigma, n+1) \succeq \tau \in U[|\sigma|]$. We have that $\sigma \not \in D_0$ because $n+1 \le |\sigma|$ (here we use our assumption that
$|f(\sigma)| <|\sigma|)$.
\end{proof}
This claim establishes that if $D_0$ is infinite, then $(D_0, f)$ refines $(C,f)$, and by the definition of $D_0$ with $n=0$, we have that $[D_0] \cap [U]= \emptyset$.

Now consider the case that $D_0$ is finite. Let $s$ be least such that $D_0$ contains no string of length $s$. Define
\[ D_1 = \{ \sigma \in C \colon (|\sigma| <s) \vee (\exists \tau \in U[s]  ( \sigma \succeq \tau))\}.\]
We will show that $D_1$ is infinite.  Take any $X \in [C]$ and let $\sigma = X\upharpoonright{s}$. As $\sigma \not \in D_0$, there is a $k \le s$ such that $f^k(\sigma) \succ \tau \in U[s]$. Hence for all $n$, $f^k(X\upharpoonright{n}) \in D_1$.

Let $l$ be such that if $|\sigma| = l$, then $|f(\sigma)| \ge s$.
Define $j:\{0,1\}^l \rightarrow \{1, \ldots, s+1\}$ by
\[j(\sigma) = \begin{cases}
1 & \sigma \not \in C \\
k& \mbox{where }k \ge 1\mbox{ is  least  such that }(\exists \tau \in U[s] )f(\sigma, k) \succeq \tau\mbox{ if }\sigma \in C.
\end{cases}\]
Note that $j(\sigma)$ is well-defined because for $\sigma \in \{0,1\}^l$ we have that $|f(\sigma)| \ge s$ and so $f(\sigma) \not \in D_0$. Hence for some $k \in \{0, \ldots, s\}$ we have that $f(f(\sigma, k))$ extends some element of $U[s]$.
Define a function $g:2^{<\omega} \rightarrow 2^{<\omega}$ by
\[g(\sigma) = \begin{cases}
\lambda &\mbox{if }|\sigma| < l\\
f(\sigma, j(\sigma \upharpoonright{l})) & \mbox{otherwise.} \end{cases}\]

It follows from the definition of $j$ and $D_1$, that if  $\sigma \in [C]$, then $g(\sigma) \in [D_1]$. Hence $(D_1, g)$ is a system. Clearly, $g$ is a piece-wise combination of iterates of $f$ and hence $(D_1,g) \le (C,f)$. Finally $[D_1] \subseteq U[s]$.
\end{proof}

The proof given of the proceeding lemma provides some more information that we will make use of in Section~\ref{sect: conservation}. We state this as the following lemma.

\begin{lemma}\label{lemma: inout} $(\msf{WKL}_0)$ \
Consider the set
\[\{ \sigma \in C \colon (\forall n \le |\sigma|)(\forall \tau \in U[|\sigma|])
(f(\sigma,n) \not \succeq \tau)\}.\]
 Case \eqref{out} of Lemma~\ref{lemma: meet or avoid} holds if this set is infinite.
 Case \eqref{in} of Lemma~\ref{lemma: meet or avoid}  holds if this set is finite, and further there is a $(D,g)$ refining $(C,f)$ such that for all $X \in [C]$ there is a $k$ with $F^k(X) \in [D]$.
\end{lemma}

We make use of full induction for the following lemma.
\begin{lemma}
Any system  $(C,f)$ contains an almost periodic point $X$ such that $X' \le_T (C\oplus f)'$.
\end{lemma}
\begin{proof} We define a sequence of systems $\{(C_e, f_e)\}_{e \in \omega}$ such that for all $e$,
$(C_{e+1}, f_{e+1}) \le (C_e, f_e)$.
Let $(C_0, f_0) = (C,f)$.
At stage $e+1$, let $U_e = \{ \sigma \in 2^{<\omega} \colon \Phi_e^\sigma(e) \downarrow\}$.
Let $(C_{e+1}, f_{e+1})$ refine $(C_e, f_e)$ such that either
$[C_{e+1}] \cap [U_e] = \emptyset$ or  $[C_{e+1}] \subseteq [U_e]$.

An  examination of the  proof of Lemma~\ref{lemma: meet or avoid} shows that this sequence can be constructed below $(C \oplus f)'$. In Lemma~\ref{lemma: meet or avoid}, $D_0$, $D_1$ and $g$ are defined uniformly from $C$ and $f$ (the definition of $D_1$ and $g$ depend on $D_0$ being finite). Further $(C\oplus f)'$, can determine whether or not $D_0$ is finite and hence decide how to refine  $(C,f)$.

By compactness, $\bigcap_e [C_e]$ is not empty. In fact $\bigcap_e [C_e]$ contains a unique point $X$, because every finite set occurs as infinitely many c.e.\ sets $U_e$. Now $X' \le_T  (C\oplus f)'$ because whether $\Phi_e^X(e)$ halts can be determined at stage $e$ of the construction. We show that $X$ is an almost periodic point of $C$. Fix $\sigma \prec X$. Now for some $e$, $[\sigma] \supseteq [C_e]$.  Thus  by Lemma~\ref{lemma: nbh ap} there is some bound $b$, such that for all $n$, there is some $k \le b$ such that
$F^{n+k}(X) \in [C_e] \subseteq [\sigma]$.  Hence $X$ is an almost periodic point of $(C,f)$.
\end{proof}

Using the standard approach, the previous proposition can be used to build an $\omega$-model of $\msf{RCA_0}$ and $\mathsf{AP}$ such that every real in the model is low. Hence we obtain the following theorem.

\begin{theorem}
\label{thm: AP does not imply ACA}
There is an $\omega$-model of $\msf{RCA}_0$ and $\mathsf{AP}$  that is not a model of $\msf{ACA}$.
\end{theorem}

\section{A Conservation Result}
\label{sect: conservation}
\begin{lemma}[$\msf{WKL_0}$]
\label{lemma: preservation induction}
Let  $(C,f)$ be a system, $P$ be a real, and $\varphi$ be a $\Delta_1$ formula. There is a system $(D,g)$ refining $(C,f)$ such that either
\begin{enumerate}
\item The set $\{m : (\exists s) \varphi(m,X\upharpoonright{s}, P)\}$ is empty for  all $X \in [D]$; or
\item There is a $b$ such that for  all $X \in [D]$, $b$ is the least element of $\{m : (\exists s) \varphi(m,X\upharpoonright{s}, P)\}$.
\end{enumerate}
\end{lemma}
\begin{proof}
For all $n$, define the following sets
\begin{align*}
&O(\le n) = \{\tau \colon (\exists m \le n) \varphi(m,  \tau, P)\},\\
&O(< n) = \{\tau \colon (\exists m < n) \varphi(m, \tau, P)\}, \\
&C(\le n) =\{ \sigma \in C \colon (\forall i \le |\sigma|)(\forall \tau \in O(\le n))(f(\sigma, i)\not \succeq \tau )\},\\
&C(< n) =\{ \sigma \in C \colon (\forall i \le |\sigma|)(\forall \tau \in O(< n))(f(\sigma, i)\not \succeq \tau )\}.
\end{align*}
Because of the uniformity in the above definitions, we have that the following set is c.e.
\[S= \{n \colon C(\le n) \mbox{ is finite}\}.\]
Hence by $I\Sigma_1$ induction $S$ is either empty or contains a least element. First we consider the case that  $S$  is empty.
Let $U= \{\tau \colon (\exists n)\varphi(n,\tau, P)\}$.
Let $(D, g)$ be a refinement of $(C,f)$ guaranteed by Lemma~\ref{lemma: meet or avoid}.
If $[D] \cap [U] = \emptyset$, then the system $(D,g)$ has the  required properties.  If $[D] \subseteq [U]$, then by applying compactness there is some bound $b$,
such that
\[[D] \subseteq [\{\tau \colon (\exists n \le b)\varphi(n,\tau)\}].\]
We now replace $(C,f)$ by $(D,g)$ and adjust the definitions accordingly. This replacement does not affect the following argument  because of the transitivity of the refinement relation. Note that now $S$ cannot be empty because it must contain $b$.

Now consider the case when $S$ has a least element $b$. By Lemma~\ref{lemma: inout}, there is a system $(E, h)$ such that
$(E, h)$ refines $(C,f)$ and $[E] \subseteq [O(\le b)]$.
Further we have that for all $X \in [C]$ there is a
$k$ such that $F^k(X) \in [E]$.
Now consider the set
\[\{\sigma \in E \colon (\forall i \le |\sigma|)(\forall \tau \in O(< b))(g(\sigma, i)\not \succeq \tau )\}.
\]
If this set is empty, then if $X \in [C]$, we have that $F^k(X) \in [E]$ for some $k$ and so for some $j$, $G^j(F^k(X))$ must extend some element of $O(<b)$. This implies that
$[C] \subseteq [O(< b)]$ as $g$ is a piecewise combination of iterates of $f$. Using compactness, this implies that $C(\le m)$ is empty for some $m$ strictly less than $b$, contradicting the minimality of $b$. Hence the set is not empty and so by Lemma~\ref{lemma: inout} there is a a system $(F,i)$ refining $(E,h)$ (and consequently refining $(C,f)$) such that $[F] \cap  [O(< b)]= \emptyset$. If $X \in [F]$, then $\{m \colon (\exists s) \varphi(m, X\upharpoonright{s},P)\}$ has least element $b$.
\end{proof}

\begin{lemma}[$\msf{WKL_0}$]
\label{lemma: nbh periodicity}
Let $(C,f)$ be a system and $i \in \omega$. There is a subsystem $(D,f)$ of $(C,f)$ and  $b\in \omega$ such that for all $X\in D$
\[(\forall n)(\exists k < b)(F^{n+k}(X) \in [X\upharpoonright{i}]).\]
\end{lemma}
\begin{proof}
Take a computable coding of finite sets of finite strings $\{E_k\}_{k\in \omega}$ such that if $E_k \subseteq E_j$, then $k\le j$. Now enumerate a c.e.\ set $W$ by adding $k$ to $W$ if
\begin{enumerate}
\item The number $k$ is a code for a finite set $E_k \subseteq \{0,1\}^i$.
\item The following tree is finite
\[\{ \sigma \in C  \colon (\forall n \le |\sigma|)(\forall \tau \in E_k)(f(\sigma, n) \not \succeq \tau)\}.\]
\end{enumerate}
As $C \ne \emptyset$, the code for $\emptyset$ is not an element of $W$. The code for $\{0,1\}^i$ is an element of $W$. Hence by $I\Sigma_1$ induction, there is a maximum element $k$ such that $E_k \subseteq \{0,1\}^i$, and for all $x \le k$ that code subsets of $\{0,1\}^i$, $x \not \in W$.  Let
\[D=\{ \sigma \in C  \colon (\forall n \le |\sigma|)(\forall \tau \in E_k)(f(\sigma, n) \not \succeq \tau)\}.\]
Hence $D$ is infinite and $[D] \subseteq [\{0,1\}^i \setminus E_k]$.
Now if $\sigma \in  \{0,1\}^i \setminus E_k$, then let $s_\sigma$ be the least number such that
\[\{ \sigma \in C  \colon (\forall n \le |\sigma|)(\forall \tau \in E_k\cup \{\sigma\})(f(\sigma, n) \not \succeq \tau)\}\]
contains no strings of length $s_\sigma$. Hence for any $X \in [D]$ there is some $n \le s_\sigma$ such that $F^n(X) \in [\sigma]$. In particular, this includes  any element of  $[D] \cap [\sigma]$. The set
$\{(\sigma, s_\sigma) \colon \sigma \in   \{0,1\}^i \setminus E_k\}$ is also c.e.\ and hence by $B\Sigma_1$ induction, there is some $b$ that bounds all elements of this set.
\end{proof}

\begin{definition}
Let $\mathcal{M}$ and $\hat{\mathcal{M}}$ be models of 2\textsuperscript{nd} order arithmetic. Call  \textit{$\mathcal{M}$  an $\omega$-submodel of $\hat{\mathcal{M}}$} if $\mathcal{M}$ is a submodel of $\hat{\mathcal{M}}$ and they share the same first order part.
\end{definition}

\begin{theorem}[Harrington -- unpublished see \cite{Simpson_2009}] Let $\mathcal{M}$ be a countable model of $\msf{RCA}_0$. Then there exists a countable model $\hat{\mathcal{M}}$ of $\msf{WKL}_0$ such that
$\mathcal{M}$ is an $\omega$-submodel of $\hat{\mathcal{M}}$.
\end{theorem}

\begin{lemma}
\label{lemma: adding ap point}
Let $\mathcal{M}$ be a countable model of $\msf{WKL}_0$ and let $(C,f)$ be a system in $\mathcal{M}$.
Then there exists a model $\hat{\mathcal{M}}$ of $\msf{WKL}_0$ such that $\mathcal{M}$ is an $\omega$-submodel of $\hat{\mathcal{M}}$ and $\hat{\mathcal{M}}$ contains an almost periodic point for the system $(C,f)$.
\end{lemma}
\begin{proof}
By Harrington's theorem it is only necessary to find a model $\hat{\mathcal{M}}$ of $\msf{RCA}_0$ such that $\mathcal{M}$ is an $\omega$-submodel of $\hat{\mathcal{M}}$ and $\hat{\mathcal{M}}$ contains an almost periodic point for the system $(C,f)$.
Let $ \mathbb{N}$ be the natural numbers inside $\mathcal{M}$ and let $ \mathbb{R}$ be the reals inside $\mathcal{M}$. From outside of the model,
let
$g: \omega \rightarrow  \mathbb{N}$ and $h:\omega \rightarrow  \mathbb{R}$ be bijections.
Define $(C_0, f_0) =(C, f) $. Now inductively define $(C_{e+1}, f_{e+1})$ as follows.\begin{enumerate}
\item If $e =2\cdot \langle n, m\rangle +1$,  then let $(C_{e+1}, f_{e+1})$ refine $(C_{e}, f_{e})$ as per Lemma~\ref{lemma: preservation induction} with $\varphi$ the $n$\textsuperscript{th} $\Delta_1$ formula and $P=h(m)$.
\item If $e=2\cdot n+2$ then let $(C_{e+1}, f_{i+1})$ refine $(C_{e}, f_{e})$ as per Lemma~\ref{lemma: nbh periodicity} with $i=g(n)$.
\end{enumerate}

Take $R \in \bigcap_e [C_e]$. Let $\hat{\mathcal{M}}$ be the model obtained by adding all reals computable in $R\oplus Y$ to $\mathcal{M}$ for any $Y \in \mathcal{M}$.
We will show that $\hat{\mathcal{M}}$ is a model of $I\Sigma_1$ induction. Let $W$ be the  $n$\textsuperscript{th} c.e.\ set relative to $R$ with parameter $h(m)$. Let $e=2\cdot \langle n, m\rangle +1$. If the outcome of
Lemma~\ref{lemma: preservation induction} was that the set
\[\{m : (\exists s) \varphi(m,X\upharpoonright{s}, P)\}\]
is empty for all $X \in [C_{e+1}]$, then the set $\{m : (\exists s) \varphi(m,R\upharpoonright{s}, P)\}$ is also empty because otherwise a path with this property could be found inside $\mathcal{M}$ using $\msf{WKL}$. Similarly if for some $b$,  the  set
\[\{m : (\exists s) \varphi(m,X\upharpoonright{s}, P)\}\]
has a least element $b$, for all $X \in [C_{e+1}]$, then the set
$\{m : (\exists s) \varphi(m,R\upharpoonright{s}, P)\}$ also has least element $b$.

We know that $R$ is an almost periodic point of $(C,f)$ because $R \in [C_{e+1}]$ for $e=2\cdot n+2$ establishes that there exists a $b$ such that for all $m$, there is a $k \le b$ such that
$F^{m+k}(R) \in R\upharpoonright{g(n)}$.
\end{proof}

\begin{theorem}Let $\mathcal{M}$ be a countable model of $\msf{RCA}_0$. Then there exists a countable model $\hat{\mathcal{M}}$ of $\msf{RCA}_0 + \mathsf{AP}$ such that
$\mathcal{M}$ is an $\omega$-submodel of $\hat{\mathcal{M}}$.
\end{theorem}
\begin{proof}
Let $\mathcal{M}$ be a countable model of $\msf{RCA}_0$. Let $g:\omega \rightarrow \omega \times \omega$ be a bijection such that for all $n$, if $(i,j) = g(n)$ then $\max\{i,j\} \le n$. (Note that $\omega$ is the real $\omega$ and $g$ exists outside the model.)

Using Harrington's theorem,  let $\mathcal{M}_0$ be a model of $\msf{WKL}_0$ such that $\mathcal{M}$ is an $\omega$-submodel of $\mathcal{M}_0$. Applying  Lemma~\ref{lemma: adding ap point}, let $\mathcal{M}_{n+1}$ be a model of $\msf{WKL}_0$, such that  $\mathcal{M}_n$ is an $\omega$-submodel of $\mathcal{M}_{n+1}$, and $\mathcal{M}_{n+1}$ contains an almost periodic point for the $i$\textsuperscript{th} system in $\mathcal{M}_j$ where $(i,j) = g(n)$. Let $\mathcal{M}_\omega = \bigcup_i \mathcal{M}_i$.

Clearly $\mathcal{M}_\omega \models PA^{-} + \mathsf{AP}$. It also models $I\Sigma_1$ induction because if not there is some function $f: \mathbb{N} \rightarrow  \mathbb{N}$ in $\mathcal{M}_\omega$ whose range has no least element. But if so, for some $n$, $f \in \mathcal{M}_n$ contradicting the fact that $\mathcal{M}_n \models \msf{WKL}_0$.
\end{proof}

We obtain the following corollary by applying the standard argument (see for example \cite[Corollary~IX.2.6]{Simpson_2009}).
\begin{corollary} $\msf{RCA}_0 +\mathsf{AP}$ is conservative over $\msf{RCA}_0$ for $\Pi^1_1$ sentences.
\end{corollary}

\section{A Subclass of PA Degrees}
\label{sect: PA degrees}
\noindent Consider the set of reals that  given any computable system $(C,f)$ can compute an almost periodic point for this system. This is an upwards-closed subclass of the $\mathrm{PA}$ degrees. By Theorem~\ref{thm: WKL does not imply AP}, we know that this is a strict subclass of the $\mathrm{PA}$ degrees.
The following corollary shows that this subclass does not coincide with those $\mathrm{PA}$ degree above $\emptyset'$

\begin{corollary}[Corollary to Theorem~\ref{thm: AP does not imply ACA}]
There is a set $X$ of $\mathrm{PA}$ degree such that $X \not \ge_T \emptyset'$ and $X$ computes an almost periodic point for every computable system.
\end{corollary}
\begin{proof}
Let $\{Z_i\}_{i\in \omega}$ be a listing of the  ideal used to separate $\msf{ACA}$ from $\mathsf{AP}$ over $\msf{RCA}_0$. Observe that no finite join of this sequence computes~$\emptyset'$.

Construct $X$ by at stage $e$ defining sufficient columns of $X$ to force that $\Phi_e^X \ne \emptyset'$, and then append $Z_e$ to an empty column of $X$.  The set $X$ bounds all elements of the ideal so $X$ is of $\mathrm{PA}$ degree.
\end{proof}

The $\mathrm{PA}$ degrees have been extensively studied. However, this subclass does not appear to have been encountered before and it merits further investigation.

\begin{question}
\label{Q1}
Are there any other characterizations of this subclass?
\end{question}

A useful  answer to Question~\ref{Q1} would give some indication as to how this subclass is dispersed in the Turing degrees. As there are computably dominated sets of $\mathrm{PA}$ degree, it is natural to ask the following question.

\begin{question}
Does this subclass have a computably dominated element?
\end{question}

For this subclass to have a computably dominated element, it is necessary that the following question has a positive answer.

\begin{question}
Does every computable system have an almost periodic point that is computably dominated?
\end{question}

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