text:
\section{Introduction}

Throughout, we will use the notation $\{ e \}^A (x)$ for the result of applying the Turing machine coded by $e$, with oracle $A$, to input $x$.  If the attempt to compute $\{ e \}^A (x)$ halts and gives output $y \in \mathbb N$, we write
\[
\{ e \}^A (x) \downarrow = y.
\]
Otherwise, we write $\{ e \}^A (x) \uparrow$ to indicate that the computation of $\{ e \}^A (x)$ never terminates.  Note the the oracle $A$ may be either a subset of $\bbn$ or a (possibly partial) function on $\bbn$.  In the event that $A$ is a partial function, we require $\{ e \}^A (x)$ to diverge if an oracle query is made for an input not in the domain of $A$.

We let $\{ e \}^A_s (x)$ denote the result of allowing the computation of $\{ e \}^A (x)$ to run for $s$ computation stages.  Finally, $\mathrm{use} (\{ e \}^A_s (x))$ will denote the largest oracle query made by the computations $\{ e \}^A_{s'}$, for $s' \leq s$.

Recall that, for sets $A,B \subseteq\mathbb N$, one says $A$ is {\em Turing reducible} to $B$ (denoted $A \leq_{\mathrm T} B$) if, and only if, there is an $e \in \mathbb N$ such that the computation of $\{ e \}^A (x)$ terminates with output 0 or 1, for each $x \in \mathbb N$, and
\[
(\forall x )(x \in A \iff \{ e \}^B(x) \downarrow = 1).
\]
In this case, we write $\{ e \}^B = \chi_A$ to indicate that the map
\[
x \mapsto \{ e \}^B (x)
\]
is the characteristic function of $A$.  Sets $A, B \subseteq \mathbb N$ are said to be {\em Turing equivalent} (written $A \equiv_{\mathrm T} B$) if, and only if, $A \leq_{\mathrm T} B$ and $B \leq_{\mathrm T} A$.  The {\em Turing degree} of $A \subseteq \mathbb N$ is the family $\{ B \subseteq \mathbb N : B \equiv_{\mathrm T} A\}$ of subsets of $\mathbb N$.

The relation of Turing reducibility gives a natural pre-order on the family of subsets of $\mathbb N$.  With this in mind, it is desirable to have a corresponding pre-order on the subsets of the Turing degrees themselves.  There are several natural ways of obtaining such a pre-order.  Extensive study has been done of two such pre-orders: {\em Muchnik reducibility} and {\em Medvedev reducibility}.

\begin{Definition}\cite{medvedev}\cite{muchnik}
Let $\mathcal A$ and $\mathcal B$ be sets of degrees.
\begin{enumerate}
\item $\mathcal A$ is {\em Muchnik reducible} (or {\em weakly reducible}) to $\mathcal B$ (written $\mathcal A \leq_w \mathcal B$) if
\[
(\forall B \in \mathcal B)(\exists A \in \mathcal A)(\exists e \in \mathbb N)\Big( \{ e \}^B = \chi_A \Big)
\]
\item $\mathcal A$ is {\em Medvedev reducible} (or {\em strongly reducible}) to $\mathcal B$ (written $\mathcal A \leq_s \mathcal B$) if
\[
(\exists e \in \mathbb N)(\forall B \in \mathcal B)(\exists A \in \mathcal A)\Big( \{ e \}^B = \chi_A\Big)
\]
\end{enumerate}
The oracle Turing machine determined by $e$ as above is called a Muchnik (resp., Medvedev) reduction of $\mathcal A$ to $\mathcal B$.
\end{Definition}

In the case that $\mathcal A \leq_s \mathcal B$, with $e$ such that
\[
(\forall B \in \mathcal B)(\exists A \in \mathcal A)\Big( \{ e \}^B = \chi_A\Big),
\]
we call $e$ a {\em code} witnessing the Medvedev reduction of $\mathcal A$ to $\mathcal B$.  In the present work, we study the arithmetic complexity of sets of codes of Medvedev reductions associated to various classes $\mathcal A$ and $\mathcal B$.  We make the following definition.

\begin{Definition}\label{D1}
Let $\mathcal A$ and $\mathcal B$ be families of subsets of $\bbn$.  We say that $e \in \bbn$ is a \emph{$\mathcal B$-universal $\mathcal A$-code} if
\[
(\forall B \in \mathcal B) (\exists A \in \mathcal A) \big( \{e\}^B = \chi_A \big).
\]
We let $\mathcal B [\mathcal A]$ denote the set of all $\mathcal B$-universal $\mathcal A$-codes.
\end{Definition}

By definition, a $\mathcal B$-universal $\mathcal A$-code determines a Medvedev reduction of $\mathcal A$ to $\mathcal B$ and $\mathcal B [ \mathcal A]$ corrresponds to the set of Medvedev reductions of $\mathcal A$ to $\mathcal B$.

As a more general case of this definition, we suppose that $\mathcal A$ and $\mathcal B$ are classes of partial functions instead of oracles.  Consider the following:

\begin{Definition}\label{D2}
Suppose that $\mathcal F, \mathcal G$ are families of (possibly partial) functions on $\bbn$.  We say that $e \in \bbn$ is a {\em $\mathcal G$-universal $\mathcal F$-code} if, and only if, the function
\[
x \mapsto \{ e \}^g (x)
\]
is in $\mathcal F$, for each $g \in \mathcal G$.  As before, we let $\mathcal G [ \mathcal F]$ denote the set of $\mathcal G$-universal $\mathcal F$-codes.
\end{Definition}

Definition~\ref{D1} may be regarded as a special case of Definition~\ref{D2} by letting $\mathcal F$ (in Definition~\ref{D2}) be the class of characteristic functions of the sets in the degrees from the class $\mathcal A$ (in Definition~\ref{D1}) and $\mathcal G$ be the class of characteristic functions of the sets in $\mathcal B$.

We will also combine the two definitions in certain case, e.g., we will consider index sets of the form $\mathcal G [ \mathcal A]$, where $\mathcal A$ is a family of sets and $\mathcal G$ is a family of functions.

Definition~\ref{D2} also reveals the reason for the choice of the terminology ``$\mathcal G$-universal $\mathcal F$-code'': if $e \in \mathcal G [ \mathcal F]$, then $\{ e \}^g$ codes a function in $\mathcal F$, regardless of the choice of oracle $g \in \mathcal G$.  The set $\mathcal G [\mathcal F]$ is therefore the set of codes, $e$, such that the behavior of the oracle machine coded by $e$ is invariant (in the sense that it is always in $\mathcal F$) under the choice of oracle $g \in \mathcal G$.

In what follows, we make use of both Definitions~\ref{D1} and \ref{D2}.  It will always be clear from context which one applies.

There are obvious codes in $\mathcal G [ \mathcal F]$, for many choices of $\mathcal G$ and $\mathcal F$.  For instance, if $\indset{tot}$ denotes the family of total functions, a code for a total  function which makes no oracle queries will be in the class $\mathcal G [ \indset{tot}]$, for any choice of $\mathcal G$.  Naturally, it is of interest when there are nontrivial elements of $\mathcal G [\mathcal F]$.  In this case, we give ``nontriviality'' a precise meaning with the following definition.

\begin{Definition}
We say that a $\mathcal G$-universal $\mathcal F$-code $e$ is \emph{non-trivial} if there are $g_0,g_1 \in \mathcal G$ such that $\{e\}^{g_0} \neq \{e\}^{g_1}$ and \emph{strongly non-trivial} if for all $g_0,g_1 \in \mathcal G$, $\{e\}^{g_0} \neq \{e\}^{g_1}$.
\end{Definition}

The nontrivial elements of $\mathcal G [ \mathcal F]$ are, therefore, the codes which do not simply ignore the oracle $g \in \mathcal G$.

In many case, the structure of the class $\mathcal G [ \mathcal F]$ can be quite complex.  In the present work, we quantify this by establishing the complexity of the $\mathcal G [\mathcal F]$ for various classes $\mathcal G$ and $\mathcal F$.  In several cases these index sets are arithmetic, but, in more than one instance, they are beyond hyperarithmetic.

Letting $\indset{inf}$ denote the family of infinite c.e.~sets and $\indset{tot}$ the family of total computable functions, we have the following arithmetic complexity result.

\begin{Proposition}
The index set $\indset{inf}[\indset{tot}]$ is $\Pi^0_3$-complete.
\end{Proposition}

Similarly, letting $\indset{fin}$ denote the set of partial computable functions with finite domain, we have

\begin{Proposition}
If $\mathcal F$ is any uniformly computable family of sets, $\mathcal F[\indset{fin}]$ has a strongly non-trivial element and is $\Pi_3^0$-complete.
\end{Proposition}

Before stating our main result, we recall a couple of standard definitions.  A total function $f: \bbn \rightarrow \bbn$ is {\em diagonally non-computable} (abbreviated, {\em dnc}) if, and only if, for each $e \in \bbn$,
\[
\{ e \} (e) \downarrow = y \implies f(e) \neq y.
\]
We let $\indset{dnc}$ denote the family of dnc functions.  A set, $A \subseteq \bbn$, is {\em immune} if it contains no infinite c.e.~set, and {\em bi-immune} if both $A$ and $\overline A$ are immune.  Let $\indset{bi}$ denote the family of bi-immune sets.  Finally, recall that a set $P \subseteq \bbn$ is $\Pi^1_1$ if it is many-one reducible to the set of Turing codes for characteristic functions of recursive trees in $\bbn^{< \omega}$, which have no infinite branches.

It is known that there is a Medvedev reduction of $\indset{dnc}$ to $\indset{bi}$ \cite{jo-le}.  In fact, the corresponding set of codes is as complicated as possible:

\begin{Theorem}
$\indset{dnc}[\indset{bi}]$ is $\Pi^1_1$-complete.
\end{Theorem}

\noindent (Note that, if $\mathcal A$ and $\mathcal B$ are hyperarithmetic, then $\mathcal B[\mathcal A]$ is at most $\Pi^1_1$.)

This latter result is a consequence of a more general theorem which, informally, states that, if $\omega^\omega$ may be ``effectively'' embedded into a hyperarithmetic class, $\mathcal A$, and $\mathcal B$ is a hyperarithmetic tail set, Medvedev reducible to $\mathcal A$, then the index set $\mathcal A [ \mathcal B ]$ is always $\Pi^1_1$-complete.

\section{Basic facts and notation}

The next Proposition makes explicit the connection between universal codes and Medvedev reductions.

\begin{Proposition}\label{medvedev-reduct}
Suppose that $\mathcal A \subseteq \baire$ and $\mathcal B$ is a family of (possibly partial) functions on $\omega$.  $\mathcal A \leq_s \mathcal B$ if and only if $\mathcal B[\mathcal A] \neq \emptyset$.  Furthermore, $\mathcal B[\mathcal A]$ is the set of codes for reductions witnessing $\mathcal A \leq_s \mathcal B$.
\end{Proposition}

\begin{proof}
Suppose $\mathcal A$ and $\mathcal B$ are as in the statement and $\mathcal B[\mathcal A] \neq \emptyset$.  If $e \in \mathcal B[\mathcal A]$, then for each $B \in \mathcal B$, not only is $A = \{e\}^B \in \mathcal A$ c.e.~in $B$, but it is computable from $B$.  Thus, $e$ codes a turing functional that uniformly computes elements of $\mathcal A$ from elements of $\mathcal B$ -- in other words, it is a Medvedev reduction.  Conversely, if $\Psi(\sigma,x) = \{e\}^{\sigma}(x)$ witnesses $\mathcal A \leq_s \mathcal B$, then for every $B \in \mathcal B$, $\Psi(B)$ is total and hence computable in $B$.

\end{proof}

In light of this theorem, we will begin our investigation of universal codes by examining the complexity of $\mathcal B[\mathcal A]$ for some classes where it is known or trivial to see that $\mathcal A \leq_s \mathcal B$.

\begin{Proposition}
The index set $\indset{inf}[\indset{tot}]$ is $\Pi^0_3$-complete.
\end{Proposition}

\begin{proof}
First of all, to see that $\indset{inf}[\indset{tot}]$ is $\Pi^0_3$, observe that
\[
e \in \indset{inf}[\indset{tot}] \iff (\forall a,n) (a \in \indset{fin} \vee (\exists s)(\forall t \geq s) (\{e\}^{W_{a,t}}_t (n) \downarrow)).
\]

Next, to see that $\indset{inf}[\indset{tot}]$ is $\Pi^0_3$-hard, let $B$ be a fixed $\Pi^0_3$ set, with $g : \mathbb N \times \mathbb N \rightarrow \mathbb N$ a total computable function such that
\[
x \in B \iff (\forall y) (g(x,y) \in \indset{fin}),
\]
for each $x \in \mathbb N$.

For an infinite set $A$, let $C^*_n(A)$ denote the set of $i$ such that $i$ is the $\langle n , k\rangle$-th element of $A$, for some $k$.  If $A$ is infinite, note that each $C^*_n (A)$ is also infinite.  Let $f : \mathbb N \rightarrow \mathbb N$ be a total computable function such that, for each $e \in \mathbb N$,
\[
\{f(e)\}^A (n) \downarrow \iff (\exists m \in C^*_n (A)) ( W_{g(e,n),m+1} = W_{g(e,n),m}).
\]
In the first place, if $e \in B$, then $g(e,n) \in \indset{fin}$, for each $n \in \mathbb N$.  It follows that $\{ f(e) \}^A (n) \downarrow$, for each $n$, since $A$ is infinite and $W_{g(e,n) , m+1} = W_{g(e,n) , m}$, for all but finitely many $m$.

On the other hand, if $e \notin B$, then there exists $n$ such that $W_{g(e,n)}$ is infinite.  Let $A \subseteq \mathbb N$ be an infinite c.e.~set such that
\[
C^*_n (A) \subseteq \{ m : W_{e,m+1} \neq W_{e,m}\}.
\]
It follows that $\{ f(e) \}^A (n)$ never converges, since there is no $m = C_n^* (A)$ such that  $W_{e,m+1} = W_{e,m}$.  In other words, $f(e) \notin \indset{inf}[\indset{tot}]$.
\end{proof}

\begin{Proposition}\label{finComp}
If $\mathcal F$ is any uniformly computable family, $\mathcal F[\indset{fin}]$ has a strongly non-trivial element and is $\Pi_3^0$-complete.
\end{Proposition}

\begin{proof}
Let $\mathcal F = \{ F_0, F_1, \ldots \}$ be a uniformly computable family.  Let $p$ be a computable function such that $p(\sigma) = i$ if $i < |\sigma|$ is least such that $\sigma = F_i \upto |\sigma|$ and $p(\sigma) = -1$ if there is no such $i$.  Define a code $e$ such that
\[
\{ e \}^A(x) = \begin{cases}
p(A\upto x) &\mbox{if $p(A\upto x) \neq p(A\upto(x-1)$}\\

\uparrow & \mbox{otherwise}
\end{cases}.
\]
If $A \in \mathcal F$, then $\{e\}^A$ has finite domain and if $A,B \in \mathcal F$ with $A \neq B$, then $\{e\}^A \neq  \{e\}^B$.

We now prove that $\mathcal F[\indset{fin}]$ is $\Pi_3^0$-complete.  To see that $\mathcal F[\indset{fin}]$ is $\Pi_3^0$, observe that
\[
\mathcal F[\indset{fin}] = \Big\{ e : (\forall i)(\exists b)(\forall x > b, s)\big( \{e\}_s^{F_i}(x)\uparrow \big) \Big\}.
\]

We use a movable markers argument to prove hardness.  Let $f: \bbn^2 \rightarrow \{0,1\}$ be a computable function such that $f_i(x) = f(i,x)$ is the characteristic function of $F_i$.  Define a computable function $g$ such that $g(e,n,s) = y$ if and only if $y$ is the $n^{th}$ element of $\overline{W}_{e,s}$.  Define a computable function $h$ such that $\{h(e)\}^{\sigma}(x) \downarrow = 1$ if
\begin{enumerate}
\item $|\sigma| \geq x$ and
\item if $i$ is least such that $\sigma \upto x = f_i \upto x$, then there is an $s > x$ such that $g(e,i,s) \neq g(e,i,x)$.
\end{enumerate}

If $e \in \indset{cof}$, then for all $i > \max(\overline{W}_e)$ and $x \in \bbn$, there is a $y > x$ such that $g(e,i,y) \neq g(e,i,x)$.  Thus, for all $i > \max(\overline{W}_e)$, $\{h(e)\}^{F_i}$ is total and $h(e) \not\in \mathcal F[\indset{fin}]$.  On the other hand, if $e \in \indset{coinf}$, then for all $i$ $\lim_{x \rightarrow \infty} g(e,i,x)$ exists.  In other words, for every $i \in \bbn$ there is an $x$ such that for all $y > x$, $g(e,i,y) = g(e,i,x)$.  We conclude that $\{h(e)\}^{F_i}$ is finite for all $i \in \bbn$ and $h(e) \in \mathcal F[\indset{fin}]$.  Thus, $h$ witnesses the desired result: $\mathcal F[\indset{fin}]$ is $\Pi_3^0$-hard.

\end{proof}

Note that while an oracle drawn from a uniformly computable family does not confer additional computational power it does affect the output of an oracle program.

\begin{Theorem}
$\Delta_2^0[\indset{fin}]$ is $\Pi_4^0$-complete and contains a strongly non-trivial element.
\end{Theorem}

\begin{proof}
First, we prove that $\Delta_2^0[\indset{fin}]$ is $\Pi_4^0$.  To this end, we say that $d \in \omega$ is a {\em $\Delta^0_2$ code} if $\{ d \}$ is a total function and, for each $x \in \omega$
\[
\lim_{s \rightarrow \infty} \{ d \} (\langle s,x \rangle)
\]
exists, where $\langle \cdot , \cdot \rangle$ is a fixed computable pairing function.  Note that the predicate ``$d$ is a $\Delta^0_2$ code'' is $\Pi^0_3$.  Supposing that $d$ is a $\Delta^0_2$ code, define the functions $d_t$, for $t \in \omega$, by
\[
d_t (x) = \{ d \} (\langle t , x \rangle)
\]
and $d_*$ by
\[
d_* ( x) = \lim_{t \rightarrow \infty} d_t (x).
\]

Define predicates $\Gamma$ and $\Sigma$ as follows:
\begin{align*}
\Gamma (e,d,x) &\iff (\forall k) (\exists t) (\mathrm{use} (\{ e\}^{d_t}_t (x)) \geq k)\\
&\iff \{ \mathrm{use} (\{ e \}^{d_t}_t (x)) : t \in \omega\} \mbox{ is unbounded}.
\end{align*}
and
\[
\Sigma (e,d,x) \iff (\forall s ) (\exists t \geq s) ( \{ e \}^{d_t}_t (x) \uparrow).
\]
It will follow that $e \in \Delta^0_2$ if, and only if,
\begin{equation}\label{E1}
(\forall d) \left( d \mbox{ is a $\Delta^0_2$ code }   \implies (\exists m) (\forall x \geq m)(\Gamma(e,d,x) \vee \Sigma(e,d,x))\right).
\end{equation}
In particular, this will show that $\Delta^0_2 [\indset{fin}]$ is $\Pi^0_4$.

Indeed, suppose that $d$ is a $\Delta^0_2$ code and $x \in \omega$.  Suppose first that $\{ e \}^{d_*} (x)$ converges.  Let $k_0 = \mathrm{use} (\{ e \}^{d_*} (x))$ and let $t_0$ be large enough that, for each $t \geq t_0$,
\[
d_t \upto k_0 = d_* \upto k_0.
\]
It follows that, for each $t \geq t_0$,
\[
\mathrm{use}(\{ e \}^{d_t}_t (x)) \leq k_0
\]
and, hence,
\[
\{ \mathrm{use} (\{ e \}^{d_t}_t (x)) : t \in \omega\}
\]
is bounded, i.e., $\neg \Gamma (e,d,x)$.  Also, if $s_0$ is large enough that $\{ e \}^{d_*}_{s_0} (x)$ has converged and $t\geq \max\{ s_0 , t_0 \}$, then $\{ e \}^{d_t}_t (x)$ converges.  Thus, $\neg \Sigma(e,d,x)$.  As $d$ and $x$ were arbitrary, this establishes the ``$\implies$'' part of (\ref{E1}).

On the other hand, suppose that $x \in \omega$ and $d$ is a $\Delta^0_2$ code such that the computation $\{ e \}^{d_*}(x)$ diverges.  There are two cases.  In the first place, suppose that $\mathrm{use} (\{ e \}^{d_*}_t)$ is unbounded, as $t \rightarrow \infty$.  Fix $k \in \omega$ and let $s$ be such that
\[
u = \mathrm{use} (\{ e \}^{d_*}_s (x) ) \geq k.
\]
Let $t_0$ be such that, for each $t \geq t_0$,
\[
d_t \upto u = d_* \upto u.
\]
If $t = \max\{ s,t_0 \}$, then
\[
\mathrm{use} (\{ e \}^{d_t}_t (x)) \geq \mathrm{use} (\{ e \}^{d_t}_s (x))  = \mathrm{use} (\{ e \}^{d_*}_t (x)) \geq k.
\]
As $k$ was arbitrary, it follows that $\{ \mathrm{use} (\{ e \}^{d_t}_t : t \in \omega\}$ is unbounded, i.e., $\Gamma (e,d,x)$.

Secondly, suppose that $\{ e \}^{d_*} (x)$ diverges, but the use of the computation is bounded, say
\[
\mathrm{use} (\{ e \}^{d_*}_t (x)) \leq u,
\]
for all $t \in \omega$.  Let $t_0$ be such that
\[
d_t \upto u = d_* \upto u,
\]
for all $t \geq t_0$.  In particular, for each $t \geq t_0$, the computation $\{ e \}^{d_t}_t (x)$ is equivalent to the computation $\{ e \}^{d_*}_t (x)$.  Hence, $\{ e \}^{d_t}_t (x)$ diverges for all $t \geq t_0$.  Hence, $\Sigma (e,d,x)$ holds, since $d$ and $x$ were arbitrary.

Combining the two cases above establishes the ``$\Longleftarrow$'' part of (\ref{E1}).

Fix a $\Pi_4^0$ predicate, $Q$, and a computable function, $g$, such that
\[
Q(e) \leftrightarrow (\forall x)(\exists y)\big[ g(e,x,y) \in \indset{inf} \big]
\]
Define a computable function, $h$, such that
\[
\{h(e)\}_s^A(n) = \begin{cases}
\langle A \upto n \rangle &\mbox{if }(\forall j \leq n)\big[ \min(C_{j+1}(A)) < s \\
	&\wedge\min(C_{j+1}(A)) > \max(W_{g(e,i,j), s}) \big]\\
\uparrow &\mbox{otherwise}
\end{cases},
\]
where $i = \min(C_0(A))$ and $\{h(e)\}^A(n) \uparrow$ for all $n\in\mathbb N$ if $C_0(A) = \emptyset$.

If $C_0(A) = \emptyset$, then $W_e = \emptyset$.  If $Q(e)$ and $C_0(A) \neq \emptyset$, let $i = \min(C_0(A))$.  Because $Q(e)$, there is a $j$ such that $g(e,i,j) \in \indset{inf}$.  Hence, for all but finitely many $s$, either $\min(C_{j+1}(A)) \geq s$ or $\min(C_{j+1}(A)) < \max(W_{g(e,i,j), s})$ and $W_e$ is finite.  We conclude that $h(e) \in \Delta^2_0[\indset{fin}]$.

Now suppose that ${}^{\neg}Q(e)$ and let $x$ be such that $(\forall y)[g(e,x,y)\in \indset{fin}]$.  Since the maximum of each $W_{g(e,x,y)}$ for $y\in\mathbb N$ can be found in the limit, there is a $\Delta^2_0$ oracle, $A$, such that $C_{j}(A) \neq \emptyset$ for all $j \in\mathbb N$, $\min(C_0(A)) = x$ and $\min(C_{j+1}(A)) > \max(W_{g(e,x,j)})$ for all $j \in \mathbb N$.  Since $W^A_{h(e)}$ is infinite and $A \in \Delta^2_0$, $h(e) \not\in \Delta^2_0[\indset{fin}]$.

Thus, $h$ reduces $Q$ to $\Delta^2_0[\indset{fin}]$.

\end{proof}

Before stating the next proposition, we recall the following standard definitions.

\begin{Definition}\label{dnc def}
A function, $f$, is said to be \emph{diagonally non-computable} if $f(e) \neq \{e\}(e)$ whenever $\{e\}(e)\downarrow$.  Let $\indset{dnc} = \{ f \in \omega^\omega : f \mbox{ is DNC} \}$.

A function, $f$, is {\em $n$ diagonally non-computable ($n$-DNC)} if it is diagonally non-computable and, additionally, $f(e) \leq n$, for each $e$.  Let $\indset{dnc}_n = \{ f \in \omega^\omega : f \mbox{ is $n$-DNC} \}$.
\end{Definition}

\begin{Theorem}
$\indset{dnc}_{n}[\indset{dnc}_{n+1}]$ is $\Pi_2^0$-complete for all $n \geq 2$.
\end{Theorem}

\begin{proof}
$e \not\in \indset{dnc}_{n}[\indset{dnc}_{n+1}]$ if and only if
\[
(\exists \sigma \in \bairestr)\Big( \mbox{$\sigma$ is a $\indset{dnc}_{n}$ string} \wedge \mbox{$\{e\}^{\sigma}\upto |\sigma|$ is not a $\indset{dnc}_{n+1}$ string} \Big).
\]
The statement ``$\sigma$ is a $\indset{dnc}_{n}$ string'' is equivalent to
\[
(\forall x < |\sigma|)\Big( (\forall s)\big(\{x\}_s(x)\uparrow\big) \vee (\exists s)\big(\{x\}_s(x)\downarrow \neq \sigma(x) < n \big) \Big),
\]
and the statement ``$\{e\}^{\sigma}\upto |\sigma|$ is not a $\indset{dnc}_{n+1}$ string'' is equivalent to
\begin{align*}
(\exists x < |\sigma|)\Big( (\forall s)\big( \{e\}^{\sigma}_s(x) \uparrow \big) \vee (\exists s,y) \big( &\{e\}^{\sigma}_s(x) \downarrow = y \\
&\wedge (y = \{x\}_s(x) \downarrow \vee y \geq n + 1) \big) \Big).
\end{align*}
Thus, ``$e \not\in \indset{dnc}_{n}[\indset{dnc}_{n+1}]$'' is $\Sigma_2^0$, i.e.,  $\indset{dnc}_{n}[\indset{dnc}_{n+1}]$ is $\Pi_2^0$.

Define a computable function $f$ such that $\{f(e)\}^{\sigma}\upto |\sigma| = \sigma \upto |W_{e,|\sigma|}|$.  If $e \in \indset{inf}$, then $\{f(e)\}^g = g$ for all $g \in \baire$.  If $e \in \indset{fin}$, then the domain of $\{f(e)\}^g$ is finite for all $g \in \baire$.  Since every $\indset{dnc}_n$ is also $\indset{dnc}_{n+1}$, $f(e) \in \indset{dnc}_{n}[\indset{dnc}_{n+1}]$ if $e \in \indset{inf}$ and not otherwise, showing that $\indset{dnc}_{n}[\indset{dnc}_{n+1}]$ is $\Pi_2^0$-hard.

\end{proof}

\section{$\Pi^1_1$-completeness}

In what follows we give a general, but somewhat technical, theorem which implies that a number of natural index sets of the form $\mathcal A [ \mathcal B]$ are $\Pi^1_1$-complete.

\begin{Definition}
Let $c$ be a finite ordinal or $\omega$.  We say that a function $f: \omega^\omega \rightarrow c^\omega$ is a {\em $\Delta^0_2$ embedding} if, and only if, there is a uniformly computable sequence, $f_s : \omega^{<\omega} \rightarrow c^{<\omega}$, of total recursive functions such that the following conditions hold:
\begin{enumerate}
	\item $f_s (\alpha) \prec f_s(\beta)$ if, and only if, $\alpha \prec \beta$, i.e., each $f_s$ is a {\em $\prec$-isomorphism}
	\item $\lim_s f_s (\alpha)$ exists, for each $\alpha \in \omega^{<\omega}$
	\item $f(x) = \bigcup_n \lim_s f_s (x \upto n)$
\end{enumerate}
Note that the union $\bigcup_n \lim_s f_s (x \upto n)$ is a well-defined function on $\omega$ since the $f_s$ all preserve proper extension
\end{Definition}

\begin{Theorem}\label{pi one one}
Suppose that $\mathcal A  \subset c^\omega$ (where $c$ is a finite ordinal or $\omega$) and $\mathcal B \subseteq 2^\omega$ are hyperarithmetic, with $\mathcal B$ a tail set containing no finite sets and no cofinite sets, and there is a $\Delta^0_2$ embedding $f: \omega^\omega \rightarrow c^\omega$ such that the range of $f$ is relatively closed in $\mathcal A$.  If there is a Medvedev reduction, $\Phi$, of $\mathcal B$ to $\mathcal A$, then the set of $\mathcal A$-universal $\mathcal B$-codes is $\Pi^1_1$-complete.
\end{Theorem}

\begin{proof}
Let $\mathcal A$, $\mathcal B$, $\Phi$ and $f$ be as above, with $f_s : \omega^{<\omega} \rightarrow c^{<\omega}$ witnessing that $f$ is a $\Delta^0_2$ embedding.  For convenience, define $f_* : \omega^{<\omega} \rightarrow c^{<\omega}$ by $f_* (\alpha) = \lim_s f_s (\alpha)$ and observe that, by the properties of the $f_s$, the map $f_*$ is also a $\prec$-isomorphism.

In the first place, it follows from the definition of $\mathcal A [\mathcal B]$ and the fact that $\mathcal A$ and $\mathcal B$ are both hyperarithmetic that $\mathcal A [ \mathcal B ]$ is $\Pi^1_1$.  To show that $\mathcal A [ \mathcal B ]$ is $\Pi^1_1$-hard, it will suffice to reduce the $\Pi^1_1$-complete set $\indset{NoPath}$ (see Corollary IV.2.16 from \cite{odifreddi}) to $\mathcal A [ \mathcal B ]$, where $\indset{NoPath}$ is the set of codes for recursive trees with no infinite branches.  More precisely, if $\rho_0 , \rho_1 , \ldots$ is a recursive enumeration of $\omega^{<\omega}$ and $e \in \omega$, with $\{ e \}$ total, define $T_e$ to be the tree generated by
\[
\{ \rho_i : \{ e \} (i) \downarrow = 1\}.
\]
Let
\[
\indset{NoPath} = \{ e : \{ e \} \mbox{ is total and } [T_e] = \emptyset\}.
\]

With this in mind, we will define Turing functionals $\Phi_e$ such that $\Phi_e$ is a Medvedev reduction of $\mathcal B$ to $\mathcal A$, if, and only if, $e \in \indset{NoPath}$.  For each $e \in \omega$ and $\sigma \in \omega^{<\omega}$, let
\[
n_{e , \sigma} = \max \{ \min \{ s , |\beta| \} : (\exists \alpha \in T_{e , |\sigma|}) (\beta \preceq f_s (\alpha) \wedge\sigma \succeq \beta\}
\]
and
\[
m_{e , \sigma} = \max \{ m : (\forall i \leq m) ( \{ e \}_{|\sigma|} (\tau_i) \downarrow)\},
\]
where $\tau_0 , \tau_1 , \ldots$ is a fixed recursive enumeration of $c^{<\omega}$.  Finally, define
\[
\Phi_e (\sigma) = (\Phi(\sigma) \cup [0,n_{e , \sigma}]) \cap [0 , m_{e , \sigma}].
\]

In the first place, if $\{ e\}$ is not total, then $m_{e,\sigma}$ is bounded as $\sigma$ varies over $\omega^{<\omega}$ and, hence, $\lim_{n \rightarrow \infty} \Phi_e (X \upto n)$ is a finite set (consequently, not in $\mathcal B$), for each $X \in \omega^\omega$.  It follows that $\Phi_e$ is not a Medvedev reduction of $\mathcal B$ to $\mathcal A$.

Therefore, assume that $\{ e \}$ is total, i.e., $m_{e , \sigma} \rightarrow \infty$, as $|\sigma| \rightarrow \infty$.

Suppose first that $[T_e] \neq \emptyset$, with $X \in [T_e]$.  The family $\mathcal B$ contains no cofinite sets (in particular, $\mathcal B$ does not contain $\omega$) and, hence, to show that $\Phi_e$ is not a Medvedev reduction of $\mathcal B$ to $\mathcal A$, it will suffice to show that $\Phi_e (X) = \bigcup_n \Phi_e (X \upto n) = \omega$.  In turn, it will be enough to show that
\[
\lim_{\substack{\sigma \prec X\\ |\sigma| \rightarrow \infty}} n_{e , \sigma} = \infty.
\]
Indeed, fix $n_0 \in \omega$ and let $\sigma_0 \prec f(X)$ be long enough that there exists an $\alpha_0 \in T_{e , |\sigma_0|}$ with $|f_* (\alpha_0)| \geq n_0$.  Next, let $s_0 \geq n_0$ be large enough that $f_s (\alpha) = f_* (\alpha)$, for each $s \geq s_0$ and $\alpha \preceq \alpha_0$.  Letting $\beta = f_* (\alpha_0) = f_{s_0} (\alpha_0)$, it follows that
\[
n_{e , \sigma} \geq \min \{ s_0 , \beta \} \geq n_0,
\]
for every $\sigma \succeq \sigma_0$, by the definition of $n_{e , \sigma}$.

Next, assume that $[T_e] = \emptyset$.  Since $\mathcal B$ is a tail set, to show that $\Phi_e$ is Medvedev reduction of $\mathcal B$ to $\mathcal A$, it will suffice to show that, for each $Y \in \mathcal A$, the set $\{ n_{e , \sigma} : \sigma \prec Y\}$ is bounded and, hence, $\Phi_e(Y) $ differs only finitely from $\Phi(Y)$, for each $Y \in \mathcal A$.

Indeed, fix $Y \in \mathcal A$.  First, suppose that $Y = f(X)$, for some $X \in \omega^\omega$.  Let $\alpha_0 \prec X$ be longest such that $\alpha_0 \in T_e$.  Let $s_0$ be large enough that $f_s (\alpha) = f_* (\alpha)$, for each $s \geq s_0$ and $\alpha \preceq \alpha_0$.  Fix $\sigma \prec Y = f(X)$, with $\sigma \succeq f_*(\alpha_0)$.  Suppose that $s \geq s_0$, $\alpha \in T_e$ and $\beta \in c^{<\omega}$, with $\beta \preceq f_s (\alpha)$ and $\sigma \succeq \beta$.  If $|\beta| > |f_* (\alpha_0)|$, then $f_s (\alpha) \succ f_*(\alpha_0) = f_s(\alpha_0)$.  Hence, $\alpha \succ \alpha_0$, since $f_s$ is a $\prec$-isomorphism.  This is a contradiction, since no extension of $\alpha_0$ is in $T_e$.  It follows that $|\beta| \leq |f_*(\alpha_0)|$.  Consequently, if $\alpha \in T_e$, $s \in \omega$ and $\beta , \sigma \in c^{<\omega}$ are such that $\beta \preceq f_s (\alpha)$ and $\sigma \succeq \beta$, then either $|\beta| \leq |f_* (\alpha_0)|$ or $s < s_0$.  It follows that
\[
n_{e , \sigma} \leq \max \{ s_0 - 1 , |f_*( \alpha_0 )| \},
\]
for any $\sigma \prec Y$.

Finally, assume that $Y \in \mathcal A$, but $Y \neq f(X)$, for every $X \in \omega^\omega$.  Since the range of $f$ is relatively closed in $\mathcal A$, choose $\sigma_0\prec Y$ longest such that $\sigma_0 \prec f(X)$, for some $X \in \omega^\omega$.  Let $\alpha_0 \in \omega^{<\omega}$ be such that $f_*(\alpha_0) \succeq \sigma_0$.  Let $s_0$ be large enough that $f_s (\alpha) = f_* (\alpha)$, for all $s \geq s_0$ and $\alpha \preceq \alpha_0$.  Fix $\sigma \prec Y$, with $|\sigma| \geq |\sigma_0|$.  If $s$, $\beta$ and $\alpha$ are such that $\beta \preceq f_s (\alpha)$ and $\sigma \succeq \beta$, then either $\beta \preceq \sigma_0$ or $s < s_0$.  Thus,
\[
n_{e , \sigma} \leq \max \{ s_0 - 1, |\sigma_0|\}
\]
and it follows that
\[
\{ n_{e , \sigma} : \sigma \prec Y\}
\]
is bounded.  This completes the proof.
\end{proof}

\begin{Definition} We define several forms of immunity and associated index sets.
\begin{enumerate}
\item $A$ is \emph{immune} if $A$ is infinite and contains no infinite c.e.~set.

\item $A$ is \emph{bi-immune} if $A$ and $\overline{A}$ are both immune.

\item $\indset{im} = \{ A \subset \mathbb N : A \mbox{ is immune} \}$.

\item $\indset{bi} = \{ A \subset \mathbb N : A \mbox{ is bi-immune} \}$.

\end{enumerate}
\end{Definition}

\begin{Corollary}\label{dnc-bi}
The index sets $\indset{dnc}[\indset{im}]$ and $\indset{dnc}[\indset{bi}]$ are both $\Pi^1_1$-complete.
\end{Corollary}

\noindent See Definition~\ref{dnc def} for the definition of $\indset{dnc}$.

\begin{proof}
An examination of the relevant definitions reveals that the sets {\sc dnc}, {\sc im} and {\sc bi} satisfy the required topological and definability properties to apply Theorem~\ref{pi one one}.  Since there are known to be Medvedev reductions of $\indset{im}$ and $\indset{bi}$ to $\indset{dnc}$, it will suffice to define a function $f: \omega^\omega \rightarrow \omega^\omega$ which is a $\Delta^0_2$ embedding into $\indset{dnc}$.

Let $\langle \cdot \rangle : \omega^{<\omega} \rightarrow \omega$ be a recursive coding of finite strings.  For each $\alpha \in \omega^{<\omega}$ and $n < |\alpha|$, define
\[
f_s (\alpha) (n) = \begin{cases}
2 \langle \alpha \upto n \rangle &\mbox{if } \{n\}_s(n) \neq 2 \langle \alpha \upto n \rangle\\
1 + 2 \langle \alpha \upto n \rangle &\mbox{otherwise}
\end{cases}.
\]
Note that $f_s (\alpha)$ is a string of integers of length $|\alpha|$ and, moreover, that each $f_s$ is a $\prec$-isomorphism.  For $X \in \omega^\omega$, let $f(X)$ be as in the definition of a $\Delta^0_2$ embedding, i.e.,
\[
f(X) = \bigcup_n \lim_{s\rightarrow \infty} f(X \upto n).
\]
First of all, note that the range of $f$ is the closed subset of $\omega^\omega$, consisting of those $X$ such that, for each $n \in \omega$,
\[
X(n) = \begin{cases}
2\langle X \upto n \rangle &\mbox{if } \{ n \} (n) \downarrow \neq 2 \langle X \upto n \rangle\\
1 + 2 \langle X \upto n \rangle &\mbox{otherwise}.
\end{cases}
\]
Furthermore, note that $f(X) \in \indset{dnc}$, for each $X \in \omega^\omega$.  We may now apply Theorem~\ref{pi one one} to conclude that $\indset{dnc}[ \indset{bi} ]$ is $\Pi^1_1$-complete.
\end{proof}

With a couple more definitions, we will be able to state another corollary of Theorem~\ref{pi one one}

\begin{Definition}
Let $\mathcal P_{\mathsf{fin}} (\omega)$ denote the family of finite subsets of $\omega$.  A {\em canonical numbering} is a total computable function $H : \omega \rightarrow \mathcal P_{\mathsf{fin}} (\omega)$ such that
\begin{enumerate}
\item each finite set is in the range of $H$,
\item the predicate ``$x \in H(e)$'' is computable, and
\item the function $e \mapsto \max H(e)$ is computable.
\end{enumerate}
\end{Definition}

Identifying $\mathcal P_{\mathsf{fin}} (\omega)$ with $2^{<\omega}$, we could alternatively characterize a canonical numbering as a total computable function $H : \omega \rightarrow 2^{< \omega}$ such that each finite set is in the range of $H$.

\begin{Definition}\cite{bkk-rodfest}
A infinite set $R \subseteq \omega$ is {\em canonically immune} if, and only if, there is a total computable function $h$ such that, for each canonical numbering $H$, and all but finitely many $e \in \omega$,
\[
H(e) \subseteq R \implies |H(e)| \leq h(e).
\]
\end{Definition}

\begin{Definition}\cite{snr-def}
A function $f: \omega \rightarrow \omega$ is {\em strongly non-recursive} if, and only if, for each total computable function $h:\omega \rightarrow \omega$, one has $f(n) \neq h(n)$, for all but finitely many $n$.
\end{Definition}

\begin{Corollary}
The index set {\sc ci[\sc snr] }is $\Pi^1_1$-complete.
\end{Corollary}

\begin{proof}
In what follows, we freely identify an element of $2^\omega$ or $2^{<\omega}$ with the subset of $\omega$ of which it is the characteristic function.

In the first place, an inspection of the proof Theorem 5.5 in \cite{bkk-rodfest} reveals that there is a Medvedev reduction of {\sc snr} to {\sc ci}.  It follows from the definitions of {\sc ci} and {\sc snr} that the requisite topological and definability properties are satisfied in order to apply Theorem~\ref{pi one one}.  All that remains is to define a $\Delta^0_2$ embedding into {\sc ci} with a relatively closed range.

To this end, we begin by defining a universal function for canonical numberings.  Let $\varphi : \omega^2 \rightarrow 2^{<\omega}$ be a universal partial recursive function.  Define
\[
D_{r,s} (e) = \begin{cases}\varphi (r,e) &\mbox{if $\varphi (r,e)$ converges within $s$ stages},\\ \langle \emptyset \rangle &\mbox{otherwise}. \end{cases}
\]
Let $D_r (e) = \lim_{s\rightarrow \infty} D_{r,s} (e)$.  It follows that each canonical numbering appears as $D_r$, for some $r$, though not every $D_r$ is a canonical numbering.  The function $D$ is itself limit computable.

Define
\[
F_n = \bigcup_{r,e \leq n} D_r (e)
\]
and note the sequence, $F_n$, of finite sets is uniformly limit computable.  Since every finite set is contained in some $F_n$, it follows that there exist $n_0 < n_1 < \ldots$ and $x_0 < x_1 < \ldots$ such that, for each $i$,
\[
F_{n_i} \setminus F_{n_i - 1} \neq \emptyset
\]
and
\[
x_i \in F_{n_i} \setminus F_{n_i - 1}.
\]
The sequence, $(x_i)_{i \in \omega}$, may be chosen to be strictly increasing and limit computable.  We may, therefore, take a computable sequence, $(x_{i,s})_{i,s \in \omega}$ such that, for each $i$,
\[
x_i = \lim_{s \rightarrow \infty} x_{i,s}
\]
and, for fixed $s$, the $x_{i,s}$ are all distinct.

For each $s \in \omega$ and $\alpha \in \omega^{<\omega}$, define $f_s (\alpha) \in 2^{<\omega}$ to have length
\[
x_{(\alpha(0) + \ldots + \alpha (|\alpha| - 1) + |\alpha|-1) , s} + 1
\]
and be such that
\[
f_s (\alpha) (j) = \begin{cases}1 &\mbox{if } (\exists i,p \in \omega) ( p < |\alpha| \wedge j = x_{i,s}  \wedge i = \alpha(0) + \ldots + \alpha(p) + p),\\ 0 &\mbox{otherwise}.\end{cases}
\]
It follows that each $f_s$ is a $\prec$-isomorphism.  Let $f : \omega^\omega \rightarrow \textsc{ci}$ be as in the definition of a $\Delta^0_2$ embedding.

It remains to verify that each $f(X)$ is canonically immune.  Indeed, suppose that $H : \omega \rightarrow 2^{<\omega}$ is a canonical numbering, with $H = D_r$.  For each $e \geq r$ observe that $D_r(e) \subseteq F_e$ and, hence, for any $X \in \omega^\omega$,
\[
f(X) \cap D_r (e) \subseteq \{ x_0 , \ldots , x_e\}.
\]
In particular, $|f(X) \cap D_r (e)| \leq e+1$.  As $H$ was arbitrary, it follows that each $f(X)$ is canonically immune, witnessed by the computable function $h(e) = e+1$.

Finally, to see that the range of $f$ is relatively closed, simply observe that the range of $f$ is the intersection of {\sc ci} with the closed set
\[
\{ Y \in 2^\omega : (\forall j) (Y(j) \neq 0 \implies (\exists i) ( j = x_i )) \}.
\]
This completes the proof.
\end{proof}

In Theorem~\ref{pi one one}, we required that the $\Delta^0_2$ embedding have relatively closed range in the class $\mathcal A$.  In fact, we can achieve the same result if we require that the map, $f$, have relatively $\Pi^0_2$ range in $\mathcal A$.

\begin{Theorem}\label{pi 2 version}
Suppose that $\mathcal A \subseteq c^\omega$ (where $c$ is a finite ordinal or $\omega$) and $\mathcal B \subseteq 2^\omega$ are hyperarithmetic, with $\mathcal B$ a tail set containing no finite sets and no cofinite sets, and there is a $\Delta^0_2$ embedding $f: \omega^\omega \rightarrow c^\omega$ such that the range of $f$ is the intersection of a $\Pi^0_2$ class, $P$, with $\mathcal A$.  If there is a Medvedev reduction, $\Phi$, of $\mathcal B$ to $\mathcal A$, then the set of $\mathcal A$-universal $\mathcal B$-codes is $\Pi^1_1$-complete.
\end{Theorem}

Note that this result is neither a generalization of Theorem~\ref{pi one one} nor {\em vice versa}, since not every closed set is a $\Pi^0_2$ class and not every $\Pi^0_2$ class is closed.  Before proceeding with the proof of Theorem~\ref{pi 2 version}, we recall the definition of a $\Pi^0_2$ class.

\begin{Definition}\label{def of pi 2}
Let $c$ be a finite ordinal or $\omega$.  A class, $P \subseteq c^\omega$, is a {\em $\Pi^0_2$ class} if, and only if, there is a total computable function $h : \omega^2 \rightarrow c^{<\omega}$ such that, for each $X \in c^\omega$,
\[
X \in P \iff (\forall n) (\exists s) (h(n,s) \prec X).
\]
\end{Definition}

\begin{proof}[Proof of Theorem~\ref{pi 2 version}]
Let $\mathcal A$, $\mathcal B$, $f$, $P$ and $\Phi$ be as in the statement of Theorem~\ref{pi 2 version}.  Let $f_s$ be as in the definition of a $\Delta^0_2$ embedding, witnessing that $f$ is such an embedding.  Again, let $f_*$ be the pointwise limit of the sequence $(f_s)_{s \in \omega}$.  Finally, let $h : \omega^2 \rightarrow c^{<\omega}$ be a total recursive function, as in Definition~\ref{def of pi 2}, witnessing that $P$ is a $\Pi^0_2$ class.  For convenience, we write
\[
U_{n,s} = \{ \sigma \in c^{<\omega} : (\exists t \leq s)(h(n,t) \preceq \sigma)\}.
\]
Each $U_{n,s}$ is, by definition, closed under extension.  With this notation, for each $X \in c^\omega$,
\[
X \in P \iff (\forall n) (\exists s) (\mbox{an initial segment of $X$ is in $U_{n,s}$}).
\]

As in the proof of Theorem~\ref{pi one one}, we reduce the $\Pi^1_1$-complete set $\indset{NoPath}$ to $\mathcal A [ \mathcal B]$.  With respect to the coding of recursive trees, we adopt notation from the proof of Theorem~\ref{pi one one}.

What follows is similar in character to the proof of Theorem~\ref{pi one one}, with the addition of some refinements to accomodate the fact that the range of $f$ may not be closed.  Given $e \in \omega$ and $\sigma \in c^{<\omega}$, let
\begin{align*}
n_{e ,\sigma} = \max\{ \min \{ s , |\beta|, n\} : (\exists \alpha &\in T_{e , |\sigma|})( \beta \preceq f_s (\alpha) \wedge \sigma \succeq \beta\\
&  \wedge (\forall k \leq n) (\beta \in U_{k,|\sigma|}))\}.
\end{align*}
As before, let
\[
m_{e , \sigma} = \max \{ m : (\forall i \leq m) ( \{ e \}_{|\sigma|} (\tau_i) \downarrow\},
\]
where $\tau_0 , \tau_1 , \ldots$ is a fixed recursive enumeration of $c^{<\omega}$.  Define
\[
\Phi_e (\sigma) = (\Phi(\sigma) \cup [0,n_{e , \sigma}]) \cap [0 , m_{e , \sigma}].
\]
As in the proof of Theorem~\ref{pi one one}, if $\{ e \}$ is not total, then $m_{e , \sigma}$ is bounded as $\sigma$ varies over $c^{<\omega}$ and, consequently, $\Phi_e (Y)$ is the characteristic function of a finite set, for every $Y \in c^\omega$.  Thus, $\Phi_e$ is not a Medvedev reduction of $\mathcal B$ to $\mathcal A$, since $\mathcal B$ contains no finite sets.

Therefore, suppose that $\{ e \}$ is total and $T_e$, the recursive tree coded by $e$, has an infinite branch, $X \in [T_e]$.  To show that $\Phi_e$ is not a Medvedev reduction of $\mathcal B$ to $\mathcal A$, it will suffice to show that
\[
\lim_{k \rightarrow \infty} n_{e , X \upto k} = \infty
\]
as $\Phi_e$ will then be the characteristic function of $\omega$.  Indeed, fix $n_0 \in \omega$.  Let $\sigma_0 \prec f(X)$, with $|\sigma_0|$ such that
\begin{itemize}
	\item $|\sigma_0| \geq n_0$ and
	\item $\sigma_0 \in U_{n_0 , |\sigma_0|}$.
\end{itemize}
Now choose $\sigma_1$ such that
\begin{itemize}
	\item $\sigma_0 \preceq \sigma_1 \prec f(X)$ and
	\item there exists $\alpha_0 \in T_{e , |\sigma_1|}$ such that $\sigma_0 \prec f_* (\alpha_0) \prec f(X)$.
\end{itemize}
Finally, let $s_0$ be such that
\begin{itemize}
	\item $s_0 \geq n_0$ and
	\item $f_{s_0} (\alpha_0) = f_* (\alpha_0)$.
\end{itemize}
It follows from the definition of $n_{e , \sigma}$ that, for each $\sigma \succeq \sigma_1$
\[
n_{e , \sigma} \geq \min\{ s_0 , f_* (\alpha_0) , n_0\} \geq n_0.
\]

Suppose now that $[T_e] = \emptyset$.  We must show that $\Phi (Y) = \lim_{j \rightarrow \infty} \Phi(Y \upto j) \in \mathcal B$, for each $Y \in \mathcal A$.  In the first place, suppose that $Y \in \mathrm{range} (f)$, say with $Y = f(X)$.  To show that $\Phi(Y) \in \mathcal B$, it will suffice to show that
\[
\{ n_{e , \sigma} : \sigma \prec f(X) \}
\]
is bounded.  Let $\alpha_0$ be longest with $\alpha_0 \in T_e$ and let $s_0$ be such that, for each $s \geq s_0$ and $\alpha \preceq \alpha_0$, we have $f_s (\alpha) = f_* (\alpha)$.  Suppose now that $\sigma \prec f(X)$, with $f_* (\alpha_0) \prec \sigma$, and $s , \beta$ are such that there exists $\alpha \in T_e$, with $f_* (\alpha_0) \prec \beta \preceq f_s (\alpha)$ and $\beta \preceq \sigma$.  If $s \geq s_0$, we have $\alpha_0 \prec \alpha$, since $f_s (\alpha_0) = f_* (\alpha_0)$ and $f_s$ is a $\prec$-isomorphism.  Hence, if $\sigma \prec f(X)$ and $s,\beta$ are such that there exists $\alpha \in T_e$ with $\beta \prec f_s (\alpha)$ and $\sigma \succeq \beta$, then either $s < s_0$ or $|\beta| \leq f_* (\alpha_0)$.  It follows that
\[
n_{e , \sigma} \leq \min \{ s_0 - 1 , |f_* (\alpha_0)|\}
\]
for $\sigma \prec f(X)$.

Finally, suppose that $Y \in \mathcal A$, but $Y \notin \mathrm{range} (f)$.  Again, we will see that
\[
\{ n_{e , \sigma} : \sigma \prec f(X) \}
\]
is bounded.  Since $Y \notin \mathrm{range} (f)$, let $n_0$ be such that, for every $n > n_0$, no initial segment of $Y$ lies in $\bigcup_t U_{n,t}$.  It follows from the definition of $n_{e , \sigma}$ that, for any $\sigma \prec Y$, we have $n_{e , \sigma} \leq n_0$.
\end{proof}

We conclude with a corollary of this result.  Recall from \cite{jo-fpff} that $\indset{dnc} <_s \indset{dnc}_n$.  Hence, there are Medvedev reductions of $\indset{im}$ and $\indset{bi}$ to $\indset{dnc}_n$, for each $n \in \omega$.  (See Definition~\ref{dnc def} for the definition of $\indset{dnc}_n$.)

\begin{Corollary}
The index sets $\indset{dnc}_n [\indset{im}]$ and $\indset{dnc}_n [\indset{bi}]$ are both $\Pi^1_1$-complete.
\end{Corollary}

\begin{proof}
In light of Theorem~\ref{pi 2 version}, it will suffice to produce a $\Delta^0_2$ embedding, $f$, of $\omega^\omega$ into $\indset{dnc}_n$, such that the range of $f$ the intersection of $\indset{dnc}_n$ with a $\Pi^0_2$ class.  Note that such an embedding cannot have a closed range, otherwise it would be a homeomorphism between $\omega^\omega$ and a compact space.

Let $\varphi : \omega^2 \rightarrow n$ be a universal partial recursive function and, for each $s \in \omega$, let $x_{0,s} < x_{1,s} < \ldots$ enumerate those $x$ such that $\varphi_x (x)$ has not yet converged in $s$ stages.  Each sequence $(x_{i,s})_{s \in \omega}$ is eventually constant, say with limit $x_i$.  Observe that $x_0 < x_1 < \ldots$ enumerate those $x$ such that $\varphi_x (x)$ diverges.

Fix a $\indset{dnc}_n$ function, $H$.  For each $s \in \omega$ and $\alpha\in \omega^{<\omega}$, define $f_s (\alpha) \in n^{<\omega}$ of length
\[
x_{(\alpha(0) + \ldots + \alpha(|\alpha| - 1)  + |\alpha|-1) , s} + 1
\]
such that
\[
f_s (\alpha) (j) = \begin{cases}
1 &\mbox{if } (\exists i,p) (j = x_{i,s} \wedge i = \alpha(0) + \ldots + \alpha(p) + p),\\
0 &\mbox{if } (\exists i) (j = x_{i,s}) \mbox{, but there is no $p$ as above},\\
H(j) &\mbox{otherwise}.
\end{cases}
\]
Let $f : \omega^\omega \rightarrow \indset{dnc}_n$ be the $\Delta^0_2$ embedding induced by the $f_s$.  It follows that, for each $Y \in \omega^\omega$ and each $j \in \omega$
\[
f(Y)(j) = \begin{cases}
1 &\mbox{if } (\exists i,p) (j = x_i \wedge i = Y(0) + \ldots + Y(p) + p),\\
0 &\mbox{if } (\exists i) (j = x_i) \mbox{, but there is no $p$ as above},\\
H(j) &\mbox{otherwise}.
\end{cases}
\]

Note that, if $G \in \indset{dnc}_n$, then $G(x) = H(x)$, for each $x \notin \{ x_0 , x_1 , \ldots \}$.  Hence,
\[
\mathrm{range} (f) = \indset{dnc}_n \cap \{ Y \in n^\omega : (\forall i) (Y(x_i) \in \{0,1\}) \wedge (\exists^\infty i) (Y(x_i) = 1) \}.
\]
Observe that, since the sequence $(x_i)_{i \in \omega}$ is limit computable, it follows that the latter set in the intersection above is $\Pi^0_2$.  In other words, $\mathrm{range} (f)$ is the intersection of $\indset{dnc}_n$ with a $\Pi^0_2$ class.  This completes the proof.
\end{proof}

redpajama_set_name: RedPajamaArXiv


text:
Billy Gilmour makes admission about Rangers and Celtic
Shane Callaghan @sugarshane1988
@sugarshane1988
Celtic and Ibrox side Glasgow Rangers both had the 18-year-old sensation as a trainee once upon a time.
Billy Gilmour has admitted The Football Daft podcast that he didn't enjoy his stint with Celtic as much as his spell with Rangers.
The 18-year-old sensation began his burgeoning career at Rangers, though he was also training with their Old Firm rivals Celtic as a much younger player.
Gilmour might be the most exciting young Scottish player around and in 2017 he secured a big move to Premier League side Chelsea, where he has since dazzled for the Under-23 side.
Reflecting on his time with the Gers and the Hoops, he said to The Football Daft podcast: "I enjoyed Rangers more obviously. I knew a lot of the boys from other teams we had been at.
""I grew up with Harry Cochrane at Rangers. We went to this one tournament in Portugal and it was amazing."
According to The Scottish Sun, Rangers only got around £500,000 for Gilmour, which is a hell of a compensation fee for a young player, but it's peanuts in comparison to what he could and probably will be worth a few years from now.
This is a player who could very easily be worth an eight-figure sum in the coming years, but next season is big for him.
Gilmour, whose dad supports Celtic, will continue with the Blues' Under-23s and if he impresses then maybe, just maybe, he'll get first-team chances.
More articles from Shane Callaghan
Shane is a Dublin-based sports writer who swapped the football pitch for the press box upon learning that he was better with a pen than with a round bit of leather. Prior to joining HITC, Shane enjoyed spells with Sports Mole and Goal.com among others, while he was also club reporter for Irish football club Drogheda United.

redpajama_set_name: RedPajamaCommonCrawl


text:
Федорково — деревня в Весьегонском муниципальном округе Тверской области России.

География
Деревня расположена в 27 км на запад от районного центра города Весьегонска.

История
В 1777 году в селе Федорково после пожара была возобновлена деревянная Успенская церковь с 2 престолами.

В конце XIX — начале XX века деревня входила в состав Любегощской волости Весьегонского уезда Тверской губернии. В 1859 году в деревне было 34 двора.

С 1929 года деревня являлась центром Федорковского сельсовета Весьегонского района Бежецкого округа Московской области, с 1935 года — в составе Калининской области, с 2005 года — в составе Ёгонского сельского поселения, с 2019 года — в составе Весьегонского муниципального округа.

Население

Примечания

Населённые пункты Весьегонского района

redpajama_set_name: RedPajamaWikipedia


It is compatible with Honda Aviator, Mahindra Gusto, TVS Jupiter, TVS Wego and more. Its higher NSD with Connected Blocks and Continuous centre Groove. Deeper Shoulder Grooves and Sipes on the tread Blocks. More rubber to wear better mass transfer and More 'biting' edges to Grip on or Off road. Longer tyre life and Better Handling and safer ride on rough and wet roads.


text:
Q: Making my controller DB calls Async or not I've inherited a MVC web application, that's using Dapper on top of ADO.NET to make DB calls in the Controller action methods. It's all pretty standard stuff - not many of the controlers are async, and all the database calls go through a repository which is fully synchronous.
I'm moving this to Azure, and I'm using SQL Azure for the database back end. I'm expecting load to be fairly standard - say 500 - 1000 hits per minute.
So, I'm wondering, should I be ploughing through this code to make all my db calls async, so that i can await then in the controllers. Doing this is going to free up my threads to serve up other requests, but I'm wondering in real terms if I'm going to notice any improvement.
I know that previously, it's been noted that if you have a single db server (as I do), then you won't really see much improvement, because the bottleneck is all on the db. However, SQL Azure is a slightly different beast, and Azure state that

Good practice demands that you use only asynchronous techniques to access Azure services such as SQL Database source

So - is this worth the effort?

A: Frankly, it's impossible to give any sort of definitive answer here. As Azure states, best practice is to use async with I/O-bound operations, including things like querying a remote database. If you were starting this application from scratch, today, I'd definitely tell you use async for your database calls.
However, this is not a new application, and it sounds like using async will require quite a bit of surgery at this point. Depending on the load you get, you may not end up seeing any gains for the work, but you might also see great gains. My recommendation is to start small. I would pick out some of the more longer-running queries you make or actions that rely on the database heavily and start with those. That way you can introduce a bit of async and judge for yourself whether it's worth pursuing it further. And, since these are likely to be the bottlenecks of your application, anyways, you gain the benefits of async where it will potentially matter the most.
Any new functionality you add should be async from the start, and then, simply when you have the time and inclination, work slowly on converting the whole application.

A: What I've learned (and verified through testing) is that you will not see a great improvement on your relatively long SQL calls.  But you will see improvement on concurrent short SQL and non-SQL related responses.  That is because there is a significant cost to initializing a thread.  So reusing the dormant threads that are waiting for SQL does increase performance.
Using async also protects you from going over "Threads Per Processor Limit" setting in IIS.  When that happens your requests get queued.  We have experimented increasing the default value of 25.  This did improve  performance under high load but we saw better improvements by changing all our controllers to async.
So I guess the answer to your question is, it depends.  If you have a significant number of concurrent requests other than your SQL calls, you should see a noticeable improvement on the response time of those concurrent requests.  But you won't see much of an improvement on the relatively long SQL calls.

redpajama_set_name: RedPajamaStackExchange


text:
A Glacier in Italy is Turning Pink. And That's Not Good News For South Asia Also
Insight So Asia World
A glacier in Italy is turning pink because of algae — a development that will make the ice melt faster, a scientist studying the phenomenon told CNN.
A glacier in Italy is turning pink because of algae — a development that will make the ice melt faster, a scientist studying the phenomenon says https://t.co/P85HqKJuNf pic.twitter.com/xsW32tEHE7
— CNN International (@cnni) July 6, 2020
Also, this giant glacier in Antarctica is melting, and it could raise sea levels by 5 feet, scientists say.
Unprecedented melting is seeing Swiss glaciers shrink 10% in five years, a report says.
Glaciers around the world are melting due to climate change, and this latest research shows the extent of the problem in Switzerland.
According to reports, Switzerland has lost more than 500 glaciers since 1900.
Down south, Pakistan's huge glacial formations are part of the whole that wrap the Himalaya's skirt (from Bangladesh in the east to Afghanistan in the west) –they are all at risk.
Pakistan's glaciers are more than anywhere except in the polar region (North and South Pole). The country's northern part Gilgit Baltistan (GB) is known as the Land of Glaciers.
However, global warming has seen the Chiatibo Glacier in Broghil National Park retreat by about 30 feet a year due to higher temperatures melting the ice.
The glacial range of Hindu Kush and the Karakoram is the region's water tower –their mountains provide water for 1.6 billion people of the region called South Asia.
The glacial range runs its westerly course through the northwestern part of Pakistan's Khyber Pakhtunkhwa province.
Here, there are more than 5,000 reasonable sized glaciers, but almost 70% of them are retreating, according to Dr. Furrukh Bashir of the Pakistan Meteorological Department.
It is one of the around 7,000 of Pakistan's 7,200 glaciers that meteorological officials say show signs of melting, citing data gathered over the last 50 years.
The first threat that brings, as these glaciers melt in the warmer air and the runoff flows downstream toward communities along riverbanks, is flooding.
The second is the depletion of the huge water supply locked away in their ice.
Gholam Murtaza, the Islamabad Bureau Chief of local national TV channel BOL says, "When glaciers melt, diseases like viral, bacteria, fossils and other biological species also wake up which could be harmful for humans."
"Are we (also) ready to deal with new type of things, syndromes, diseases etc. also. Has this angle been explored?".
There will be recovery from the pandemic, but at terrible cost, according to Noam Chomsky. (However) "there will be no recovery from the melting of the polar ice sheets, the glaciers in the Himalayas or the other dire effects of global warming – which will render much of South Asia uninhabitable if the world persists on its present course, and not in the distant future," warns Chomsky.
According to a Pakistani analyst, "Unfortunately we will not be able to do much about it; it needs global effort or at least a regional approach."
That's not good news for the region and for the Himalayas where a quarter of humanity are dependent on these glaciers.
Scientific communities have agreed global warming is causing glacial retreat. "These glaciers are retreating and telling you that climate change is real, global warming is real." A quarter of humanity are dependent on these glaciers", Bashir said.
The most recent scientific studies estimate that the world may be reaching that level (of global warming) in 50 years, "on our present course", Chomsky has warned.
How the region can participate in changing the course?
"China, India and Pakistan will have to come together, but the current and emerging situation makes chances of such a cooperation remote," says the analyst.
Independent observers, experts and professionals we talked to, say the three nations have to heed to Chomsky's warnings and come together on it from civilizational standpoint. A more severe Pandemic than COVID-19 has been predicted: Glaciers turning pink in Italy may be the optic we all need to shake us up from our cold perceptions to the hot reality of climate change.
Second of a series of reports on melting glaciers challenges for Pakistan
Melting Glaciers of the 3rd Pole
Another, Probably More Severe Pandemic Has Been Predicted: Noam Chomsky
Perfect Timing? China Helps Pakistan Build…
Chanakya Twist to Pink Himalayan Salt: Mined in…
MBS, Khan Say Hello, Discuss Green White and Green
People With Better Education Live Longer in Same…
Conversation in a WhatsApp Forum: Bangladesh,…
U.S. is in Middle East 'to Protect Israel' Not For…
DNA Inherited From Neanderthals May Increase Risk of Covid-19


May I take a nap now? It's been busy around here. Real busy.
First there was a colonoscopy last Tuesday, then cooking like mad beginning Wednesday, followed closely by Thanksgiving and Hanukkah all at the same time.
My granddaughter and her boyfriend were in from college and grandson dearest was hanging around annoying his little sister, with their mother — my daughter — looking on and giggling (thanks a lot). Did I mention the children's friends coming and going?
Did you know college students really do bring their dirty laundry home?
My computer decided to glitch at the worst possible time: Right when I needed my recipes, which are stored on my computer. Thank goodness for friends! One needed her recipe for chicken bulgogi, which I had in an old-fashioned recipe box. (It's a Korean dish.) She also had my recipe for pumpkin cake. So we traded. Sometimes it's good to share your recipes.
A couple of other recipes I needed were printed out and the rest of the dishes could be made from memory. Good thing, because the computer wasn't fixed until yesterday evening.
The December bulletin I do for the synagogue isn't finished because absolutely everybody missed their deadline. By the time I had the content, I was out of town for a colonoscopy, then cooking and celebrating Thanksgivukkah and enjoying the children's company.
This afternoon I get to go to the ophthalmologist — and I'm way behind on that. He broke his ankle, then I was in the hospital in another state, then I had a cold the day of my next appointment, complete with watery eyes. And nobody wanted my germs, anyway. The last time my eyes were checked, I had a tiny spot of background retinopathy in each eye. I hope that hasn't gotten worse.
Finals week is upon us, so my husband is busy grading papers and preparing exams and isn't around much. He is not going to be happy to take a break for a social function one evening next week. I'd be happy to skip it myself, except I'm in charge.
Where is all this leading? My blood glucose, which was running low last week, has now reversed itself and is running high. Yes, there were two holidays and the resultant traditional foods, but I don't think it's all food. I do enjoy myself on a holiday, but I'm also very good at getting back on track the next day.
However, if food were the only thing that affected our blood glucose, diabetes wouldn't be any "fun" at all! Nope. Stress can also whack out your BGs.
Case in point: I'm afraid of heights and there is this toll road around Chicago called the Skyway. I do everything I can to avoid the Skyway. In Indiana (I think — it's around the border between Indiana and Illinois), it begins with a climb up a bridge that does look as if you're going into the sky. I know this because I accidentally got on the Skyway once. When I got a chance to check my glucose, it was in the 300's.
I'm talking physical stress, such as an illness, and emotional stress — both good and bad — such as getting married or, at the other end of the spectrum, divorced.
What happens when you get stressed is that the "fight or flight" hormones such as epinephrine and cortisol kick in to raise your blood glucose to give you the energy you need at that time.
Longer-lasting stressors, such as having the flu or getting a divorce — or perhaps one thing after another that just keeps going and going and…well, just like the Energizer Bunny — wear you down and can wreak havoc with glucose levels.
Some have a tendency to stress eat, and I mean doughnuts and potato chips; not celery and reduced-fat cottage cheese. That doesn't help. Or they forgo physical activity because of "no time," or they're just too stressed to get up and moving.
Speaking of physical activity, it's a good way to get your blood glucose down. Check your BGs before and after taking a walk or roller skating or dancing or something and you'll see for yourself.
The lesson? Next time your glucose goes high, don't just blame it on the food, especially if you've been "good" in that department. Look around and see what else is going on in your life.
Oh, another stress buster — for me, at least — is to write it out. Now that I feel calmer, and my glucose is not all that bad…I think I'll take a nap. After all, the ophthalmologist should have clear baby blues to look into, not the sleep-deprived ones I have now.


text:
Q: How to align Logo Image and Menu items in one line without using position absolute? Background: The current design that I am working on, requires a logo followed by menu in one line. What I want is that the logo and the menu (which consist of two parts which are two unordered list here) to appear in one row or as one section.
Parts of the header: The header consist of Logo Image, menu left and menu right. The menu left and menu right are unordered list items in code.
Problem faced: I am trying to add the logo image and menus in one div element to bring them together in one line. But the image is appearing on top and followed by the menu element. I have tried using the display:inline to bring the logo image on one side and menu starting after the logo image but it didnot work.I am sharing my code. Can the image be positioned and aligned along with the menu items without using position absolute? There is only HTML5 and CSS3 in the code. There is no javascript in the code.

.header{
 position:relative;
 margin:0;
 padding:0;
}

.top{
 height:20%;
 border:1px solid #000;
}

.logo{
 height:100%;
 display:inline;
}
.element{
 clear:both;
 border:1px solid #000;
 float:left;
}
.leftc{
 display:inline;
}
.leftc li{
 list-style:none;
 display:inline-block;
}

.rightc{
 display:inline;
}
.rightc li{
 list-style:none;
 display:inline-block;

}

  <title>Home</title>
  <link rel="stylesheet" type="text/css" href="style.css"/>

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A: Explanation in comments of CSS where the changes are made.

*

*Give height to header

*Making height of top to 100% as it to cover the whole header height can be adjusted accordingly.

*Properties to be added to div with class log
.top>.log{
height: 100%;
margin: 0;
width: 200px;
display: inline;
float: left;
}

.header{
 position:relative;
 margin:0;
 padding:0;
    height:30vh;/*giving height to header*/
}

.top{
 height:100%;/*making it to cover complete header*/
 border:1px solid #000;
}

.logo{
 height:100%;
 display:inline;

}
/* properties to be added to log div*/
.top>.log{
    height: 100%;
    margin: 0;
    width: 200px;
    display: inline;
    float: left;
}
.element{
    /*removing the clear property*/
 border:1px solid #000;
 float:left;
}
.leftc{
 display:inline;
}
.leftc li{
 list-style:none;
 display:inline-block;
}

.rightc{
 display:inline;
}
.rightc li{
 list-style:none;
 display:inline-block;

}

  <title>Home</title>
  <link rel="stylesheet" type="text/css" href="style.css"/>

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text:
Struggle or Starve
Stories of everyday heroism between the wars
Edited by Carol White, Sian Rhiannon Williams
'A delightful book...moving, poignant, funny and a very good historical record' Emeritus Professor Deirdre Beddoe, University of Glamorgan
Anthology, Autobiography & Memoir, Non-Fiction
Struggle or Starve quantity
HONNO VOICES – bringing the past to life
The past is different country, and there were far fewer options open for women then… Olive was never meek; she was dauntless. She took the cards that life dealt out to her – on the whole it was a pretty lousy hand – and played it by the rules and made the very best of it.
Olive left school to run the household for her alcoholic mother – making sure that there was food on the table and that she extracted the housekeeping from her father at the mine gates before it hit the bookie's hand. And Olive was typical of the indomitable women whose voices come alive in these pages…
In order to keep food on her mother's table Ivy James scoured the spoil tips of the local colliery for small lumps of coal to sell – this back-breaking task was paid so poorly that Ivy took on night work as a cleaner and chucker out at nearby hotels. Hard physical and demeaning work for anyone, never mind a woman. Others took in laundry, had broods of children only to lose their lives before their children were raised. Each of their stories brings an insight into their struggle for dignity, recognition and wider opportunities.
First published in 1998, with new editions in 2002 and 2009.
"[These accounts] should surely have a wider audience… those fascinated by working class literature, politics and autobiography"
Professor Angela V. John, University of Greenwich
"fascinating, transporting and bound to capture the imagination of anyone with even a passing interest in social history."
Molly James, New Welsh Review
"A marvellous compilation of reminiscences by Welsh women" Time Out
"This is a book to treasure" The Big Issue Cymru
"There are some very sad stories here... However, there are also many happy and amusing memories recounted... I laughed and cried quite often at different points in my reading."
Luci Davin, www.thebookbag.co.uk
Categories: Anthology, Autobiography & Memoir, Non-Fiction
First Published by Honno: 1st January 1998


text:
LeagueHome
My Draft Rankings
Detail: Ryan Goins
2B/SS
2nd at SS, 1st at 2B
R Roto Transactions
Roto Status
Pro Status
RBOE
2019 White Sox 52 144 36 6 1 2 0 13 10 0 1 17 1 1 0 .250 .303
2018 Royals 41 115 26 8 1 0 0 10 6 0 0 4 0 0 0 .226 .250
2017 Blue Jays 143 418 99 21 1 9 4 37 62 3 2 31 0 0 5 .237 .291
2016 Blue Jays 76 183 34 9 2 3 0 13 12 1 1 9 0 1 0 .186 .234
2013 Blue Jays 34 119 30 5 0 2 0 11 8 0 0 2 0 0 0 .252 .267
Total MLB 541 1536 353 71 12 22 7 150 158 6 6 107 1 3 11 .230 .276
Consensus 116 346 87 18 2 7 0 38 32 5 2 22 1 2 2 .252 .294
MORPS 123 438 116 22 2 10 0 52 44 7 3 28 0 2 2 .265 .310
Marcel 0 242 63 0 0 6 0 28 25 2 1 16 1 1 1 .260 .288
RV Current 144 409 96 18 3 6 0 44 38 5 2 22 1 2 2 .235 .271
RV Pre-Australia 127 429 110 21 1 9 0 49 40 6 2 26 2 3 2 .256 .296
RotoValue 143 392 95 17 2 7 0 42 31 6 2 26 1 2 2 .242 .284
Steamer 81 311 74 16 1 4 0 32 29 4 3 17 0 1 3 .238 .269
2014 Game Log
2014-09-28 v Bal 2B 3 1 0 1 0 0 0 0 0 0 0 0 0 0 .333 .533
2014-09-25 v Sea SS 4 0 0 0 0 0 0 1 0 0 0 0 0 0 .000 .000
2014-09-24 v Sea 2B 3 2 0 0 0 0 0 1 0 0 0 0 0 0 .667 .600
2014-09-23 v Sea 2B,SS 3 0 0 0 0 0 0 1 0 0 0 0 0 1 .000 .000
2014-09-21 @ NYY 2B 3 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-09-17 @ Bal 2B 4 1 1 0 0 0 0 0 0 0 0 0 0 0 .250 .312
2014-09-14 v TB 2B 2 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-09-10 v ChC 2B,SS 4 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-09-09 v ChC 2B 2 1 1 0 0 0 1 0 0 0 0 0 0 0 .500 .625
2014-09-07 @ Bos 2B 3 0 0 0 0 0 0 0 0 0 1 0 0 0 .000 .175
2014-09-05 @ Bos 2B 1 1 1 0 0 0 0 0 0 0 0 0 0 0 1.000 1.250
2014-09-04 @ TB 2B 2 1 0 0 0 0 0 0 0 0 0 0 0 0 .500 .450
2014-09-02 @ TB 2B 1 1 0 1 0 0 1 1 0 0 0 0 0 0 1.000 1.600
Total 59 10 3 2 0 0 4 4 0 0 1 0 0 1 .169 .199
2014-08-10 v Det 2B 2 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-08-03 @ Hou 2B 3 1 0 0 0 0 0 1 0 0 0 0 0 0 .333 .300
2014-08-01 @ Hou 2B 3 0 0 0 0 0 0 0 0 1 0 0 0 0 .000 -0.167
Total 25 3 0 0 0 0 2 2 0 1 0 0 0 0 .120 .088
2014-07-28 @ Bos 2B,SS 5 4 1 0 0 0 1 4 0 0 0 0 0 0 .800 .790
2014-07-27 @ NYY PH,2B 1 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-07-24 v Bos 2B 4 2 1 0 0 0 1 1 0 0 0 0 0 0 .500 .537
2014-04-20 @ Cle 2B 2 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-04-18 @ Cle SS,2B 3 1 0 0 0 0 1 0 0 0 1 0 0 0 .333 .400
2014-04-17 @ Min 2B 3 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-04-17 @ Min SS 3 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-04-13 @ Bal SS,2B 4 1 0 0 0 0 1 0 0 0 1 0 0 0 .250 .320
2014-04-12 @ Bal SS 2 1 0 0 0 0 0 0 0 0 0 0 0 0 .500 .450
2014-04-10 v Hou SS 0 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-04-09 v Hou SS,2B 3 0 0 0 0 0 0 0 0 0 1 0 0 0 .000 .175
2014-04-08 v Hou 2B 0 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-04-06 v NYY PR 0 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-04-05 v NYY 2B 3 1 1 0 0 0 0 0 0 0 0 0 0 0 .333 .417
2014-04-03 @ TB SS 3 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
2014-04-02 @ TB PR,SS 1 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
Total 3 0 0 0 0 0 0 0 0 0 0 0 0 0 .000 .000
Mon Sep 16 2019 22:37 Chicago White Sox-Minnesota Runs
Sun Sep 15 2019 01:07 Chicago White Sox-Seattle Runs
Sun Sep 8 2019 17:51 L.A. Angels-Chicago White Sox Runs
Sat Sep 7 2019 23:01 L.A. Angels-Chicago White Sox Runs
Thu Sep 5 2019 16:00 Chicago White Sox-Cleveland Runs
Wed Sep 4 2019 23:30 Chicago White Sox-Cleveland Runs
Fri Aug 30 2019 23:17 Chicago White Sox-Atlanta Runs
Thu Aug 29 2019 17:47 Minnesota-Chicago White Sox Runs
Wed Aug 28 2019 23:42 Minnesota-Chicago White Sox Runs
Tue Aug 27 2019 23:42 Minnesota-Chicago White Sox Runs
Sun Aug 25 2019 17:42 Texas-Chicago White Sox Runs
Wed Aug 21 2019 16:46 Chicago White Sox-Minnesota Runs
Tue Aug 20 2019 23:25 Chicago White Sox-Minnesota Runs
Mon Aug 19 2019 23:55 Chicago White Sox-Minnesota Runs
Sun Aug 18 2019 02:04 'Savages' again: Yankees erupt at umpires, beat Indians 6-5
Sun Aug 18 2019 01:04 Chicago White Sox-L.A. Angels Runs
Sat Aug 17 2019 01:33 Chicago White Sox-L.A. Angels Runs
Fri Aug 16 2019 01:13 Chicago White Sox-L.A. Angels Runs
Mon Aug 5 2019 00:58 Bieber pitches Indians past Angels 6-2
Sun Aug 4 2019 20:48 Bieber pitches Indians past Angels 6-2
2014 Games by Position
MORPS projections courtesty of Tim Oberschlake.
The Marcel projections model was developed by Tom Tango.
Steamer projections courtesy of Jared Cross, Dash Davidson, and Peter Rosenbloom.
Depth charts are provided by Sportradar US.
Fielding Independent Pitching or FIP, was developed by Tom Tango based on DIPS work by Voros McCracken.
Weighted On Base Average, or wOBA, was developed by Tom Tango. RotoValue uses Version 2, but ignoring reached base on error before 2015.


Sailing World was looking for some sailing writing. So I helped.
Below is a story I relayed about the time I bought a sailboat and tried to race her for a summer on Narragansett Bay, Obviously, it did not go as well as I had dreamed, but it was a pretty fun time nonetheless. This wonderful artwork is provided by Carlo Giambarresi and Morgan Gaynin, and the original post on Sailing World can be found here.
Previous Post Day 19: Virginia.


Common Reasons Why People Immigrate to the U.S.
The movement of people from one geographic region to another is something that has been occurring for over thousands of years. Over the last five decades, immigration to the United States has been a hot-button political issue. To begin to understand immigration in this country, immigration lawyers in San Diego look at the underlying reasons. The following figures are from surveys conducted among undocumented immigrants by the National Association of Latino Elected and Appointed Officials in 2013.
When it comes to working and earning wages for families back home, more women risk becoming undocumented immigrants than men. Specifically, 41 percent of women and 37 percent of men. What is surprising is that more than half of these immigrants enter the U.S. legally, which means the majority were approved for some sort of visa and they then decided to overstay.
Many immigrants arrive in the U.S. seeking a better life for their families. To this effect, 40 percent of men indicate this reason compared to 36 percent of women. Quite a few of these families fail to find a decent quality of life as undocumented migrants and return to their countries after a few years.
Immigrants who arrive in the U.S. seeking economic opportunity and quality of life may attempt to bring over relatives. Family reunification is cited as a main immigration reason by 16 percent of women and eight percent of men.
Many immigrants fleeing oppressive regimes or political persecution are able to file for asylum or refugee status in the U.S. Nonetheless, five percent of undocumented men and three percent of men cite such reasons. The U.S. War on Drugs has thoroughly destabilized entire communities in El Salvador, Honduras and Guatemala, which means that this reason is bound to increase in the near future.
Learn more about U.S.A immigrations services in San Diego by calling KS Visa Law at (858) 874-0711 today. We will answer any questions you have about the immigration process and help you with your visa or citizenship.


text:
<?php
class Template_Validate_LoginExists extends Zend_Validate_Abstract
{
	const LOGIN_NOTVALID = 'msgLogin';

	protected $_messageTemplates = array(
        self::LOGIN_NOTVALID => "Login is not exsist",
    );

	public function isValid($value, $context = null)
	{
		if($this->loginExists($value)) {
			$this->_error(self::LOGIN_NOTVALID);
			return true;

		} else {
			return false;
		}
	}

	private function loginExists($value)
	{
		$userMapper = new Template_Mapper_My_UserMapper();
		return($userMapper->loginExists($value));
	}

}

redpajama_set_name: RedPajamaGithub


Kim thinks her kid's teacher needs to stop telling kids they're smart or handsome when they're 'stupid and fat'.
UPDATE: We had so many comments this weekend about Kim, and now everyone's on her side!
Buckle up, y'all. This one's a doozy. Kim doesn't know how to handle her 3rd grader's teacher when he's coddling the students so much. She says a 'fattie' shouldn't be called handsome and a stupid kid shouldn't be told he's 'well-spoken'.


Arts Committee shall consist of 12 members appointed by the Board of Aldermen.
Recommend to the Board of Aldermen a policy for the selection of art to be displayed in or on public facilities.
Coordinate arrangements to have art work displayed on town property, in conjunction with the staff of the Economic & Community Development Department.
Work collaboratively with the community on projects related to the arts. Members of the Arts Committee may assist in community fund raising activities, but any funds received by the committee or by members acting in their official capacity must be accounted for, and all expenditures by the committee must be made, in accordance with the provisions of the Local Government Budget and Fiscal Control Act. Such funds may be received and expended by the town only for purposes for which the town is statutorily authorized to expend public funds.
Report to the Board as requested and perform such additional duties as the Board may request from time to time.


Hospice provides emotional, medical, and spiritual support for people who are terminally ill. The goal is to improve patient's quality of life by making them comfortable in their homes. The focus of care is on symptom management rather than curing the disease. Hospice helps families cope with the care of their ill family member and acts as an advocate during this time of illness.
86% of the American public would choose to die at home.
9 out of 10 patients with a life limiting diagnosis would benefit from a consultation with Hospice to compliment the care of their own physician.
They want control over the many decisions regarding their care.
Their Doctor certifies that they have an estimated life expectancy of six months or less if their disease follows a typical course.
They want to be at home or in a home like setting and reside within our service area.
They are no longer willing or able to pursue curative measures and have shifted their concern to the quality of the time remaining.
They want to be able to come off hospice if their situation improves.
They are now concerned with remaining as alert, pain free and comfortable as possible.
They are concerned that family be supported through their final months and beyond.
Ask your physician or call Hospice of South Texas. You can assist your physician by indicating that you want to be told when and if it is time to consider hospice. You can also tell your physician that you are ready to consider hospice. Call us early. We will help you find the best resources to meet your specific needs.
Hospice is available to all terminally ill individuals and their families. Although the majority of patients admitted into hospice have conditions related to cancer, other frequent admission diagnoses include heart and lung disease, AIDS, and neurological disorders such as Alzheimer's disease and ALS.
Terminal prognosis, certified by attending physician and hospice medical director, of six (6) months or less if disease takes its normal course. Patient no longer seeking curative treatment; has chosen palliative care.
Who can refer patients for admission to Hospice of South Texas?
How does hospice care differ from other kinds of care?
Patients and families are a vital part of the decision-making process. Hospice of South Texas works to help patients maintain control of their lives whenever possible, enabling people with a terminal illness to make decisions about how and where they want to spend the rest of their lives.
We offer palliative care rather than curative treatment, using sophisticated methods of pain and symptom control that allow the patient to live as fully and as comfortably as possible.
We emphasize quality, rather than the length of life. Hospice helps the patient to make the most of each hour and each day of remaining life by providing comfort and relief from pain and suffering.
We care for the person, not the disease, focusing on the care, not a cure. The hospice movement stresses human values that go beyond the physical needs of the patient. Physical, mental, emotional, financial, and spiritual issues are addressed. We affirm life and regard dying as a normal process.
We provide on-call availability, 24 hours a day, allowing the patient and family to reach the nurse at any time, night or day, seven days a week.
Hospice provides continuing contact and support for family and friends for a year following the death of a loved one.
Does Hospice of South Texas serve only cancer patients?
Definitely not! Hospice of South Texas provides care to patients of all ages with life-limiting illnesses such as end stage Alzheimer's, congestive heart failure, renal failure, multiple sclerosis, COPD(chronic obstructive pulmonary disease), AIDS, cancers, Parkinson's disease, ALS (Lou Gehrig's disease), muscular dystrophy, cystic fibrosis, and more.
If I accept hospice care, won't I be giving up hope?
Hospice of South Texas focuses on the hope that every day will be the best it can be rather than that the disease will be cured. The goal of hospice is to ease the suffering of the terminally ill and their families and friends.
Can I come off of hospice services if I change my mind and want curative treatment again?
Yes. Hospice care will be discontinued when patient and family, physician and/or members of the hospice team decide hospice is no longer necessary or appropriate. When hospice initiates the discharge, every effort is made to resolve any problems and assist in referring the patient to appropriate facilities or agencies. Patients are always welcome back on hospice services when the time comes and when appropriate.
Will Hospice help end life to relieve suffering?
No. Hospice does not support euthanasia. The Hospice staff supports people through the natural dying process.
Must I sign a DNR order (Do Not Resuscitate) before I can have hospice services?
No. Do Not Resuscitate (DNR) Orders are encouraged but are not required.
Does this mean there is nothing more we can do?
We cannot add days to your life but we can add life to your days.
Hospice is not the end of treatment – it is a shift to intensive palliative care that focuses on allowing the patient to live his or her life to the fullest.
In Hospice, there is always more that can be done to relieve the pain and suffering, and bring comfort to the patient and family.
Can I keep my doctor when I come on hospice?
Yes. Hospice encourages participation by the patient's attending physician in coordinating the medical care of the patient through ongoing communications between the attending physician and the Hospice of South Texas team.
How does Hospice of South Texas receive reimbursement for its services?
Since 1985, community support from grants, memorials, gifts, contributions, and volunteers have made it possible for Hospice of South Texas to provide full services to patients and families regardless of ability to pay.
What are other sources of financial support?
An important source of financial support for Hospice of South Texas are donations from friends and loved ones of former patients. Another is the community support of fundraising activities. Churches, clubs and businesses contribute as well to help Hospice of South Texas provide non-funded services. Donations are made to: Hospice of South Texas, 605 E. Locust Ave., Victoria, Texas 77901.
Where do patients live while they are receiving hospice services?
Patients can receive care wherever they live: at home, in the home of relatives or friends, in nursing homes, in group living facilities.
What about continuous care in the home?
Hospice may provide continuous care in the home for patients who cannot be moved, but whose acute symptoms require intense monitoring or treatment. Continuous care shall be determined by the interdisciplinary group and needs approval by the Executive Director or Director of Nursing due to staff availability. The care must be ordered by the physician and for shot time frames only/and until the crisis is resolved.
What about medicines and medical equipment?
The hospice patient also receives all prescribed medications for his life-threatening illness and for the alleviation of distressing symptoms. These drugs are paid for by Hospice of South Texas at 100%.
Exceptions: Medications for prior conditions other than the life-threatening illness (e.g. prior heart disease, diabetes, hypertension, etc.) and over the counter medications are not covered.
How can I help Hospice of South Texas?
Make sure that friends, neighbors, and family members are aware of their right to this special kind of care.
Send a letter to your physician thanking him/her for allowing Hospice of South Texas to care for your loved one.
How can I give a gift to Hospice of South Texas?


Uniglaze India Pvt. Ltd. (UIPL) is not only economic entity that is in pursuit of profit through air competition, but aspire to become Social, Ethical, Environmental Credible and Trusted entity.
We have established a set of principles to be carried out by each individual or entities affiliated to the group.
Will always offer product and services by seeing Customer's point of view.


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/ First published in Tom Dispatch, 22 December 2015.
"What did you do at school today, Seamus?" It's a question I ask him everyday.
"Well," my proud preschooler begins, "we did not have a lockdown drill today." And that's about as far as he gets in the art of storytelling. Sometimes I'll get something about "bim" (gym) or how "Bambi" (Jeremy) pinched him during free play. But the thing that preoccupies my precocious three year old every single day he goes to school is the lockdown drill he and his classmates had in their first month of school.
"Does he talk about it at home?" she asks.
She informs us that the drills happen about once a month, and that Seamus remains easily startled long after they're over, running for shelter between an adult's legs whenever he hears loud noises in the classroom.
At that moment — not exactly one of my proudest — I burst into tears. I just couldn't square my son's loving exuberance and confidence in the people around him with the sheer, teeth-hurting terror of children being stalked by an armed killer through the halls of The Friendship School. How, after all, do you practice for the unthinkable? This is a subject that's been on my mind since I was hardly older than he is now. I look over at him playing contently with his sisters, Madeline, almost two, and Rosena, almost nine, so proud to share his classroom with them.
"At home," I tell the teacher through my tears, "we chant 'Gun Control, Not Lockdown Drills!' whenever he talks about them." And then I add, "It makes me so angry that he and his friends have to go through this trauma and the big men get to keep their right to bear assault weapons. He should be scared of lockdown drills. They sound terrible. He shouldn't have to practice surviving a mass killing episode at one of his favorite places in the whole wide world." I wipe my tears away, but they just keep coming.
Our kids ask us all sorts of questions. Why? Why? Why? They are tiny existentialists. Why is the sky blue? Why do people die? Why does grass grow? They regularly demand that we explain the world to them. Good luck!
His teacher is so earnest and so young and I feel so brittle and so extreme as I cry, folded into one of the small seats at a quarter-sized table in her cheerful classroom. "I am sorry," I finally say.
I wonder, of course. I know that so much of this is based on fears — not quite irrational but blown out of all proportion — that have been woven into our American world. My husband reminds me of how his parents' generation had to practice surviving a nuclear attack by doing "duck and cover" drills under their desks. I was too young to duck and cover, but my parents were ardent anti-nuclear activists with no inhibitions about describing to a child just what such a war would mean so I learned to be terrified of nuclear war at a very young age.
I came to believe that the only thing keeping Soviet and American intercontinental ballistic missiles from decimating our cities was the activism, organizing, and witness of my parents and their small band of friends and fellow travelers. We would stand in front of the Pentagon — this was in the late 1970s and throughout the 1980s — holding up signs with slogans like "You can't hide from a nuclear bomb" and the old symbol for a fallout shelter printed below it. I was taught that there could be no security, no safety in a world full of nuclear weapons, that the only way to be safe was to get rid of them.
Imagine how I feel all these years later in a world still chock-a-block full of such weapons. These days, I wonder why the fear of them has disappeared, while the weaponry remains. Is that better or worse for Seamus's generation? And what about our present set of fears? What about our twenty-first-century whys?
Assuming there are more Adam Lanzas out there (and there obviously are), that more gun shops will sell ever more implements of rapid-fire death and destruction, and that more gun lobbyists and promoters will continue to cling to this "God-given, constitutionally enshrined right," my son does need to endure more lockdown drills.
The consensus of school security experts is certainly that the massacre at Sandy Hook Elementary School in Newtown, Connecticut (only 80 miles from our house), would have been much worse if the students and teachers hadn't been practicing for exactly the nightmare scenario that struck on December 14, 2012.
But how can I explain any of this to my little boy when it makes no sense to me? When it makes no sense, period?
Why? Why? Why? As a kid, I got an earful every time I asked that question. My parents were comfortable exposing my brother, sister, and me to the horrors of our world. In first or second grade, my activist parents involved me in a UNICEF slide show about world hunger. We would go to churches and schools where I would recite the script, full of sad (and still, sadly, largely on the mark) statistics about how children throughout the world suffer from malnutrition. I could tell you why kids were hungry all over the world, since my mom had tacked on a conclusion to the slide show that lay the blame squarely on the U.S. military-industrial complex.
So instead of Knight Rider or The Cosby Show, we watched black and white documentaries about Hiroshima and Nagasaki projected onto our living room wall. I couldn't tell you about the latest plot twists on Full House, but I could tell you why nuclear weapons were wrong. Those grainy images of destroyed cities, burnt skin, and scarred faces were etched into my young brain by the age of five. My heroines were two young anti-nuclear activists. Sadako Sasaki was a Japanese girl who contracted leukemia after the atomic bombing. She folded hundreds of paper cranes as a prayer for healing and peace before dying at the age of 12. Samantha Smith, a young girl from Maine, wrote to Soviet leader Yuri Andropov with a plea for peace. He, in turn, invited her to tour the Soviet Union where she connected deeply with young Russians. She died in a plane crash at the age of 13.
I wonder now about my childhood fears. They helped me support and believe in the anti-nuclear work of my parents. But nightmares, morbid fascinations with young martyrs, a fixation on the tick-tockings of the Bulletin of the Atomic Scientists' Doomsday Clock — these are not things that I want to pass on to the next generation. I guess I'm happy that they don't know what nuclear weapons are (yet) and it's one more thing I'm not looking forward to explaining to them.
The questions are already coming fast and furious these days and they are only going to multiply. We have to try — I have to try — to answer them as best we — I — can. It's a precious facet of parenting, the opportunity to explain, educate about, and even expound upon the wonders and horrors of this world of ours, and it's a heavy responsibility. Who wants to explain the hard stuff? But if we don't, others surely will. In these early years, our kids turn to us first, but if we can't or won't answer their questions, how long will they keep asking them?
Why do we practice lockdown drills? Why do people kill kids? Why is there war? Why are all those weapons, the nuclear ones and the assault rifles alike, still here?
At some preschools, it's protocol to explain lockdown drills in terms of preparing in case a stinky skunk gets into the building. No one wants to get sprayed by a stinky skunk, do they?
Somehow, and I can't tell you quite why, this seems to me almost worse than the truth. At Seamus' school, they don't talk explicitly about an armed intruder, but they do make a distinction between fire drills where they evacuate the building and "keeping safe from a threat" by "hiding" in it.
In the month since our parent-teacher meeting, Seamus has endured another lockdown drill and our country has continued to experience mass shooting events — San Bernardino and Colorado Springs being just the most horrific. While at breakfast, Patrick and I read the news about healthcare offices and social service agencies turned into abattoirs, and yet we speak about such things only in code over granola and yogurt. It's as if we have an unspoken agreement not to delve into this epidemic of gun violence and mass shootings with our kids.
Still, it's strange not to talk about this one subject when we talk openly in front of our children about so much else: Iraq and Afghanistan, the Syrian refugee crisis, hunger and homelessness, Guantánamo and climate change. We usually welcome their whys and jump over each other to explain. Patrick is much better at talking in a way that they can all take in. I forget myself easily and slip into lecture mode (next slide, please).
After the police killings of Lashano Gilbert (tased to death in our town of New London, Connecticut), Eric Garner, Michael Brown, and Freddie Gray, we took the kids to candlelight vigils and demonstrations, doing our best to answer all Seamus's questions. "Why do the police kill people?" followed, of course, by "Are they going to kill me?" Then we somehow had to explain white privilege to a three year old and how the very things that we encouraged in him — curiosity, openness, questioning authority — were the things black parents were forced to discourage in their sons to keep them from getting killed by police.
And then, of course, came the next inevitable "Why?" (the same one I'm sure we'll hear for years to come). And soon enough, we were trying desperately to untangle ourselves from the essentially unintelligible — for such a young child certainly, but possibly the rest of us as well — when it came to the legacy of slavery and racism and state violence in explaining to our little white boy why he doesn't need to cry every time he sees a police officer.
And then, of course, there's the anxiety I have about how he'll take in any of this and how he might talk about it in his racially diverse classroom — the ridiculous game of "telephone" that he could play with all the new words and fragments of concepts rattling around in his brain.
My stepdaughter Rosena was a kindergartner when Adam Lanza killed those 20 little kids and six adults in their school just 80 miles west of us. Her school upped its security protocols, instituted regular drills, and provided parents and caregivers with resources on how to talk to their children about what happened. For five and six year olds, they advised not initiating such a conversation, nor allowing them to watch TV or listen to the radio news about the massacre. (Not exactly the easiest thing in our 24/7 media moment.) They also suggested responding to questions only in the most general terms. Basically, we were to sit tight and hope our kids didn't get enough information to formulate a why.
Good luck on that these days, but sometimes I do wish the same for myself. No news, sit tight, and pretend nothing's going on. After all, like so many of our present American fears, the fear that my kids are going to be gunned down in their classrooms is pretty irrational, right? Such school shootings don't exactly happen often. Just because one did occur relatively near here three years ago doesn't mean pre-schools and elementary schools are systematically under attack, yes?
Unlike so many people on this planet, we don't live in a war zone (if you put aside the global destructiveness of nuclear weapons). And given the yearly figures on death-by-vehicle in this country, my kids are unbelievably safer in school, any school, than they are in the back seat of my own car any day of the week, right?
Of course, there's another problem lurking here and it's mine. I'm not there. My three-year-old son is having scary experiences and I'm not there to walk him through them. And then there are those lockdown drills and what they are preparing him for. They couldn't be creepier. They're a reminder not just to our children but to their parents that, after a fashion, we may indeed be living in a kind of war zone. In 2013, according to the Centers for Disease Control, 33,636 people were killed by guns in this country; in that same year, 127 American soldiers were killed in Afghanistan.
Why is the sky blue? I have no idea, but it takes only a minute of Googling to find out that it has something to do with the way air molecules scatter more blue light than red light. Why do people die? Because no one can live forever, because they get sick and their bodies get old and their organs don't work any more and then we cry because we miss them and love them, but they live on, at least until our own memories go. Why does grass grow? Well, Google it yourself.
The problem, however, is with the most human of questions, the ones that defy Googling and good sense — or any sense we may have of the goodness of humanity. And maybe, kids, we just have to wrestle together with those as best we can in this truly confusing world.
And keep one thing in mind: the very same litany of questions our kids never stop asking and that we struggle to answer, or wonder whether to answer at all, is always running like some strange song through our own adult heads as well, largely unanswered.
Why this particular world? Why this particular way? Why now?


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