subspace topology in a metric space

Suppose $X$ is a topological space whose topology is induced by a metric $d$ , and suppose $Y\subseteq X$ is a subset. Then the subspace topology in $Y$ is the same as the metric topology when by $d$ restricted to $Y$ .

Let $d^{\prime}\colon Y\colon Y\to\mathbb{R}$ be the restriction of $d$ to $Y$ , and let

	$\displaystyle B_{r}(x)$	$\displaystyle=$	$\displaystyle\{z\in X:d^{\prime}(z,x)<r\},$
	$\displaystyle B^{\prime}_{r}(x)$	$\displaystyle=$	$\displaystyle\{z\in Y:d^{\prime}(z,x)<r\}.$

The proof rests on the identity

$B^{\prime}_{r}(x)=Y\cap B_{r}(x),\quad x\in Y,r>0.$

Suppose $A\subseteq Y$ is open in the subspace topology of $Y$ , then $A=Y\cap V$ for some open $V\subseteq X$ . Since $V$ is open in $X$ ,

$V=\cup\{B_{r_{i}}(x_{i}):i=1,2,\ldots\}$

for some $r_{i}>0$ , $x_{i}\in X$ , and

	$\displaystyle A$	$\displaystyle=$	$\displaystyle\cup\{Y\cap B_{r_{i}}(x_{i}):i=1,2,\ldots\}$
		$\displaystyle=$	$\displaystyle\cup\{B^{\prime}_{r_{i}}(x_{i}):i=1,2,\ldots\}.$

Thus $A$ is open also in the metric topology of $d^{\prime}$ . The converse direction is proven similarly.

Title	subspace topology in a metric space
Canonical name	SubspaceTopologyInAMetricSpace
Date of creation	2013-03-22 15:17:44
Last modified on	2013-03-22 15:17:44
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	5
Author	matte (1858)
Entry type	Theorem
Classification	msc 54B05