# subspace topology in a metric space

###### Theorem 1.

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

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 SubspaceTopologyInAMetricSpace 2013-03-22 15:17:44 2013-03-22 15:17:44 matte (1858) matte (1858) 5 matte (1858) Theorem msc 54B05