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\begin{document}
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\section*{II Basic topological properties}
\subsection* {5. Axioms of countability }


\begin{defn}
A space $X$ satisfies the \textbf{first axiom of countability(first
countable)} if for all $x\in X$, there exists a countable collection
of open sets $\mathcal{O}=\{O_n\}$ which satisfies the condition
that for all open neighborhood
$O$ of $x$, there exists an $O_n\in\mathcal{O}$ s.t. $x\in O_n\subset O$.\\
This $\mathcal{O}$ is called \textbf{a basis at $x$}.
\end{defn}

\begin{ex}
A metric space is 1st countable.
\end{ex}

\begin{proof}
\{balls with rational radius\} forms a basis at each point.
\end{proof}

\begin{defn}
A space $X$ satisfies the \textbf{second axiom of countability(second
countable)} if $X$ has a countable basis.
\end{defn}

\begin{note}
Second countable\,\,$\Rightarrow$\,\,First countable.
\end{note}

\begin{ex}
\item 1. $\re^{n}$ is second countable: $\{B_{r}(q) | \: q \in \mathbb{Q}^{n},
r \in \mathbb{Q}\}$ is a countable basis of $\re^{n}$.\
\item 2. A discrete space is first countable. An uncountable discrete space
is not second countable.\
\item 3. $\re_l$ is $\re$ with half open interval topology, i.e., the
topology is generated by $\{(a,b]|\,a<b\}$. Then $\re_l$ is 1st
countable ($(r,x], r\in\mathbb{Q}$ is a countable basis at $x$)
 but not 2nd countable.\\
\begin{proof}
Suppose $\re_l$ is 2nd countable, then there exists a basis of the
form $\{(a_n,b_n]|n\in\mathbb{N}\}$ by Lemma 1. Choose $b$ with $b\neq
b_n$ for all $n\in\mathbb{N}$. Then $(a,b]$ is not a union of the
intervals $(a_n,b_n]$
and this is contradiction.
\end{proof}
\end{ex}

\begin{lem}
If X is 2nd countable, each basis for X has a countable
subcollection which also forms a basis for X.
\end{lem}
(Proof is a homework.)

\begin{prop}
A subspace of 1st countable space(2nd countable respectively) is 1st
countable(2nd countable respectively).
\end{prop}
Proof is clear.\

\begin{prop}
X, Y: 1st countable\,\,$\Rightarrow X\times Y$:1st countable.\\
X, Y: 2nd countable\,\,$\Rightarrow X\times Y$:2nd countable.
\end{prop}

\begin{proof}
$\mathcal{O}_X,\mathcal{O}_Y$ : countable basis of X,Y,
respectively.\\
$\Rightarrow\mathcal{O}_X\times\mathcal{O}_Y=\{O_X\times
O_Y|O_X\in\mathcal{O}_X, O_Y\in \mathcal{O}_Y\}$ is a countable
basis for $X\times Y$\\
Similarly for the 1st countability.
\end{proof}

\begin{ex}
In general, not all product preserves the property.\\
Let $X=\prod_{\alpha\in A}I_{\alpha}$, where $A$ is an uncountable
index set and
$I_{\alpha}=[0,1]$. Now we claim that X is not 2nd countable:\\
If $X=\{x:A\rightarrow I=[0,1]\}$ is 2nd countable, the standard basis for the product topology
has a countable subcollection
$\mathcal{U}={p_{\alpha_1}^{-1}(O_{\alpha_1})\bigcap\cdots\bigcap
p_{\alpha_n}^{-1}(O_{\alpha_n})}$ which is also a basis by the
previous lemma. Choose an index $\alpha\in A$ which does not appear
in any basic open set in $\mathcal{U}$. Such an index exists since $A$ is uncountable.
Then for $x\in P_{\alpha}^{-1}(0,1/2)$, there is no $U\in\mathcal{U}$ s.t. $x\in
U\subset P_{\alpha}^{-1}(0,1/2)$ since $P_\alpha(U)=I$.
\end{ex}

\begin{hw}
Is the above example first countable?
\end{hw}


\begin{defn}
A sequence in a space X is a function $x : \mathbb{N} \rightarrow X$
usually written as $(x_{n})_{n=1}^{\infty}$ where $x_{n}=x(n)$.
\end{defn}

\begin{thm}
When X is 1st countable, the following statements hold.\\
(1) When $A\subset X$, $x\in\overline{A} \Leftrightarrow \exists\, a\,
sequence\,
(a_n)\, in\, A \text{ s.t. } a_n\rightarrow x$.\\
(2) $A\subset X \text{ is closed } \Leftrightarrow \exists
a_n\rightarrow x \text{ with } a_n\in A \text{ implies } x\in A$.\\
(3) $f:X\rightarrow Y \text{ is continuous } \Leftrightarrow
x_n\rightarrow x \text{ implies } f(x_n)\rightarrow f(x)$.\\
\end{thm}

\begin{proof}
(1) $(\Rightarrow)$
Since X is 1st countable, $\forall x\in X$, we can
construct a decreasing sequence of basic open neighborhoods of $x$.
Indeed if we let $\mathcal{U}=\{U_n\}$ be a countable basis at $x$,
then $\mathcal{V}=\{V_n|V_n=U_1\cap\cdots\cap
U_n,n=1,2,\cdots\}$ is clearly a decreasing sequence of open neighborhoods of $x$.
If $x\in A$, let $a_n=x$.
If $x\nin A$, then $x\in A'$. Choose a point $a_n\in V_n\cap A$ and then
$(a_n)$ is a sequence converging to $x$.\\
$(\Leftarrow)$
$a_n\rightarrow x$ $\Rightarrow$ for any neighborhood
of $x$ it contains $a_n$'s for large n.
Then either $x\in A$ or $x\in A'$.\\
Hence $x\in \overline{A}$. (We do not need 1st countability of $X$.)\\
(2) $(\Rightarrow)$
$a_n\rightarrow x$ with $a_n\in A \Rightarrow x\in
\overline{A}=A$. (We do not need 1st countability of $X$.)\\
$(\Leftarrow)$
Show $\overline{A} \subset A$:\\ 
$x\in \overline{A}$\\
$\Rightarrow \exists\,(a_n)$ in $A\text{ s.t. } a_n\rightarrow x$ by
(1)\\
$\Rightarrow x\in A$.\\
(3) $(\Rightarrow)$
For any open neighborhood U of $f(x)$, $f^{-1}(U)$
is an
open neighborhood of $x$\\
$\Rightarrow x_n\in f^{-1}(U)$ for large n\\
$\Rightarrow f(x_n)\in U$ for large n.\\
Hence $f(x_n)\rightarrow f(x)$. (We do not need 1st countability of $X$.)\\
$(\Leftarrow)$
Show $f^{-1}(closed\,set)$ is closed:\\
$x\in \overline{f^{-1}(C)}$ where $C$ is closed
set.\\
$\Rightarrow\exists$ a sequence $(a_n)$ in $f^{-1}(C)$ s.t.
$a_n\rightarrow x$ by (1)\\
$\Rightarrow f(a_n)\rightarrow f(x)$ by the hypothesis and
$f(a_n)\in C$\\
$\Rightarrow f(x)\in C$ by (2)\\
$\Rightarrow x\in f^{-1}(C)$.\\
Hence $f^{-1}(C)$ is closed.
\end{proof}
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