|
|
|
Viewing Version
2
of
'compact metric spaces are second countable'
|
[ view 'compact metric spaces are second countable'
|
back to history
]
| Title of object: |
compact metric spaces are second countable |
| Canonical Name: |
CompactMetricSpacesAreSecondCountable |
| Type: |
Theorem |
| Created on: |
2007-04-28 18:45:39 |
| Modified on: |
2007-04-28 18:48:03 |
| Classification: |
msc:54D70 |
| Keywords: |
compact, metric space, metrizable, open ball, basis, open cover, countable |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amsthm}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
%\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\theoremstyle{plain}
\newtheorem*{thm}{Theorem}
\newtheorem*{lem}{Lemma}
\newtheorem*{cor}{Corollary}
\newtheorem*{prop}{Proposition}
|
Content:
\begin{prop}
Every compact metric space is second countable.
\end{prop}
\begin{proof}
Let $(X,d)$ be a compact metric space, and for each $n\in\mathbb{Z}_+$ define $\mathcal{A}_n=\{B(x,1/n):x\in X\}$, where $B(x,1/n)$ denotes the open ball centered about $x$ of radius $1/n$. Each such collection is an open cover of the compact space $X$, so for each $n\in\mathbb{Z}_+$ there exists a finite collection $\mathcal{A}_n^\prime\subset\mathcal{A}_n$ that covers $X$. Put $\mathcal{B}=\bigcup_{n=1}^\infty\mathcal{A}_n^\prime$. Being a countable union of finite sets, it follows that $\mathcal{B}$ is countable; we assert that it forms a basis for the metric topology on $X$. The first property of a basis is satisfied trivially, as each set $\mathcal{A}_n^\prime$ is an open cover of $X$. For the second property, let $x,x_1,x_2\in X$, $n_1,n_2\in\mathbb{Z}_+$, and suppose $x\in B(x_1,1/n_1)\cap B(x_2,1/n_2)$. Since both $B(x_1,1/n_1)$ and $B(x_2,1/n_2)$ are open in the metric topology on $X$, there exists $\epsilon>0$ such that $B(x,\epsilon)\subset B(x_1,1/n_1)\cap B(x_2,1/n_2)$. Select $N\in\mathbb{Z}$ such that $1/N<\epsilon$. There must exist $x_3\in X$ such that $x\in B(x_3,1/2N)$ (since $\mathcal{A}_{2N}^\prime$ is an open cover of $X$). To see that $B(x_3,1/2N)\subset B(x_1,1/n_1)\cap B(x_2,1/n_2)$, let $y\in B(x_3,1/2N)$. Then we have
\begin{equation}
d(x,y)\leq d(x,x_3)+d(x_3,y)<\dfrac{1}{2N}+\dfrac{1}{2N}=\dfrac{1}{N}<\epsilon\text{,}
\end{equation}
so that $y\in B(x,\epsilon)$, from which it follows that $y\in B(x_1,1/n_1)\cap B(x_2,1/n_2)$, hence that $B(x_3,1/2N)\subset B(x_1,1/n_1)\cap B(x_2,1/n_2)$. Thus the countable collection $\mathcal{B}$ forms a basis for a topology on $X$; that the topology induced by $\mathcal{B}$ is in fact the metric topology follows readily from the definition of the metric topology. Thus $X$ is second-countable.
\end{proof}
Note that, because a countable union of countable sets is countable, it would have been sufficient to assume that $(X,d)$ was Lindel\"{o}f. |
|
|
|
|
|
