Danny DeVito was dead on as the Lorax. So much so, that the producers at Universal couldn´t imagine anyone else playing his part so well in the Spanish, Italian, German and Russian versions. They decided that not only should DeVito speak for the trees as the Lorax, but also for those of who watch the dubbed versions. Universal asked him if he would do those versions-knowing fully well he wasn´t fluent in any of those languages!-and DeVito graciously agreed to take the multilingual plunge! He actually had to do five versions (not including the English one!) because he did both the Castillian and the Mexican Spanish ones!
We asked him all about this experience and why he would even go through the trouble of doing it. Watch this video where you can see him with the Spanish voice coach and hear his español; then read on for an entertaining excerpt of our interview.
How long did it take you to learn the four languages?
Danny DeVito: Oh, well, here's the thing about that. I didn't learn the languages. I'm the guy in school who went and took Latin forever and just couldn't even say open the door. I speak very little Italian from just the home slang. And it wouldn't ever, ever fly as a dub job for Italy.
So, the way I did it was I had two translator expert people who know the language and all the inflections and the proper way to say the words. And then, I had another person from the country in the booth. We had the picture in English with my voice. And what I wanted to do was make sure I got the performance big time, and also, they were there to police the pronunciation so that the inflections were right.
So, I did Spanish twice because one of them, you speak with it. They don't even say Thneed. They say something else because Thneed, it would be some word you don't want to say probably.
And then, the Italian was really fun. And the German turned out to be one of the most challenging right away.
But they're going to appreciate it. And it wound up sounding really good. I've heard a lot of it. I play it back a lot. I'm very, I don't know what you call it – I guess a perfectionist. But, you really want to make sure you're doing it.
And I had these two poor people on the other side of me repeating it and saying the lines over and over again and me looking at the phonetic spelling.
Why did you decide to go through the trouble?
Danny DeVito: First of all, I like to do different things. I like to challenge myself. And I knew it wasn't going to be a piece of cake.
Once I got it in my head that I was going to do it, I said, I'm not going back out now. I'm going to just go.
You've taken your clothes off and you're ready to dive in the river. You're not going to back off.
My whole family was saying, boy, that's really great. Imagine going to dinner one night and thinking, I've got cold feet, I don't really want to do it, it's going to take too long.
But, I had great people with me and I stuck to it, and it worked out good. And I'm going to see it in Russia and in March I'm going to go see it in Russia and Spain and Germany and Italy.
Dr. Seuss' The Lorax (El Lórax) hits theaters this Friday, March 2nd. You won´t want to miss this one, trust me. Watch and share The Lorax trailer on Facebook, plus watch clips, play games, get free mobile downloads and download printables.
Check out our giveaway for a chance to win a prize bundle which includes a copy of Dr. Seuss' The Lorax book in Spanish, plus other fabulous goodies.
What dialect was he using? I noticed when he was saying "ahora me llamas" he said it like a Colombian.


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text:
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redpajama_set_name: RedPajamaCommonCrawl


text:
YANKS RELEASE PHIL NIEKRO, DEALS BAYLOR
Craig Barnes, Staff WriterSUN-SENTINEL
The New York Yankees grabbed baseball headlines all over Friday, releasing 46-year-old knuckleballer Phil Niekro in a move that somewhat overshadowed an earlier trade that saw them deal designated hitter Don Baylor to Boston for the Red Sox' designated hitter, Mike Easler.
The trade of the 36-year-old Baylor, who hit .231 with 23 homers and 91 RBI in 1985, for the 35-year-old Easler, who hit .262 with 16 homers and 74 RBI in 1985, came as no surprise.
Niekro, however, was stunned by his release, which came one day before the Yankees would have been bound to pay him his entire 1986 contract of $350,000.
The club tried to get Niekro, a 20-year veteran, to sign a waiver releasing it of any guarantee until the season started, meaning if the right-hander had consented and eventually been cut, the club would have owed him only 45 days severance pay, $62,145.
"It had not been a good spring for Phil until (Thursday) night," said Woody Woodward, vice president for baseball administration, "and we wanted the few extra days to evaluate him."
For the spring, Niekro had allowed 14 hits and eight runs in 13 innings, pitching in four games.
Niekro, who won 16 games in each of the last two seasons, said he was signed by the Yankees "as bait" to get his brother Joe to re-sign with the team. He also said he wouldn't have made the team if "I had pitched 25 scoreless innings."
Woodward denied both charges.
"The decision wasn't strictly a front-office decision," manager Lou Piniella said. "It wasn't a dead issue, but I couldn't promise Phil anything. I asked him to stay. I thought he could still pitch, but decisions had to be made."
Both Niekros were visibly angry about the release.
"It's like they wanted me as an ace up their sleeve," said Phil, who won his 300th major league game on the final day of the 1985 season. "They knew when I came down here I wasn't going to make the team."
Phil said he couldn't understand why his birthday was so important to the Yankees and not his productivity.
"They can talk about age, but they're not going to find anybody who wants to win more than I do," he said.
Phil was particularly upset because he decided not to take the Yankees to arbitration in the offseason.
"I probably saved them $400,000 by not going," he said. "I thought I was doing it in good faith when I was burying myself all the time.
"They're not dealing with a piece of meat. They're dealing with a human being."
Phil, who will be 47 Tuesday, signed his new contract, and then Joe, who was a free agent at the end of 1985, agreed to a three-year contract with the Yankees in January.
"I think I was used so the Yankees could secure Joe," Phil said. "Once he had signed, I was expendable. All I wanted was honesty and truth. I hate to be suckered in like that. It pulls all the air out of you, all the time them knowing I wasn't going to make the team."
Joe, 41, said he was bitter about the way his brother was treated. He stopped short of requesting a trade but emphasized he wouldn't be happy pitching for the Yankees.
"I signed a three-year contract to pitch for the organization, and I want to win for guys like (Dave) Winfield, (Don) Mattingly, (Willie) Randolph and the others," he said, "but if that's the way they treat winners, why should I want to be here?
"Phil and I are more than brothers. We are friends. What happens to him, happens to me. I will pitch my tail off for this team, but it doesn't mean I have to do it with a smile on my face, and I won't. I don't understand what they did."
If the move was a financial one, Joe had a solution. "I told them to take part of my salary ($2.5 million for three years)," he said. "I believe the man deserved a chance to start with the team. If he failed later, then release him."
Joe, a right-hander who is 204-180 during his 17-year career, said he expected Phil to be with the Yankees when he signed.
"Why wouldn't you think he would be here?" Joe said. "Spring training shouldn't be a determining factor for him, especially based on one bad outing in wind and rain. He's proved what he could do.
"How can you release a 16-game winner? I don't know the answer."
Phil didn't have the answer either. He took his uniform, stirrup socks, hat and cleats to Woodward's office when he notified him of his decision not to sign the waiver. He was uncertain about his future.
"There are still 10 days until the start of the season," he said. "I don't know if anyone is interested. My career is obviously over as far as (the Yankees) are concerned.
"It just hurts because this is a good bunch of guys. I'd gotten close to Lou, the coaches, the players, the city and the fans. They're so damn close to a pennant. It makes it more difficult to swallow, especially when you don't understand."
Phil was wearing a belt buckle inscribed with "I'd Rather Be Fishing," and with his wife, Nancy, and their three sons arriving in town for spring break, fishing is all he has to do today.
Husband of killer clown suspect: Cops 'had nothing before, they have nothing now'


text:
Akhil's mother on hunger strike
By AT News | Tuesday, Mar 4, 2014 (10:03 am)
Akhil Gogoi's mother Priyoda Gogoi is on a hunger strike near Jorhat on Tuesday demanding immediate release of her son who was arrested on Sunday in connection with the recent self immolation case.
Flanked by 50 local KMSS workers, Gogoi began her hunger strike at her own residence in Teok from Tuesday morning.
Several local organizations also have visited her and promised help.
Assam Times Staff. [email protected]
Priyodi
akhil
Padma to join BJP!
21 Mar 2014 - 10:02am | AT News
Yet another AGP leader is likely to join BJP. He is none other than Padma Hazarika who is yet to make up his mind.Hazarika, who was suspended from the party for anomalies in the Rajya Sabha voting...
Bulti Roy: Will she get justice?
22 Feb 2009 - 9:38pm | Dibya J Borthakur
It is beyond imagination for one how a teacher threw off the future of his own student until one meets Bulti.Bulti Roy, a 17 year old girl from Dighliati village under Roha police station was going...
A tribute to Shantanu Goswami
30 Jan 2012 - 12:08am | editor
As a result of a long battle with a deadly disease Shantanu Goswami passed away on the January 19 at Gangaram Hospital, New Delhi. Shantanu served as Research Assistant in the Department of...


text:
This story is a part of Focal Level iPhone 2022, CNET's assortment of stories, ideas and recommendation round Apple's hottest product.
Again in 2007, David Barnard's plans for getting his first iPhone have been easy. He was going to stroll as much as AT&T retailer through the weekend and purchase one. That is it.
However his brother Sam satisfied him in any other case.
It took a easy cellphone name to steer David to drive to the Apple Retailer in San Antonio and wait in line, and be one of many very first individuals to purchase one.
That iPhone launch has at all times been a treasured reminiscence for David, who was captured by a San Antonio Specific-Information photographer as he, his brother and sister-in-law have been standing on the entrance of the road and as they entered the shop. David's response shortly after getting his fingers on the coveted system ended up within the newspaper.
The iPhone launches took on an extra layer of which means after Sam handed away from most cancers in 2015.
"We fought like cats and canines as children, after which we began to form of bond round being Mac nerds," David says.
Fifteen years later, David continues to be a self-professed "hopeless fanboy," however a lot of all the things else has modified. Because the iPhone went mainstream, Apple turned from an underdog to a titan of business. Its gross sales jumped greater than 15-fold to $366 billion final 12 months, up from $24 billion for the entire iPhone's launch 12 months in 2007.
See additionally: iPhone 14 Professional and 14 Professional Max Assessment: Welcome to Apple's Dynamic Island
The tech business's grown alongside Apple as nicely. Again in 2007, billions of individuals have been on-line and utilizing computer systems, however in the present day the size is far bigger. Fb, now the world's largest social community with greater than 2.9 billion customers logging every month, counted fewer than 100 million again then. And the worldwide smartphone market was lower than 10% the dimensions it's in the present day.
But these iPhone traces aren't what they was.
Business watchers, historians and analysts agree there are various causes the traces now not materialize like in years previous. One motive is e-commerce, which in the present day permits individuals like Barnard to purchase their iPhones on-line and simply have them shipped to their doorstep. Smartphones have additionally change into so mainstream, there is not as a lot cachet in being the first in your block with the most recent gadget.
Now we have extra advanced emotions concerning the tech business than we did again then too. The previous few years, tech corporations have been awash with controversies concerning how they deal with (or mishandle) our private privateness, how they've helped create huge authorities surveillance networks, or how they've allowed surprising quantities of hate and violence to spring from their merchandise.
See additionally: iPhone 14 Assessment: A Good Improve for Most Individuals
Right this moment, we're solely simply starting to reckon with the aftereffects of placing internet-connected supercomputers in our pockets.
"The smartphone itself is a tool loaded with optimistic and detrimental affiliation," mentioned Margaret O'Mara, a historical past professor on the College of Washington and creator of The Code: Silicon Valley and the Remaking of America.
To make sure, Apple is just one firm, even when it's the world's most extremely valued one at about $2.45 trillion. Regulators and lawmakers all over the world are typically extra targeted on reining in friends like Fb father or mother Meta, Google and YouTube father or mother Alphabet, Amazon and Twitter, whose platforms and companies have helped embolden individuals in search of to tear down trendy democracy itself.
Nonetheless, O'Mara says, even when there are fewer traces outdoors Apple Shops, the diehards have been nonetheless there for the iPhone 14 launch Friday. Others, in the meantime, have moved on-line, to social networks and stay streaming platforms, the place they share, debate, focus on and obsess.
"There are nonetheless very intense and passionate fandoms with an eagerness to be the primary in line, so to talk, or to be engaged," she mentioned.
A line of keen iPhone 14 prospects outdoors the Apple Retailer on fifth Ave in New York Metropolis.
Joseph Kaminsky/CNET
Getting excited
Bob O'Donnell by no means stood in line for an Apple system, however he did go to e-book launch events for the Harry Potter collection along with his children. "It was an occasion," he mentioned.
A longtime business analyst and now founder of Technalysis Research, O'Donnell mentioned it is simply more durable to generate these ranges of pleasure for lots of issues, not to mention a tech gadget. "Actually, everybody has a smartphone now, and so now it is not as particular or distinctive," he added.
Nonetheless, he says, Apple might be able to draw these traces once more if it ever will get round to releasing its long-rumored headset, significantly as a result of digital actuality has struggled to stay as much as its hype.
John Maeda says Apple goggles might not deliver out the crowds, however the even-longer-rumored Apple Automotive would. The technologist and creator, who's labored at MIT Media Lab and Silicon Valley VC agency Kleiner Perkins Caufield & Byers, amongst different locations, mentioned what's helped Apple stand out is its capacity to construct well-designed merchandise with equally well-written software program to energy them. "Corporations that may do each are uncommon," he mentioned.
Apple's newest iPhone launches Friday.
It is also why individuals like David Barnard are nonetheless enthusiastic about iPhones 15 years after the primary one landed on retailer cabinets. Shortly after the iPhone launched, Barnard started shifting his profession to app growth, which finally led him to his present job as a developer advocate at app gross sales platform RevenueCat. Barnard eagerly preordered the iPhone 14 Professional per week earlier than its debut, and mentioned he is trying ahead to attempting the Dynamic Island, a brand new approach to swap between apps on the prime of the display screen.
And if he hadn't been capable of get an iPhone delivered to his residence, Barnard mentioned you'd in all probability have discovered him in keeping with a pair mates outdoors an Apple Retailer.
"I would complain on Twitter, however I might do it," he mentioned. "And I might be completely happy and excited to do it, as a result of it's an expertise."
Will YouTube's pledge to combat on-line extremism make a distinction? Verdict


text:
1st Step to Heaven is a 1986 album by Deutsch Amerikanische Freundschaft, released on Dean Records via Ariola in Europe.

It is their only album in English rather than German. Singer Gabi Delgado-López later noted: "So we wanted to break our own rules and said: OK, so now we sing in English, now we don't wear black. (laughing) With purpose. Because we wanted to break our own rules."

The album charted for one week on the Swedish album charts at No. 46. The first single from the album, "Voulez-vous coucher avec moi ce soir" (an original, though the title is a quote from the lyrics of "Lady Marmalade"), reached No. 47 on the German singles chart (charting for 7 weeks) and No. 17 on the Belgian singles chart (charting for 5 weeks). Another single, "Brothers", reached No. 23 on the Belgian singles chart (charting for 4 weeks).

The band reunited to make the album but split again soon after its release.

A variant track listing was released in the US as Hitz Blitz in 1989.

Track listing

 LP (Dean 207 435)
Side A:
"Voulez-vous coucher avec moi" 	5:04
"Opium" 	1:58
"Brothers" 	3:53
"Voulez-vous coucher avec moi Part II" 	5:13
Side B:
"Sex Up" 	3:26
"Pure Joy" 	3:30
"Party" 	3:23
"1st Step to Heaven" 	5:53

 CD (Dean 257 435)
 "Voulez-vous coucher avec moi - Part I" (2:25)
 "Pure Joy" (4:47)
 "Blond Hair Dark Brown Hair" (3:10)
 "Sex Up" (3:15)
 "Absolute Bodycontrol" (5:18)
 "Voulez-vous coucher avec moi - Part II"  (5:28)
 "Crazy Crazy" (3:08)
 "Opium" - Mix (2:10)
 "Brother/Brother" (3:37)
 "1st Step to Heaven" - Mix (5:46)

 Hitz Blitz CD (JCI JCD-9027)

 "The Gun" ("Powder Keg Mix") 	8:22
"Voulez-vous coucher avec moi Part 2" 	5:28
"Pure Joy" 	4:47
"Brothers" 	3:37
"Absolute Bodycontrol" 	5:18
"Sex Up" 	3:15
"Voulez-vous coucher avec moi Part 1" 	2:25
"1st Step to Heaven Medley" 	8:07
"Opium Mix" 	2:10
"Crazy Crazy" 	3:08
"Blond Hair Dark Brown Hair" 	3:10
"1st Step to Heaven Mix" 	5:46
"The Gun" ("Bang Bang Mix") 	6:23

References

1986 albums
Deutsch Amerikanische Freundschaft albums

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}
\normalsize

A metric space $(X,d)$ is {\it ultrametric} if the following {\it
ultrametric inequality}
$$
d(x,y)=d(x,z)\vee d(z,y)
$$
holds for all $x,y,z\in X$, where we put
$$
d(x,z)\vee d(z,y)=\max\{d(x,z),d(z,y)\}.
$$

The ultrametric spaces and, in particular, the finite ultrametric
spaces play an important role for the modern physics, data analysis,
mathematical taxonomy and many others branches of the knowledge (see, for example, \cite{L2}).
In particular, the following fact seems to be fundamental in the theory of
hierarchical models:
\begin{theorem}[see \cite{CM}]\label{th1.1}
For every dendrogram $\mathcal{D}$ (= nested
family of partitions) over a finite set $X$ there is an ultrametric
$d$ on $X$ representing this $\mathcal{D}$.
\end{theorem}
We do not define more exactly the sense of the word ``representing'' in Theorem~\ref{th1.1}
but note that the ``dual form'' of this theorem will be given in Theorem~\ref{th3.11} below.

There exist also the significant relations between (finite) ultrametric
spaces and many mathematical constructions.
The well known example is the canonical representation of finite
ultrametric spaces by the weighted rooted trees. Note that this
representation can be generalized for the case of representation of
quasi-ultrametric finite spaces by the weighted digraphs, \cite{GV}.
It is also shown in \cite{GV} that finite ultrametric spaces can be
viewed as $k$-person positional games. An isomorphism of categories
of ultrametric spaces and complete, atomic, tree-like, real graduated
 lattices was proved in \cite{L1}.

Some basic results in the theory of (finite) ultrametric spaces were
obtained quite recently.
The isometry groups of finite ultrametric spaces are completely
characterized in \cite{S}.
A set of necessary and sufficient conditions under which the Cartesian
 products of metric spaces are ultrametric was found  in \cite{DM}. The
 counterpart of Gromov-Hausdorff metric in the ultrametric spaces was studied in
 \cite{Z}, \cite{Q}. The infinitesimal structure of pointed metric spaces
 having ultrametric ``tangent'' spaces  was described in \cite{DD}.

In the present paper we solve the following two problems.

Let $(X,\rho)$ be a  metric space and let $A$ be a
subset of $X$. Recall that the {\it diameter} of $A$ is the quantity
$$
\diam A=\sup\{\rho(x,y): x,y\in A\}.
$$
We shall say that a two-element subset $\{a,b\}$ of $A$ is a $\rho$-diametrical pair
for $A$ if $\rho(a,b)=\diam A$. Write $\mathbb{R}:=(-\infty,\infty)$, $\mathbb{R}^+:=[0,\infty)$ and
$\mathcal{F}(Y)$ for the set of all nonempty finite subsets of a set $Y$.

(a) Under what conditions on a function
$\tau:\mathcal{F}(Y)\rightarrow\mathbb{R}$ there is an ultrametric
$\rho:Y\times Y\rightarrow\mathbb{R}^+$ such that
$$
\diam A=\tau(A)
$$
for every $A\in\mathcal{F}(Y)$?

(b) Characterize the structure of simple graphs $G=(V,E)$ for which there is an ultrametric
 $\rho$ defined on the vertex set $V$ such that the edge set $E$ coincides with
the set of $\rho$-diametrical pairs.

\section{Diameter in ultrametric spaces}

The following theorem describes the characteristic properties of
diameters of finite subsets in ultrametric spaces.

\begin{theorem}\label{t2.1}
Let $X$ be a nonempty set. The following two statements are
equivalent for every function
$\tau:\mathcal{F}(X)\rightarrow\mathbb{R}$.

\begin{itemize}
\item[$(i)$] For all $A,B,C\in\mathcal{F}(X)$ the function $\tau$
satisfies the conditions:
\begin{itemize}
\item[$(i_1)$] $(\tau(A)=0)\Leftrightarrow(\card A=1)$;
\item[$(i_2)$]
\begin{equation}\label{eq2.1}
\tau(A\cup B)\leqslant\tau(A\cup C)\vee\tau(C\cup B).
\end{equation}
\end{itemize}

\item[$(ii)$] There is an ultrametric $\rho:X\times
X\rightarrow\mathbb{R}^+$ such that
\begin{equation}\label{eq2.2}
\tau(A)=\diam A
\end{equation}
for every $A\in\mathcal{F}(X)$.
\end{itemize}
\end{theorem}

\begin{remark}
Conditions (i$_1$) and (i$_2$) are closely related to the similar
conditions from the paper of P.~Balk \cite{Balk}. In fact, Balk's paper
was a starting point in our consideration.
\end{remark}

\begin{remark}\label{r2.3}
Let (ii) hold. Properties (i$_1$) and (i$_2$) imply that the map
$\tau:\mathcal{F}(X)\rightarrow\mathbb{R}$ is isotonic and
nonnegative. Indeed, substituting of $A$ for $B$ into \eqref{eq2.1}
we see that
$$
\tau(A)\leqslant\tau(C)
$$
if $A\subseteq C$. Putting $A=\{c\}$ in the last inequality, where $c$
is an arbitrary element of $C$, and using (i$_1$) we obtain
$\tau(C)\geqslant 0$ for $C\in\mathcal{F}(X)$.
\end{remark}

\begin{proof}[\it Proof of Theorem \ref{t2.1}]
(ii)$\Rightarrow$(i). Let $\rho$ be an ultrametric on $X$ such that
$$
\tau(A)=\diam A,\quad A\in\mathcal{F}(X).
$$
Property (i$_1$) is evident. To prove (i$_2$) suppose that
$$
\underset{x,y\in A\cup B}{\max}\rho(x,y)=\underset{x,y\in
A}{\max}\rho(x,y).
$$
It implies
$$
\diam(A\cup B)=\underset{x,y\in
A}{\max}\rho(x,y)\leqslant\underset{x,z\in A\cup
C}{\max}\rho(x,z)
$$
$$
\leqslant\underset{x,z\in A\cup C}{\max}\rho(x,z)\vee \underset{y,z\in B\cup
C}{\max}\rho(y,z)=\diam(A\cup C)\vee\diam(C\cup B).
$$

Similarly we obtain \eqref{eq2.1} if
$\underset{x,y\in A\cup B}{\max}\rho(x,z)=\underset{x,y\in B}{\max}\rho(x,y)$.

Suppose now that $\underset{x,y\in A\cup
B}{\max}\rho(x,y)=\underset{x\in A,y\in
B}{\max}\rho(x,y)=\rho(a,b)$ where $a\in A,\,b\in B $. Then for
every $c\in C$ we have
$$
\diam(A\cup B)=\underset{x\in A,y\in B}{\max}\rho(x,y)= \rho(a,b)
\leqslant\rho(a,c)\vee\rho(b,c)
$$
$$
\leqslant
\underset{x\in A,z\in C}
{\max}\rho(x,z)\vee \underset{y\in B,z\in C}{\max}\rho(y,z)
\leqslant\diam(A\cup
C)\vee\diam(C\cup B).
$$
Property (i$_2$) follows.

(i)$\Rightarrow$(ii). Let $\tau:\mathcal{F}(X)\to\mathbb{R}$ be a
function satisfying  (i$_1$) and (i$_2$). Let us define a
function $\rho$ on $X\times X$ by the rule
$$
\rho(x,y):=\tau(\{x,y\}), \qquad x,y\in X
$$
where $\{x,y\}$ is the set with the elements $x$ and $y$. Property
(i$_1$) implies that $\rho(x,y)=0$ if and only if $x=y$. Moreover,
$\rho$ is nonnegative (see Remark \ref{r2.3}). The ultrametric
inequality $\rho(a,b)\leqslant\rho(a,c)\vee\rho(c,b)$
follows from \eqref{eq2.1} with $A=\{a\}$, $B=\{b\}$ and $C=\{c\}$.
Thus, $\rho$ is an ultrametric. It still remains to prove
\eqref{eq2.2} for  $A\in\mathcal{F}(X)$ with $\card A \geq 3$. Let $A=\{x_1,\ldots,
x_n\}$. Then for $j=1,\ldots,n$ we obtain
$$
\tau(A)=\tau(\{x_1,\ldots,x_{n-1}\}\cup\{x_n\})\leqslant\tau(\{x_1,\ldots,x_{n-1}\}\cup\{x_j\})\vee\tau(\{x_n,x_j\})
$$
$$
=\tau(\{x_1,\ldots,x_{n-1}\})\vee\rho(x_n,x_j)\leqslant\tau(\{x_1,\ldots,x_{n-1}\})
\vee(\underset{1\leqslant j\leqslant n}{\max}\rho(x_n,x_j)).
$$
The repetition of this procedure gives
$$
\tau(A)\leqslant\underset{1\leqslant i\leqslant j\leqslant n
}{\bigvee}\rho(x_i,x_j)=\diam A.
$$
As was shown in Remark \ref{r2.3} $\tau$ is an isotonic map.
Consequently, if $x_i,x_j$ is a diametrical pair of points of $A$, then
$$
\tau(A)\geqslant\tau(\{x_i,x_j\})=\rho(x_i,x_j)=\diam A.
$$
Thus, \eqref{eq2.1} holds for every $A\in\mathcal{F}(X)$.
\end{proof}

\begin{remark}\label{r2.4}
The second part of Theorem \ref{t2.1} shows, in particular, that for
every function $\tau:\mathcal{F}(X)\to\mathbb{R}$ satisfying
(i$_1$)--(ii$_2$) there is a {\it unique} ultrametric $\rho:X\times
X\to\mathbb{R}^+$ such that $\rho(x,y)=\tau(\{x,y\})$ for all $x,y\in X$.
\end{remark}

Theorem \ref{t2.1} can be generalized to the case of functions $\tau$ defined also on
infinite subsets of $X$.

Let $\mathfrak{Y}(X)$ be the set of nonempty subsets of $X$.

\begin{theorem}\label{t2.5}
The following two statements are equivalent for every function
$\tau:\mathfrak{Y}(X)\to\mathbb{R}$.
\begin{itemize}
\item[$(i)$] For all $A,B,C\in\mathfrak{Y}(X)$ the function $\tau$
satisfies conditions (i$_1$) and (i$_2$) from Theorem 2.1 and
the following condition
\begin{itemize}
\item[$(i_3)$]
\begin{equation}\label{eq2.3}
\tau(A)\leqslant\sup\{\tau(B): B\in\mathcal{F}(A) \}.
\end{equation}
\end{itemize}

\item[$(ii)$] There is an ultrametric $\rho:X\times
X\to\mathbb{R}^+$ such that $\tau(A)=\diam A$ for every
$A\in\mathfrak{Y}(X)$.
\end{itemize}
\end{theorem}

This theorem can be obtained as a consequence of Theorem \ref{t2.1}.
Indeed, since $\mathcal{F}(X)\subseteq\mathfrak{Y}(X)$, condition
(i) of Theorem \ref{t2.5} implies condition (i) of Theorem
\ref{t2.1}. Consequently, the restriction $\tau|_{\mathcal{F}(X)}$
is a diameter of finite sets for some ultrametric $\rho:X\times
X\to\mathbb{R}^+$. If $\card A=\infty$, then the equality
$\tau(A)=\diam A$  follows from the same equality with
$\card A<\infty$ by application of condition (i$_3$). The details of the
proof we leave to the reader.

\begin{example}\label{ex2.6}
Let $X$ be an infinite set. Define a function
$\tau:\mathfrak{Y}(X)\to\mathbb{R}$ as
$$
\tau(A)=
\begin{cases}
0 \quad\text{if } \card A=1, \\
1 \quad\text{if } 1<\card A<\infty, \\
2 \quad\text{if } \card A=\infty.
\end{cases}
$$
It is easy to prove that condition (i$_3$) does not hold but (i$_1$)
and (i$_2$) take place for this $\tau$. Thus, (i$_1$) and (i$_2$)
do not imply (i$_3$).
\end{example}

The key property of ultrametric spaces is that their balls are
either disjoint or comparable w.r.t. the set-theoretic inclusion.
Let us extend this property to the ``balls'' in $\mathcal{F}(X)$. Let
$\tau$ be a real-valued function on $\mathcal{F}(X)$ satisfying
(i$_1$) and (i$_2$). For $r>0$ and $A\in\mathcal{F}(X)$ define a
``closed'' ball
\begin{equation}\label{eq2.3**}
B_r(A):=\{C\in\mathcal{F}(X) : \tau(A\cup C)\leqslant r \}
\end{equation}
with a center $A$ and radius $r$.

The following properties are evident:

\begin{itemize}
\item[$\bullet$] $B_r(A)=\varnothing$ if and only if $r<\tau(A)$;

\item[$\bullet$] Every nonempty subset $C$ of the set $A$ belongs to the ball $B_r(A)$ if this ball is nonempty;

\item[$\bullet$] $B_{r_1}(A)\supseteq B_{r_2}(A)$ if $r_1\geqslant r_2$.

\end{itemize}

\begin{proposition}\label{prop2.7}
Let $X$ be a nonempty set and $\tau:\mathcal{F}(X)\to\mathbb{R}$ be
a function satisfying condition (i) of Theorem \ref{t2.1}. Then for every ball
$B_r(A)$ and every $C \in \mathcal{F}(X)$ we have
\begin{equation}\label{eq2.5*}
B_r(C)=B_r(A)
\end{equation}
if $C\in B_r(A)$, furthermore, for each pair $B_{r_1}(A_1)$,
 $B_{r_2}(A_2)$ with $r_1\geqslant r_2$ we have either
\begin{equation}\label{eq2.4}
B_{r_1}(A_1)\cap B_{r_2}(A_2)=\varnothing \mbox{ or } B_{r_2}(A_2)\subseteq B_{r_1}(A_1).
\end{equation}

\end{proposition}

\begin{proof}
Let $C$ belong to $B_r(A)$. To prove the inclusion
\begin{equation}\label{eq2.6}
B_{r}(C)\subseteq B_{r}(A)
\end{equation}
suppose that $Y\in B_r(C)$, i.e. $\tau(Y\cup C)\leqslant r$. Since
$C\in B_r(A)$, we have $\tau(A\cup C)\leqslant r$. Using (i$_2$) we obtain
$\tau(A\cup Y)\leqslant\tau(A\cup C)\vee\tau(C\cup Y)$.
Consequently, $\tau(A\cup Y)\leqslant r$ holds, so \eqref{eq2.6} follows.
 To prove \eqref{eq2.5*} it remains to show that
\begin{equation}\label{eq2.6*}
B_{r}(A)\subseteq B_{r}(C).
\end{equation}
For every $Z\in B_r(A)$ we have the inequality
\begin{equation}\label{eq2.7}
\tau(A\cup Z)\leqslant r.
\end{equation}
Since $C\in B_r(A)$, we also obtain
\begin{equation}\label{eq2.8}
\tau(A\cup C)\leqslant r.
\end{equation}
Inequalities \eqref{eq2.7}, \eqref{eq2.8} and condition (i$_1$) imply
$$
\tau(C\cup Z)\leqslant\tau(C\cup A)\vee\tau(A\cup Z)\leqslant r\vee
r=r.
$$
Thus $Z\in B_r(C)$, that implies \eqref{eq2.6*}.

Suppose now that $B_{r_1}(A_1)$ and $B_{r_2}(A_2)$ be some balls
such that $r_1\geqslant r_2$. If $B_{r_1}(A_1)\cap B_{r_2}(A_2)\neq\varnothing$,
 then there is $C\in B_{r_1}(A_1)\cap B_{r_2}(A_2)$. As we already know,
the equalities $B_{r_1}(A_1)=B_{r_1}(C)$ and $B_{r_2}(A_2)=B_{r_2}(C)$
hold. Consequently, the inequality $r_1\geqslant r_2$ implies
\eqref{eq2.4}.
\end{proof}

Using Proposition~\ref{prop2.7} and the characteristic property of bases
for topological spaces \cite[p. 21]{En} we see that the family
$$
\textbf{B}(X):=\{B_r(A):A \in \mathcal{F}(X), r\geqslant \tau (A)\}
$$
is a base for some topology on $\mathcal{F}(X)$. It is well known that
the family of closed balls in each ultrametric space forms a clopen base for
the space, but, in general, it is not the case for the base $\textbf{B}(X)$.

Indeed, suppose that $a,b,c,d$ are some points of $X$ such that
\begin{equation}\label{eq2.12}
\tau (\{a,b\})<\tau(\{c,d\})=\sup\{\tau(A):A\in\mathcal{F}(X)\}
\end{equation}
and the ball
$$
B_{r_1}(\{a,b\}),\quad r_1:=\tau(\{a,b\})
$$
is closed in the topology generated by the base $\textbf{B}(X)$.
The inequality in~(\ref{eq2.12}) and~(\ref{eq2.3**}) imply
$$
\{c,d\}\in \mathcal{F}(X)\backslash B_{r_1}(\{a,b\}).
$$
By the supposition the set $\mathcal{F}(X)\backslash B_{r_1}(\{a,b\})$ is open. Since
$\textbf{B}(X)$ is a base, there is $B_{r_2}(A)\in \textbf{B}(X)$ such that
\begin{equation}\label{eq2.13}
\{c,d\}\in B_{r_2}(A)\subseteq\mathcal{F}(X)\backslash B_{r_1}(\{a,b\}).
\end{equation}
By Proposition~\ref{prop2.7}, we have the equality
$B_{r_2}(A)=B_{r_2}(\{c,d\})$.
The smallest $r\in\mathbb{R^+}$ for which $\{c,d\}\in B_r(\{c,d\})$ equals $\tau(\{c,d\})$.
Consequently, the inequality $r_2\geqslant \tau(\{c,d\})$ holds.
The last inequality and the inclusion in~(\ref{eq2.13}) show that
\begin{equation}\label{eq2.14}
\{a,b\} \notin B_{r_3}(\{c,d\})
\end{equation}
where $r_3:=\tau(\{c,d\})$. The equality in~(\ref{eq2.12}) and~(\ref{eq2.3**}) imply
$A \in B_{r_3}(\{c,d\})$
for every $A\in \mathcal{F}(X)$, contrary to~(\ref{eq2.14}).

\section{Diametrical pairs of points in ultrametric spaces}

To formulate the main results of the present section we recall the basic
definitions from the graph theory  (see \cite{BM} for all terms that
will be used without explicit formulations).

A graph $G$ is an ordered pair $(V,E)$ consisting of a set $V=V(G)$ of
{\it vertices} and a set $E=E(G)$ of {\it edges}. In this paper we consider
only {\it simple} graphs, so each edge is an unordered pair of distinct vertices
and $E(G)$ is a set of two-element subsets of $V(G)$.

\begin{definition}\label{def3.1}
Let $G$ be a simple graph and let $k$ be a cardinal number. The graph $G$ is
$k$-partite if the vertex set $V(G)$ can be partitioned into $k$ nonvoid disjoint
subsets, or parts, in such a way that no edge has both ends in the same part. A $k$-partite
graph is complete if any two vertices in different parts are adjacent.
\end{definition}

We shall say that $G$ is a {\it complete multipartite graph} if there is
 $k \geqslant 1$ such that $G$ is complete $k$-partite, cf. \cite[p. 14]{Di}.

\begin{remark}\label{rem3.2}
It is easy to see that  1-partite graph $G$ is empty in the sense that no two
vertices are adjacent.
\end{remark}

To simplify the formulations  we consider the cases of finite and
infinite ultrametric spaces separately.
\begin{theorem}\label{th3.3}
Let $(X,\rho)$ be an ultrametric space with $\card X = \infty$ and let $G$ be a graph such
that $V(G)=X$ and
\begin{equation}\label{eq3.1}
(\{u,v\}\in E(G))\Leftrightarrow (\{u,v\} \mbox{ is a }\rho\mbox{-}\mbox{diametrical pair
of points for } X).
\end{equation}
 Then $G$ is complete multipartite graph.
Conversely, if $G=(V,E)$ is a complete multipartite graph  with $\card V=\infty$, then
 there is an ultrametric $\rho: V\times V \rightarrow \mathbb{R}^+$ such that
 $E$ is the set of $\rho$-diametrical pairs of points for $V$.
\end{theorem}

\begin{theorem}\label{th3.4}
Let $(X,\rho)$ be an ultrametric space with $2\leqslant \card X < \infty$ and let $G$ be a graph such
that $V(G)=X$ and condition~(\ref{eq3.1}) holds. Then $G$ is a complete $k$-partite graph with $k\geqslant 2$.
Conversely, if $G=(V,E)$ is a finite ($\card V < \infty$) complete multipartite graph  with $k \geqslant 2$, then
 there is an ultrametric $\rho: V\times V \rightarrow \mathbb{R}^+$ such that
 $E$ is the set of $\rho$-diametrical pairs of points for $V$.
\end{theorem}

The proof of these theorems based on the following three lemmas.
\begin{lemma}\label{lem3.5}
Let $(X,\rho)$ be an ultrametric space with $\card X \geqslant 2$. Let
us define the binary relation $\equiv$ on $X$ by the rule
\begin{equation}\label{eq3.2}
(x \equiv y) \Leftrightarrow (\rho(x,y)<\diam X).
\end{equation}
Then $\equiv$ is an equivalence relation on $X$.
\end{lemma}
\begin{proof}
It is evident that $\equiv$ is symmetric and reflexive. Suppose now that we have $x\equiv y$ and
$y\equiv z$ for $x,y,z \in X$. Then, by the ultrametric inequality, we obtain.
$$
\rho(x,z)\leqslant \rho(x,y)\vee \rho(y,z)<(\diam X)\vee (\diam X) =  \diam X.
$$
Hence, $\equiv$ is transitive. Thus, $\equiv$ is an equivalence relation.
\end{proof}
Let us recall that the {\it complement} $\overline{G}$ of a graph $G=(V,E)$ is a
graph whose vertex set is $V$ and whose edges are the pairs of distinct
nonadjacent vertices of $G$.
\begin{lemma}\label{lem3.6}
Let $G=(V,E)$ be a graph. The following properties are equivalent:
\begin{itemize}
\item [(i)] $G$ is a complete multipartite graph;
\item [(ii)] Every connected component of the complement $\overline{G}$ is a complete graph.
\end{itemize}
\end{lemma}
\begin{proof}
Suppose $G$ is a complete multipartite graph. Let $\{Y_i:i\in \mathcal{I}\}$ be a partition
of the vertex set $V$ with an index set $\mathcal{I}$ such that the properties described in
Definition \ref{def3.1} hold. Let $u,v \in V$, $u\neq v$. Then $u$ and $v$ are adjacent
 in $\overline{G}$ if they belong to the same part $Y_i$ and nonadjacent in the opposite case.
Hence, for every $i \in \mathcal{I}$ the subgraph of $\overline{G}$ induced by $Y_i$ is a connected
component of $\overline{G}$ and this component is a complete graph. Implication
(i)$\Rightarrow$(ii) follows. The proof of implication (ii)$\Rightarrow$(i) is also
straightforward and we omit this here.
\end{proof}
\begin{lemma}\label{lem3.7}
Let $X$ be a nonempty set and let $\equiv$ be an equivalence relation on $X$.
If $\card X =\infty$, then there is an ultrametric $\rho:X\times X \rightarrow \mathbb{R}^+$
such that~(\ref{eq3.2}) holds. For finite $X$ with $\card X \geqslant 2$, an ultrametric $\rho:X\times X \rightarrow \mathbb{R}^+$
satisfying~(\ref{eq3.2})  exists if and only if the relation $\equiv$ is different
from the universal relation $\{(x,y):x,y\in X\}$.
\end{lemma}
\begin{proof}
Consider first the case where $\card X=\infty$. If $\equiv$ is the universal relation on $X$, then
every ultrametric $\rho:X\times X\rightarrow \mathbb{R}^+$ satisfying the inequality
\begin{equation}\label{eq3.3}
\rho(x,y)<\diam X
\end{equation}
for all $x,y \in X$ satisfies also~(\ref{eq3.2}). Suppose that $\equiv$ is different from the universal
relation, then the partition of the set $X$ on the equivalence classes
$$\{y:y\equiv x\}, \quad x\in X$$
contains more than one class. Let us define a function $\rho$ on $X\times X$ as

\begin{equation}\label{eq3.4}
\rho(x,y):=\left\{
\begin{array}{l}
\mbox{ }0\quad \mbox{if } x=y,\\\
\frac{1}{2}\quad\mbox{if } x\neq y\mbox{ and } x\equiv y,\\\
1\quad\mbox{if }x\not\equiv y.\\
\end{array}
\right.
\end{equation}
We claim that $\rho$ is an ultrametric on $X$ and~(\ref{eq3.2}) holds for this $\rho$.
To prove that $\rho$ is an ultrametric it is suffice to show that
\begin{equation}\label{eq3.5}
\rho(x,z)\leqslant\rho(x,y)\vee\rho(y,z)
\end{equation}
for pairwise distinct $x,y,z \in X$. The proof is very simple. Since $x,y$ and $z$ are pairwise
distinct, inequality~(\ref{eq3.5}) does not hold if and only if $\rho(x,z)=1$ and
$\rho(x,y)=\rho(y,z)=\frac{1}{2}$. Hence, by~(\ref{eq3.4}),  we obtain
$$
x \not\equiv z, \quad x\equiv y\mbox{ and } y\equiv z,
$$
contrary to the transitivity of $\equiv$.
Suppose now that $\card X <\infty$. If there is an ultrametric $\rho:X\times X\rightarrow \mathbb{R}^+$
such that~(\ref{eq3.2}) holds, then using the finiteness of $X$ we can find $x,y \in X$ such that
$$
\rho(x,y)=\diam X.
$$
Consequently, $\equiv$ is different from the universal relation. The remaining
part of the proof can be obtained by a repetition of the arguments given above.
\end{proof}
\begin{remark}\label{rem3.7*}
In the proof of Lemma~\ref{lem3.7} we use the existence of an ultrametric
$\rho$ satisfying~(\ref{eq3.3}) for all $x,y \in X$. The ``explicit'' construction
of such type metric for arbitrary $\card X$ we shall give in Example~\ref{ex3.10}.
\end{remark}

\begin{proof}[Proof of Theorem~\ref{th3.3}]
Let $(X,\rho)$ be an ultrametric space, $\card X =\infty$ and let $G$  be a graph
with $V(G)=X$ and $E(G)$ defined by~(\ref{eq3.1}). By Lemma~\ref{lem3.6}, $G$ is complete
multipartite if  every connected component of $\overline{G}$ is complete. If
$x$ and $y$ are distinct points, then
$$
(\{x,y\}\in E(\overline{G}))
\Leftrightarrow
 (\{x,y\}\notin E(G))
 \Leftrightarrow
(\rho(x,y)<\diam X)
 \Leftrightarrow
(x\equiv y),
$$
(see~(\ref{eq3.2})).
By Lemma~\ref{lem3.5}, the relation $\equiv$ is an equivalence on $X$.
The equivalence classes induced by $\equiv$ form the set of the connected
components of $\overline{G}$. All these components are complete because $\equiv$ is
transitive.

Conversely, if $G=(V,E)$    is a complete multipartite graph with $\card V =\infty$, then,
by Lemma~\ref{lem3.6},  every connected component of $\overline{G}$ is a complete graph.
Let us define the relation $\equiv$ on $V$ as
$$
(x\equiv y)\Leftrightarrow (x \mbox{ and } y \mbox{ belong to the same connected component of } \overline{G}).
$$
Since every component is complete, the relation $\equiv$ is an equivalence relation
on $V$. By Lemma~\ref{lem3.7}, there is an ultrametric
$\rho:V \times V\rightarrow \mathbb{R}^+$ such that
$$
(x\equiv y)\Leftrightarrow (\rho(x,y)<\diam V)
$$
for $x,y \in V$. Consequently, we obtain
$$
(x\not\equiv y)\Leftrightarrow(\{x,y\}\in E(\overline{\overline{G}}))
\Leftrightarrow(\{x,y\}\in E(G))
\Leftrightarrow (\rho(x,y)=\diam V).
$$
Thus, $E(G)$ is the set of $\rho$-diametrical pairs of $(V,\rho)$.
\end{proof}
Theorem~\ref{th3.4} can be proved in much the same way as Theorem~\ref{th3.3}.
\begin{remark}\label{rem3.8}
The second part of Lemma~\ref{lem3.7} is, in fact, a special particular case of Theorem~\ref{th1.1}.
Lemma~\ref{lem3.6} provides a transition of results from complete multipartite graphs to equivalence
relations and back.
\end{remark}

Let $(X,d)$ be a nonempty metric space. Following  paper~\cite{DLPS}
we shall say that the set
$$
\spec(X,d)=\{d(x,y):x,y\in X\}
$$
is the {\it spectrum} of $(X,d)$.

Let $A$ be a subset of $\mathbb{R}^+$. If $0 \in A$, then there is an ultrametric
space $(X,\rho)$ such that $\spec (X,\rho)=A$.
This result was proved in \cite{DLPS} by the following elegant method. Define
$d: A \times A\rightarrow \mathbb{R}^+$, setting
\begin{equation}\label{eq3.6}
d(x,y)=
\left\{
\begin{array}{l}
x\vee y \quad \mbox{if } x \neq y, \\\
0 \quad\mbox{otherwise. }\\
\end{array}
\right.
\end{equation}
Then $(A,d)$ is an ultrametric space and $\spec (A,d)=A$.

Using ultrametric~(\ref{eq3.6}) we can simply fill the gap in the proof of Lemma~\ref{lem3.7}
mentioned in Remark~\ref{rem3.7*}.
\begin{example}\label{ex3.10}
Let $X$ be an infinite set, $A_c$ a countable infinite subset of $X$, $A$ a countable
infinite subset of $\mathbb{R}^+$, $0\in A$,
\begin{equation}\label{eq3.7}
\sup A \notin A
\end{equation}
and let $f:A_c\rightarrow A$  be a bijection. For an arbitrary $\varepsilon>0$
define a function $\rho:X\times X\rightarrow \mathbb{R}^+$ as
\begin{equation}\label{eq3.8}
\rho(x,y):=\left\{
\begin{array}{l}
0 \quad\mbox{if } x=y, \\\
\varepsilon\quad \mbox{if } x\neq y \mbox{ and } x,y\in X\backslash A_c, \\\
\varepsilon+d(f(x),f(y))\quad \mbox{if } x\neq y \mbox{ and } x,y\in  A_c, \\\
\varepsilon+f(x)\quad \mbox{if } x\in A_c \mbox{ and } y\in  X\backslash A_c, \\\
\varepsilon+f(y)\quad \mbox{if } x\in X\backslash A_c \mbox{ and } y\in  A_c \\
\end{array}
\right.
\end{equation}
where $d$ is defined by~(\ref{eq3.6}). It is simple to see that $\rho$ is symmetric
and nonnegative. To prove the ultrametric inequality it is sufficient to show that
\begin{equation}\label{eq3.9}
\rho(x,y)\leqslant\rho(x,z)\vee\rho(z,y)
\end{equation}
for $x\in A_c$, $y\in X\backslash A_c$ and $x\neq z\neq y$.
Using these relations,~(\ref{eq3.6}) and~(\ref{eq3.8}) we obtain
$$
\rho(x,y)=\varepsilon+f(x), \quad \rho(x,z)\geqslant \varepsilon+f(x) \quad \mbox{and} \quad \rho(z,y)\geqslant \varepsilon,
$$
so~(\ref{eq3.9}) follows. Hence, $\rho$ is an ultrametric.
\end{example}

 Note that~(\ref{eq3.8}) and~(\ref{eq3.7}) imply~(\ref{eq3.3}). Lemma~\ref{lem3.7} is
 completely proved now.

Using the idea of spectrum of ultrametric spaces we can give the following ``dual form'' of
Theorem~\ref{th1.1}.
\begin{theorem}\label{th3.11}
Let $X$ be  a finite set of points, $\card X \geqslant 2$, $n$ be an integer number,
 $1\leqslant n \leqslant \card X -1$, and let $\{G_1,...,G_n\}$ be
a  set of complete multipartite graphs with $V(G_1)=...=V(G_n)=X$ satisfying the following conditions:
\begin{itemize}
  \item [($i_1$)] $G_1$ is complete and $G_n$ is nonempty in the sense that $\{x,y\}\in E(G_1)$
  for all distinct $x,y \in X$ and $E(G_n)\neq\varnothing$;
  \item [($i_2$)] $E(G_{i+1})$ is a proper subset of $E(G_i)$ for $i=1,...,n-1$.
\end{itemize}
Then there is an ultrametric space $(X,\rho)$ with
\begin{equation}\label{eq3.10}
\spec(X,\rho)=\{a_0,a_1,...,a_n\}, 0=a_0<a_1<...<a_n,
\end{equation}
such that for $x,y \in X$
\begin{equation}\label{eq3.11}
(\{x,y\} \in E(G_i))\Leftrightarrow (\rho(x,y)\geqslant a_i), \quad i=1,...,n.
\end{equation}
Conversely, if $(X,\rho)$ is an ultrametric
space with spectrum~(\ref{eq3.10}) and if $\{G_1,...,G_n\}$ is a set of graphs such
that $V(G_1)=...=V(G_n)=X$ and~(\ref{eq3.11}) holds,  then all $G_i$ are complete
multipartite graphs meeting conditions ($i_1$) and ($i_2$).
\end{theorem}
This theorem lies beyond our considerations of diameters in ultrametric spaces
so we omit the proof here.

\section{Finite ultrametric spaces with minimal sets of diametrical pairs of points}
Let $(X,d)$ be a nonempty finite ultrametric space. Let us denote by $\dip(X)$
the set of all {\bf ordered} pairs $(x,y)$, $x,y\in X$, for which $d(x,y)=\diam X$, i.e.
$$
\dip(X)=\{(x,y)\in X\times X: d(x,y)=\diam X\}.
$$

It is almost evident that the set $\dip(X)$ cannot be a ``small'' subset of  $X\times X$.
In this section we describe finite ultrametric spaces $(X,d)$ for which $\card \dip(X)$
is smallest among all ultrametric spaces having the same number of elements.

We start from the following simple example.
\begin{example}\label{ex4.1}
Let $(X,d)$ be a nonempty ultrametric space, let $a$ be a point such that $a\notin X$
and let $t$ be a real number for which $\diam X<t$. Write $X_a:=X\cup\{a\}$ and define
\begin{equation*
d_{a}(x,y):=\left\{
\begin{array}{l}
d(x,y)\quad \mbox{if } x,y \in X, \\\
0\quad \mbox{if } x=y=a, \\\
t\quad \mbox{if } x\in X, y=a   \mbox{ or } y\in X, x=a.\\
\end{array}
\right.
\end{equation*}
It is easy to see that $(X_a,d_a)$ is an ultrametric space and
$$
(d_a(x,y)=\diam X_a)\Leftrightarrow ((x=a \mbox{ and }y \in X) \mbox{ or } (y=a \mbox{ and }x \in X))
$$
and $\card DP(X_a)=2\card X$.
\end{example}

\begin{theorem}\label{th4.2}
Let $(X,\rho)$ be a finite ultrametric space with $\card X\geqslant 2$. Then the inequality
\begin{equation}\label{eq4.2}
\card \dip(X)\geqslant 2(\card X-1)
\end{equation}
holds. Equality in~(\ref{eq4.2}) is attained if and only if there is an ultrametric
space $(Y,d)$ such that $(X,\rho)$ is isometric to  $(Y_a,d_a)$.
\end{theorem}
We divided the proof of Theorem~\ref{th4.2} into next two lemmas.

\begin{lemma}\label{lem4.4}
Let $(X,\rho)$ be a finite ultrametric space with $\card X\geqslant 2$. Then inequality~(\ref{eq4.2})
holds.
\end{lemma}
\begin{proof}
Let $x_1$ and $x_2$ be fixed points of $X$ such that $\rho(x_1,x_2)=\diam X$.
The ultrametric inequality implies that $\rho(x_1,y)=\diam X$ or $\rho(x_2,y)=\diam X$
for every $y\in X\backslash\{x_1,x_2\}$. Consequently, the number of diametrical
pairs of points is greater or equal than $1+\card (X\backslash \{x_1,x_2\})=\card X-1$. Inequality~(\ref{eq4.2}) follows.
\end{proof}

\begin{lemma}\label{lem4.3}
Let $(X,\rho)$ be a finite ultrametric space with $\card X\geqslant 2$. If
\begin{equation}\label{eq4.6}
\card \dip(X) = 2(\card X-1),
\end{equation}
then there is  $x_0 \in X$ such that
\begin{equation}\label{eq4.7}
\diam X=\rho(x_0,x)
\end{equation}
for every $x\in X\backslash \{x_0\}$ and
$\diam X>\rho(x,y)$ for all $x,y \in X\backslash \{x_0\}$.
\end{lemma}
\begin{proof}
Suppose that~(\ref{eq4.6}) holds. By Theorem~\ref{th3.4} the set $X$ together
with the set of $\rho$-diametrical pairs of points  form a complete  $k$-partite
graph $G$ with $k \geqslant 2$. We must show that $G$ is bipartite ($k=2$) with
a bipartition $(X_1,X_2)$,
\begin{equation}\label{eq4.7*}
X_1\cup X_2=X, \quad X_1\cap X_2=\varnothing
\end{equation}
such that $\card X_1=1$ or $\card X_2=1$.
Let $X_1,...,X_k$ be a partition of $X=V(G)$. Write
$$
m_i=\card X_i,\quad i=1,...,k
$$
and $n=\card X$. If $k\geqslant 3$, then using Lemma~\ref{lem4.4}
and the equality $\sum\limits_{i=2}^{k}m_i=n-m_1$ we obtain
$$
\card DP(X)=\sum\limits_{i=1}^{k}m_i(n-m_i)>m_1(n-m_1)
+\sum\limits_{i=2}^{k}m_i(n-\sum\limits_{i=2}^{k}m_i)
$$
$$
=m_1(n-m_1)+(n-m_1)(n-(n-m_1))\geqslant 2(n-1).
$$
Hence,~(\ref{eq4.6}) does not hold if $k\geqslant 3$. Thus, $k=2$, i.e.
$G$ is a complete bipartite graph. Let us consider bipartition~(\ref{eq4.7*}). Then $n=m_1+m_2$ and
$$
\card \dip(X)=m_1(n-m_1)+m_2(n-m_2)=2m_1(n-m_1).
$$
Using~(\ref{eq4.6}) we obtain
\begin{equation}\label{eq4.10}
2(n-m_1)m_1=2(n-1).
\end{equation}
Quadratic equation~(\ref{eq4.10}) has the roots $1$ and $n-1$. Thus, $\card X_1=1$ or
$\card X_2=1$ as required.
\end{proof}

Theorem~\ref{th4.2} can be reformulated in the graph theory language.
\begin{corollary}\label{cor4.5}
Let $G=(V,E)$ be a finite complete $k$-partite graph with $k \geqslant 2$.
Then
\begin{equation}\label{eq4.12}
\card E \geqslant \card V-1.
\end{equation}
Equality in~(\ref{eq4.12}) is attained if and only if $G$ is a star.
\end{corollary}

\textbf{Acknowledgment.} The second author is thankful to the Finnish
Academy of Science and Letters for the support.

redpajama_set_name: RedPajamaArXiv


text:
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Gilmour Space completes first of four tests for world's first interplanetary CubeSat propulsion system
Four Japanese companies form a joint venture to develop small rockets
Japanese company Interstellar launches sounding rocket Momo, aborts operation mid-flight
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Where to buy CubeSats online
Kymeta and Sky Perfect JSAT Sign Strategic Partnership Agreement
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TeamIndus Selects Lab2Moon Challenge Winners to Develop Payloads for its Rover
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Rocket Lab to Conduct Inaugural Launch at the World's First Private Launch Site
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ISRO Contracts out Satellite Manufacturing to Private Consortium for its Navigation Constellation
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Nanosatellite Operator Sky and Space Global Completes 3 Diamonds Testing Ahead of May Launch
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Google Lunar XPRIZE extends deadline, offers $4.75 million in additional rewards
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Rocket Lab Successfully Launches Electron, Making New Zealand a Spacefaring Nation
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Today, China Aerospace Science and Technology Corporation (CASC), the main contractor for the Chinese space programme, announced that it will provide commercial launch services using its Long March 11 rocket. Long...
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Utilizes appropriate resources to plan and provide nursing services that are safe, effective, and financially responsible. Physical Requirements Work requires walking, standing, sitting, lifting, reaching, stooping, bending, pushing, and pulling. Must be able to lift and support the weight of 35 pounds in handling patients, medical equipment, and supplies. Must speak English in good, understandable terms. Intact sense of sight, hearing, smell, and touch. Finger dexterity. Critical thinking and ability to concentrate. Must be able to respond quickly to changes in patient and/or unit conditions. Physical Abilities Testing required. Additional department specific physical requirements may be identified for unique responsibilities within the department by the department leader. Education, Experience and Certifications Graduate from an accredited School of Nursing required; BSN preferred. Current Basic Life Support for Healthcare Provider status according to American Heart Association. Current RN license or temporary license as a Registered Nurse Petitioner in the state in which you work and reside or; if declaring a National License Compact (NLC) state as your primary state of residency, meet the licensure requirements in your home state; or for Non-National License Compact states, current RN license or temporary license as a Registered Nurse Petitioner required in the state where the RN works. Additional education, training, certifications, or experience may be required within the department by the department manager.


text:
Solid Happiness is an oil on canvas painted in Italy by the Chilean artist Roberto Matta in 1997. Germana Ferrari Matta, the painter's last companion and the cataloguer of his oeuvre, certified the work on 28 April 2001.

Origin and history of the painting
The painting went from the Diana Lowenstein gallery in Buenos Aires to the Galería Almirante in Madrid and from there to Sotheby's auction house in New York, before being acquired by Fundación Carlos Slim for the Latin American Vanguard and Art collection of Museo Soumaya.

Barbara Eschenburg has written: "...Matta feels obliged to transmit messages in his work that come from the subconscious through the psyche, combining perceptions, reflections, and knowledge. Through them he formulates large questions..."

The painting was included in the exhibition El año de los 000 organized by the Fundación Telefónica in Santiago de Chile from April 7 to July 2, 2000.

References

1997 paintings
Chilean paintings


"Cultivators of the earth are the most valuable citizens. They are the most vigorous, the most independent, the most virtuous and they are tied to their country and wedded to its liberty and interests by the most lasting bands."
These words of Thomas Jefferson describe the vast majority of patriots who fought in the American Revolution. They worked small hardscrabble farms, many carved by hand out of an unrelenting wilderness. Today it is difficult to understand what a mammoth effort it was to live off the land in colonial times. But as these men went to war against the British, their knowledge of the land and its plants played a vital role in the birth of our nation.
among the best foragers for edible fruits and berries.
The Revolutionary War soldier rarely came from such large plantations as Thomas Jefferson's famous Virginia home.
Yarrow came to the New World with English colonists.
The Oswego tribe of New York used monarda for tea, which was adopted by many colonists after the Boston Tea Party.
The petals of the calendula pot marigold were age-old battlefield dressings when cloth and cotton were unavailable.
The patriotic garden, as you see, is far more than merely a collection of red, white and blue flowers. It is a testament to survival in the wilderness.
And while Jefferson grew a huge range of plants at his plantation, Monticello, it was a far cry from the world of the soldier-farmer. These soldier-farmers and their families held generations of accumulated knowledge, some inherited from their British ancestors and still more gleaned from Native Americans.
This knowledge of wild and useful plants recognized by every hunter-gatherer came into play as starving soldiers foraged through the countryside.
Certainly, battlefield plants were vital to surviving skirmishes with Redcoats. Herbs with antibacterial and coagulant properties were hoarded for the inevitable flood of the wounded.
Three coagulants came out of rural farm gardens, brought from the Old World as a living medicine chest. Yarrow, a European herb named for Achilles, has been carried into battle since ancient Greece. Periwinkle (Vinca major), also brought to America as a medicinal plant, was similarly valued for its ability to help staunch the flow of blood. Marigold (Calendula officinalis) was used as an antiseptic field dressing, and was still in use as late as World War I.
These farmer-soldiers also knew which plants yielded food and in which season. Sallet and other pot greens offered foliage that could be eaten. But, more importantly, the knowledge of plants that did not make good pot herbs was invaluable. (British soldiers foraging in the Jamestown colony so poisoned themselves with datura that it was forever called Jimson weed.) Many other pot herbs were unpalatable in their raw form, but quite tasty when stewed.
It was known that a diet required a range of foodstuffs in order to avoid symptoms of unpleasant vitamin deficiency such as scurvy. A majority of these plants are today considered mere weeds, but to knowledgeable foraging armies they were the essence of life.
Succulent purslane, relished by Henry David Thoreau, already was well-known in colonial times. This plant was eaten fresh as a salad or stewed. The tender young leaves of wild dock, a close relative to common sorrel of England, made their way into soldiers' stew pots. Dandelion would be the first greens to emerge following the bitter winter at Valley Forge, serving as a vitamin-rich spring tonic whether fresh or stewed. Even nettle was a favored pot herb.
Fruits held into the lean seasons fed the soldier with a sharp eye for plants. Wild rose hips, crabapples, elder, wild plums and many small berries tossed into the stew pot could add flavor, sugar and fiber.
With coffee and tea worth their weight in gold, soldiers sought out alternative plants to flavor their hot winter drinks. Oswego tea, a New York native herb known as bee balm or Monarda didyma, became a colonial favorite after the Boston Tea Party. Chicory also naturalized throughout New England.
The manicured gardens of Williamsburg or the flag-colored "patriotic" flower gardens of today offer few links to the valiant struggles of the American Revolution.
Our truest patriot gardens are a collection of the native and the introduced, the ragged and the cultivated. But, most of all, they express the ingenuity and resourcefulness of a people who fought for independence from tyranny.


This Sunday is the last day of the "Mice to Monsters" show, and I'll be teaching how I draw and paint penguins. The child who masters my technique gets to illustrate the sequel to BABY PENGUINS EVERYWHERE!!
Big thanks to all who came out to 440 Gallery this weekend. It was great to read my book in public for the first time (with my trusty studio assistant by my side)!
I also led a workshop to show how I made the book. First we looked at source material.
I shared a handful of the hundreds of sketches I made while inventing the look of the penguins and their world. Even grown illustrators use trial and error!
Finally we turned the kids loose on the art supplies!
At this age, process is more important than product. That said, the results were nothing short of spectacular!
What a great day. Kids, go make your own books!


With six published mystery novels under my belt, I've hosted a few book launches in my time. The first one was a party in my home with lots of food and wine. Even from the comfort of home, it was a nerve-wracking experience. Later launches, held at community centers and the library, were just as nerve-wracking, as they are for most writers. You wonder if anyone will come, if there will be enough food, whether your talk will go well or be a fumbling disaster, and whether you'll sell a single book. Happily, my events have turned out well and left me feeling greatly relieved.
Yet, these types of launches have lost their appeal over time. Maybe it's the repetition, the growing dread that your supporters have done enough and aren't likely to make the trip again, or that I've said it all before.
But things have changed. Over the past couple of years I've found myself experimenting with the novella. I've found a new dynamic publisher who embraces technology and different ways of launching books. To that end, the launch of my first novella, Dead Man Floating, won't be a physical launch but a virtual one. I'm hosting a Facebook Release Party on September 12th and I'm pretty excited about it.
I'll be offering ebook prizes (starting a new post for each contest) donated by writing colleagues and a $10 Amazon gift card. I'll be interacting with guests through comments and discussing a little about how the book came to being. As Dead Man Floating is inspired by my own work in security, I'll talk a little about my experiences. I'll offer security tips and ask questions of attendees and fellow writers to keep the discussion going.
Honestly, I don't know what to expect, but I'll come prepared and I'll let you know how it goes on my blog on September 13th. You're welcome to join the party (see link above). By the way, Dead Man Floating is now available for pre-ordering!
This is one of the many questions most fiction writers must ask in most projects. I'm told that the coziest of cozy mysteries eschew all profanity, but it sometimes happens that the villain is a roughneck. What then?
Personally, as a reader, I can take it or leave it. I generally don't notice it when it's missing. On the other hand, I can tolerate a moderate amount of bad language -- more when I'm reading to myself than when I'm reading aloud to Mom. When I'm reading aloud to Mom, I generally redact most if not all of the F-bombs and pretty much all of the references to anatomical naughty bits.
I once began reading a paranormal mystery by a writer I had recently met and had found to be a very nice person. The profanity was so thick, I stopped reading after a couple of pages. It felt like a verbal assault. I was willing to concede that these characters, in this situation, would talk exactly like that; I just found the volume to be too loud for comfort.
Part of the problem was the repetitiveness. There was just enough variety to show me that the writer could mix it up, but not enough to make it really interesting. I have a friend from Chicago who used to be an artist with bad language. Every sentence popped and sizzled; he seemed to have an innate sense of the rhythm of his words. It was like listening to good bluegrass: edgy, but delicious.
Knowing these things as a reader, what do I do as a writer?
I generally make the decision to limit the bad language I use. My husband separates profanity into two kinds: irreverence and vulgarity. Irreverence is using words that would be fine if a preacher used them in a sermon. Vulgarity is using words referring to sexuality, body parts, bodily functions, or lower eliminations. I say "lower" because, insofar as I know, nobody considers "snot," "spit," or "earwax" profane.
It seems to me that limiting the bad language you use in your writing -- if you're writing about people who use profanity -- makes the profanity you do choose to use more powerful. If the usage is so common as to be unremarkable ... part of what writing is about is eliminating the unremarkable from your writing, yes? So I would go with just using enough to give a flavoring without crossing the line into a waste of ink.
Following up on last week's blog about becoming a writing entrepreneur, I want to focus on something I started doing in earnest about five years ago, which is to sell my six titles at farmer's markets and craft fairs.
It didn't take much research to learn that there are many opportunities to sell books beyond the traditional stores. A couple of crafting friends encouraged me to try at my local Curves' Christmas fair back then. I enjoyed it so much that they gave me the names of other fairs.
Handselling at these types of events has been much more fun than standing behind a table in a bookstore trying to draw people in. Craft fair customers are never in rush and take the time to chat, to ask about my books. Some of my customers include teachers who buy copies for their school library, others belong to book clubs. I've landed a few great speaking engagements from customers who've invited me to speak at their organizations.
In addition to an attractive table covering and book display, I've found other useful tools along the way. First, professional business cards and bookmarks are a must. Growing numbers of customers want to know if my books are available on Kindle or Kobo. Taking away a reminder could definitely result in future sales.
Second, have an eye-catching banner made. Many of the fairs and trade shows provide curtains to attach banners to, and I constantly see customers looking at those signs before approaching a table. Third—and this one has really improved sales—use Square or something similar to conduct credit card transactions.
Square was recommended to me a couple of years ago and is easy to use on an ipad or my iphone. The first winter I used it was at a huge Christmas craft fair. One woman bought $75.00 worth of books on her card. Without Square that sale would have been lost.
Today, I was at a smaller, four-hour event. Over 50% of my sales were made through Square. I hate to think where I'd be without it. They're inexpensive to buy (in fact, your $10 investment is refunded by Square) and easy to set up even for luddites like me. Square takes a tiny percentage of each sale, but money from those transactions (for which Square takes a tiny percentage) is always in my account by Monday morning.
If you want to sell print copies of your books, don't rely just on the traditional means. There are plenty of other ways that can be fun and profitable.
In my 30+ years of writing, I've gone from a struggling newbie agonizing over every word and more drafts than I can remember to a published author. I've learned the business side of things, how bookstores operate, plus the ins and outs of contracts. I've read numerous articles on self-publishing and traditional publishing, all of which have helped me to view my work from a business perspective as well as a creative one.
Although a minority of writers make a living from their work, plenty of them are willing to share their experiences and give advice. One blog from author Joanna Penn made me realize that publishing lots of books in popular genres is only part of the strategy to earning a good living from writing. You see, Joanna has become a writing entrepreneur.
Her six-figure income isn't just about writing and publishing. It's a combination of sales from teaching courses, professional speaking, affiliate sales, and podcast sponsorship. It's really quite something and I encourage you to read her blog if you're interesting in building a writing business. She's broken down the percentage of income from different sources quite nicely.
It's also interesting to note that she writes both nonfiction and fiction. As you'll see when you read the blog, the key is to diversify. Many writers think that diversification is about writing lots of books in different genres and then turning them into audio books. But that's only the beginning for the writing entrepreneur.
I've been lucky enough to take on a few paid speaking engagements this year. I'm also paid to facilitate a creative writing program from time to time. But Joanna's blog has made me realize that there is so much more I could do.
Surprise! My Ebooks are On Sale Right Now!
I used to be pretty good at keeping track of my books' sales ranking on amazon, but then life became unbelievably hectic and I lost track until last week. As it turns out, the normally $7.99 price for my first Casey Holland mystery, The Opposite of Dark has been reduced to $1.99 on amazon.com!
The story begins when the police tell Casey that her that her father's been murdered in his West Vancouver home. But Casey buried her dad three years earlier, and he never owned a home in West Van, although he'd always wanted one. The story is about unraveling the truth about one's past, and how the childhood's we thought we knew weren't the whole picture.
The second installment, Deadly Accusations, is also on sale which is great because the book really is a follow-up to all of the family drama Casey experiences in the first book.
The story picks up the story four months later, when Casey now finds herself with a boyfriend, Lou, eleven-year-old ward, Summer, and still reeling with the aftermath of family shocks. Compounding her problems is the murder of an irritating new coworker who, for some reason, didn't like Casey, and another colleague likes Lou a little too much.
And last, but not least, my latest, The Deep End, a story about youth at risk (inspired by my volunteer work in a youth detention centre) has also been reduced to $3.99.
For all you Kobo users, The Opposite of Dark has been reduced by at least 50% there as well to $4.99! Bargains, bargains, bargains! Ya gotta love 'em!


text:
<?php

/**
 * @file
 * Contains \Drupal\Core\StackMiddleware\PageCache.
 */

namespace Drupal\Core\StackMiddleware;

use Drupal\Core\DrupalKernelInterface;
use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpKernel\HttpKernelInterface;

/**
 * Executes the page caching before the main kernel takes over the request.
 */
class PageCache implements HttpKernelInterface {

  /**
   * The wrapped HTTP kernel.
   *
   * @var \Symfony\Component\HttpKernel\HttpKernelInterface
   */
  protected $httpKernel;

  /**
   * The main Drupal kernel.
   *
   * @var \Drupal\Core\DrupalKernelInterface
   */
  protected $drupalKernel;

  /**
   * Constructs a ReverseProxyMiddleware object.
   *
   * @param \Symfony\Component\HttpKernel\HttpKernelInterface $http_kernel
   *   The decorated kernel.
   * @param \Drupal\Core\DrupalKernelInterface $drupal_kernel
   *   The main Drupal kernel.
   */
  public function __construct(HttpKernelInterface $http_kernel, DrupalKernelInterface $drupal_kernel) {
    $this->httpKernel = $http_kernel;
    $this->drupalKernel = $drupal_kernel;
  }

  /**
   * {@inheritdoc}
   */
  public function handle(Request $request, $type = self::MASTER_REQUEST, $catch = TRUE) {
    $this->drupalKernel->handlePageCache($request);

    return $this->httpKernel->handle($request, $type, $catch);
  }

}

redpajama_set_name: RedPajamaGithub


text:
Q: less time used with larger number in timeit.timeit()? import timeit
t=timeit.timeit('x=map(lambda x:x*10,range(32))')
print(t)

which shows:
0.4678139

but
t=timeit.timeit('x=map(lambda x:x*10,range(32))',number=100)
print(t)

gives:
4.309999999999731e-05

I suppose number=100 should takes 100 times 0.4678139, so what's wrong here?

A: The default arguments for timer.timeit include number=1000000, hence your first result is quite close to 1e4 times the second.

redpajama_set_name: RedPajamaStackExchange


text:
Anders Leonard  Zorn (Mora, Dalarna, 18 de fevereiro de 1860 — Mora, Dalarna, 22 de agosto de 1920) foi um pintor sueco que ganhou reputação como um artista de retratos.
Em Mora, cidade onde o artista nasceu, existe um museu dedicado aos seus trabalhos - o Zornmuseet (Museu de Zorn).

Ele se tornou famoso por suas pinturas do povo de Dalarna, região de onde veio, e suas pinturas de nudez em espaço aberto, além de suas vívidas representações da água.

Realizou sete viagens aos Estados Unidos da América e entre seus trabalhos encontram-se retratos de três diferentes presidentes estadunidenses, entre eles o retrato do presidente Grover Cleveland em 1899.

A Paleta Zorn
Zorn é conhecido por usar uma paleta de cores básica consistindo de branco de chumbo (alvaiade), amarelo-ocre, vermelhão e preto de marfim. Esta paleta de cores limitada mostra uma gama enorme em termos de mistura de cores. Uma grande variedade de faixas tonais com as quais são possíveis se misturarem, e é considerada como um desenvolvimento muito importante para a pintura de retratos. No entanto, a paleta de cores também pode ser usada em naturezas mortas e pinturas de paisagens sob certas circunstâncias. O aspecto mais marcante é que um tipo de cor verde-oliva pode se obter misturando preto de marfim e amarelo-ocre, já que o preto de marfim é azulado por natureza.

Galeria

Ver também
Impressionismo
Pintura na Suécia

Ligações externas

Pintores da Suécia
Retratistas
Naturais de Mora (comuna)


text:
Death Pranks & Comic Book Grief | 4M #68
Funeral Industry News Morticians' Monday Morning Mashup January 1, 2023
Welcome to the sixty-eighth edition of Morticians' Monday Morning Mashup, 4M #68, where we'll serve up bite-sized, easily-digestible nuggets of the deathcare news you need to crush conversations in the week ahead. Bon appetit!
Readin', writin', and … thanatology?
A funeral home owner in India would like to expand the educational system in his country to include the study of death. "'Death literacy' has to be propagated via Thanatology as a subject or at least a topic that is integrated into the education system," Rishabh Jalan, CEO & Co-Founder, Last Journey Funeral Services wrote in a recent op-ed. Jalan compares the potential inclusion of death studies in the curriculum to the recent addition of sex education. In his detailed argument, Jalan argues that ignoring thanatology in primary and secondary education is an "uncanny paradox" in such an "emotionally-driven" country with elaborate death rituals and "some of the world's best forensic science curricula."
#Deathtok shouldn't claim this
Deathcare professionals are doing amazing things on TikTok, especially when it comes to educating the public on what you guys actually do. But there's a new death-related TikTok trend that — like most TikTok trends — has become controversial, theoretically "dividing the Internet" (yes, again). Apparently, the hashtag "celebrity death prank" is trending. The prank, which is being called "disquieting, strange, pathetic" and "sick" involves people filming themselves telling someone else that a beloved celebrity or family member (!) has died, and waiting until the person reacts to tell them it's a joke. Just … no.
If you're fed up with the frankly boring names of most deathcare establishments, why not take a cue from this Singapore funeral service provider: Casket Fairprice? Perhaps something is lost in translation, but it looks like this funeral home's claim to fame is — you guessed it — fair prices. Casket Fairprice's website promises "transparent pricing" and that "all prices listed are nett!" [sic]. So yes, apparently in Singapore listing your prices online isn't such a big deal, and neither is bragging about it.
Does this venue make me look dead?
Some New Jersey bridezillas — and groomzillas, for that matter — aren't happy with the recent renovations in the venue they booked for their future weddings. The "once-golden room with ornate chandeliers" at the Madison Hotel has been "transformed into a black-walled and blue-carpeted space" that couples are calling "gothic beach house" and "beautiful … for a funeral home." You can judge the aesthetic for yourself — and perhaps glean some design inspiration — from the venue's Instagram post, below.
Grieving in the MCU
As someone who ugly cries when a beloved character dies in a TV show or movie (or when ESPN's Gene Wojakowski narrates a super sad story about a dying athlete on College GameDay), I appreciate the fact that someone has taken the time to tie each of Kubler-Ross' seven stages of grief to events in Phase Four of the Marvel Cinematic Universe (MCU). From shock and disbelief (stage one) in Black Widow to reconstruction (stage seven) in Black Panther: Wakanda Forever, it's quite fascinating to see how grief and loss is accurately portrayed in comic adaptations.
It's 2023 and time for a change
It's more than likely there is no course on search engine optimization (SEO) in mortuary school. But that's ok, because the cool kids at MemoryShare know all about it — and they know how to improve yours … while at the same time making streaming services super simple for your staff. Get ready for a little mortuary math here (also probably not a class):
SEO + superior streaming services + super simplicity + safety and security =
The MemoryShare not-so-secret sauce for success
MemoryShare's exclusive app makes streaming super easy; you can record your full service with the touch of a button. By sharing these streams with your community, you showcase your excellent service and create name recognition with the viewers. Aaaannnd that's not all, folks. Because MemoryShare videos are embedded on your own website (and not on the site of a separate streaming provider) your site gets more views, which results in street cred with the search engines.
Two things funeral homes need in 2023 are visibility and increased call volume. MemoryShare provides both of these things as well as a team of REAL people you can talk to if you ever run into trouble. So what are you waiting for? Learn more about MemoryShare today!
Morticians' Monday Morning Mashup


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text:
UTA Closure: What It Means
Live Aug 02, 2017 by David Farrell Comments
It is difficult to peg a figure on what the music agency billed in 2016 but a ballpark number is $50 million, so that is between $5 and $7 million in commissions United Talent Agency stands to lose by closing the Canadian office.
It means that a lot of headlining acts and their managers will be reviewing their agreements to see if their key man clause holds up, and for many more acts without top draw clout the immediate future looks very uncertain.
We have reported that Canadian CEO Jack Ross has all but said he's starting a new shop, the guess is that partner Ralph James will take partial retirement but continue to work with Nickelback who have been with him since day one and contines to represent the band exclusively for Canada, Australia and Asia.
Premier agent Rob Zifarelli already operates out of Toronto and New York. It's hard to guess where his allegiance lies.
The closure affects a wide swath of Canadian acts, from the aforementioned Nickelback to Marianas Trench, Skydiggers, Bahamas, 54:40, Billy Talent, City and Colour, Alexisonfire, Sam Roberts Band, Doc Walker, Trailer Park Boys, and Walk Off The Earth.
There's going to be some gold digging here by the Feldman and Paquin agencies and more. And then there's Jack, Ross, who is fielding phone calls, making plans, setting himself up for another agency start-up. He's been there before, done that, and has impeccable credentials, deep loyalties and an inspiring track record in working with acts.
The news hit many like an unexpected thunderbolt; others had heard rumblings, a few had speculated that the UTA's acquisition of Neil Warnock's The Agency Group in 2015 was just that: An acquisition, and that this wasn't a marriage made in heaven, and the purchase necessarily meant a shoring up of overheads.
It also gave the combined entity a roster of close to 2,000 clients. Not everyone was going to get the special treatment, and the agents would be pressed hard to pay for the buy.
The live business isn't all that it's cracked up to be in the mainstream media, either. The competition is intense, costs for travel, venues, general overheads and agent fees have ballooned, and gate receipts are haphazard. Festivals have helped to offset declines in the number of available club venues featuring live talent, but the festival industry is showing stress marks and fatigue. Everything's got to be bigger, better, more fantastic—and the up-front gambles by event promoters more speculative each time.
And then there's the weather: Once, surely predictable; with climate change not so much.
For Ron Kitchener, he has four acts with UTA, but they are all represented by Nick Meinema out of the Nashville office.
Julien Paquin wasn't shocked to hear the news, but he was surprised. He also sees an opening ahead: "There is no doubt some opportunity for Paquin. UTA has a lot of talented people and A list artists – from developing artists to headliners," he emailed FYI yesterday, adding that "some will need new homes."
Sam Feldman offers a bit of sangfroid, opining that "the inevitable corporate dynamic meant that the acquisition would be reviewed on numbers, not people."
"It's really too bad," he added, "because there are some good ones there," he said in an email yesterday afternoon.
"The Canadian market is unique in terms of opportunities well beyond the arenas in seven major markets and the extraordinary talent that comes out of here," Feldman said, noting that "after 40 plus years here I'd like to think we know where the bodies are buried and I'm sure that the staff there will be treated properly and will land on their feet."
He's probably right. But there's gonna be a lot of upheaval in the market before the dust settles.
The game has changed, and so too have the players.
Sam Feldman
Julien Paquin
Ron Kitchener
David Farrell
David Farrell has worked at the forefront of Canadian music journalism for over 30 years, founding and editing The Record in the 1980s, before launching FYI Music News...
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