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\begin{document}
\begin{center}
 { \bfseries
Math 123 \emph{OPTIONAL} Homework Assignment on Nets} \\
Not not turn in.
\end{center}


  \begin{ques}
    Suppose that $X$ is a first countable space.  Show that each
$x\in X$ has a neighborhood basis of open sets $\{\,U_n\,\}_{n=1}^\infty$
such that $U_{n+1}\subseteq U_n$.
  \end{ques}

  \begin{ques}
    Let $X$ and $Y$ be $1^{\hbox{st}}$ countable spaces.
    \begin{enumerate}
\item[(a) ] Show that ${\cal O} \subseteq X$ is open in $X$
if and only if every {\it sequence\/} converging to some
$x\in {\cal O}$ is eventually in ${\cal O}$.
\item[(b) ] Show that $F\subseteq X$ is closed if and only if 
every convergent {\it sequence\/} in $F$
converges to a point in $F$.
\item[(c) ] Show that $f:X\to Y$ is continuous if and only if
whenever $\xn$ coverges to $x\in X$ then $\{f(x_n)\}_{n=1}^\infty$
converges to $f(x)\in Y$.
\item[(d)$^{\displaystyle*}$] Let
$\xn$ be a sequence in $X$.
Show that if $\xn$ has a convergent {\it subnet\/},
then $\xn$ has a convergent {\it subsequence}. (Hint: if $\xn$ has an
accumulation point, then it must have a convergent subsequence.)
\item[(e)$^{\displaystyle*}$] Show that $1^{\hbox{st}}$
countability is required in part (d).
(Hint: let $\ell^\infty$ denote the set of {\it bounded\/}
sequences.
If $\alpha=\{\alpha_n\}_{n=1}^\infty\in\ell^\infty$,
let $I_\alpha$ be any closed bounded interval in {\bf R}
such that $\alpha_n\in I_\alpha$ for all $n$.
Set 
$$Z=\prod_{\alpha\in\ell^\infty}I_\alpha .$$
Consider the sequence $\xn$ in the compact space $Z$
defined by $x_n(\alpha)=\alpha_n$.)
\end{enumerate}
  \end{ques}

  \begin{ans}
    \noindent(a)\enspace ($\Longleftarrow$) This is easy.

\noindent($\Longrightarrow$) Just as in lecture,
if ${\cal O}$ fails to be open, then there
is a $x\in {\cal O}$ such that
every 
\def\nh{neighborhood}
\nh\ of $x$ intersects ${\cal O}^c$.
Choose a \nh\ basis at $x$ of the form
$\{U_n\}_{n=1}^\infty$ with $U_{n+1}\subseteq U_n$.
Now choose $x_n\in U_n\cap {\cal O}^c$.
We have $x_n\to x$ while $\xn\subseteq {\cal O}^c$,
a contradiction.

\noindent(d)\enspace Let $\{x_{n_\beta}\}_{\beta\in B}$ be
a subnet converging to $x\in X$.
Let $\{U_n\}$ be a \nh\ basis at $x$ as above.

We will define a subsequence inductively.
Since $\{x_{n_\beta}\}$ is eventually in $U_1$, there is
a $n_1$ so that $x_{n_1}\in U_1$.
Now suppose we have choosen $n_1<n_2<\cdots<n_k$ with
$x_{n_j}\in U_j$ for $j=1,2,\ldots,k$.
Using the definition of a subnet, there is a $\beta_0\in B$
so that $\beta\ge\beta_0$ implies that $n_\beta\ge n_k$.
On the other hand, since our subnet converges to $x$, there
is a $\beta_0'$ such that $\beta\ge\beta_0'$ implies that
$x_{n_\beta}\in U_{k+1}$.  Thus if $\beta\ge\beta_0$ and
$\beta\ge\beta_0'$, then we may take $n_{k+1}={n_\beta}$.
This suffices.

\noindent(e)\enspace
Since $Z$ is compact, $\xn$ must have a convergent subnet. 
Suppose that $\xn$ had a convergent subsequence: say $\set{
x_{n_k}}_{k=1}^\infty$ converges to $x\in Z$.  Now let
$\alpha\in\ell^\infty$ be defined as follows:
$$\alpha_n=
\begin{cases}
  (-1)^{k}&\text{if $n=n_k$ for some $k\ge1$, and,}\\
0&\text{otherwise.}
\end{cases}
$$
In other words, $x_{n_k}(\alpha)=(-1)^k$.  But we must have
$x_{n_k}(\alpha)$ converging to the number $x(\alpha)$.  This
is a contradiction.
  \end{ans}

  \begin{ques}
    Let $\H$ be a separable infinite dimensional (complex) Hilbert
    space with orthonormal basis $\set{e_{n}}_{n=1}^{\infty}$.  Let
    $S=\set{\sqrt n e_{n}}_{n=1}^{\infty}$, and let $W$ be the weak
    closure of $S$ in $\H$.
    \begin{enumerate}
    \item Show that $0\in W$.
    \item Observe that no \emph{sequence} in $S$ converges weakly to $0$.
    \end{enumerate}
(Hints: A basic weak neighborhood of $0$ is of the form
\begin{equation*}
  U=\set{h\in \H:\text{$|(h\mid k_{j})|<\epsilon$ for $j=1,\dots,n$}}.
\end{equation*}
If $\sqrt n e_{n}\notin U$, then
\begin{equation*}
  \sum_{j=1}^{n}|(e_{n}\mid  k_{j})|^{2}\ge \frac{\epsilon^{2}}n.
\end{equation*}
Observe that this can't happen for all $n$.

For the second part, observe that the principal of uniform boundedness
implies that any weakly convergent \emph{sequence} is bounded.  Of
course, no bounded sequence from $S$ can converge weakly to $0$.)
  \end{ques}
  \begin{ans}
    As in the hint, if $\sqrt n e_{n}\notin U$, then
    \begin{equation}
      \label{eq:1}
      \sum_{j=1}^{n}|(h\mid k_{j})|^{2}\ge \frac\epsilon n.
    \end{equation}
On the other hand,
\begin{equation*}
  \sum_{j=1}^{n}\|k_{j}\|^{2}=\sum_{j=1}^{n}\sum_{n=1}^{\infty}
  |(e_{n}|k_{j})|^{2} =\sum_{n=1}^{\infty}\sum_{j=1}^{n}|(e_{n}|k_{j})|^{2}<\infty.
\end{equation*}
Clearly, the above can't hold if \eqref{eq:1} holds for all $n$.
Hence $S\cap U\not=\emptyset$.  Since $U$ is arbitrary, we have shown
that $0\in W$.

For the second part, suppose that $h_{n}\to h$ weakly.  Let
$\phi_{h_{n}}$ be the linear functional $\phi_{h_{n}}(k):=(k\mid
|h_{n})$.  Then $\|\phi_{h_{n}}\|=\|h_{n}\|$.  Clearly,
$\set{\phi_{h_{n}}(k):n\in\mathbf N}$ is bounded for each $k$.  Then
the principal of uniform boundedness implies that $\set{\|h_{n}\|:n\in
\mathbf N}$ is bounded.  It then follows that if $\set{h_{n}}$ is a
sequence in $S$ converging to $0$ weakly, then there is a $M$ such
that $\|h_{n}\|\le M$ for all $n$.  This is clearly nonsense.  So no
sequence from $S$ can converge weakly to $0$.
\end{ans}

\end{document}

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