subspace topology in a metric space

Theorem 1.

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