Inclusion Mapping on Metric Space is Continuous

Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Let $\left({H, d_H}\right)$ be a metric subspace of $M$.

Then the inclusion mapping $i_H: H \to A$ is continuous.

Proof

Let $a \in H$.

Let $\epsilon \in \R_{>0}$.

Let $\delta = \epsilon$.

Then:

 $\displaystyle d_H \left({a, y}\right)$ $<$ $\displaystyle \delta$ for some $y \in H$ $\displaystyle \implies \ \$ $\displaystyle d \left({i_H \left({a}\right), i_H \left({y}\right)}\right)$ $=$ $\displaystyle d \left({a, y}\right)$ $\displaystyle$ $=$ $\displaystyle d_H \left({a, y}\right)$ $\displaystyle$ $<$ $\displaystyle \delta$ $\displaystyle$ $=$ $\displaystyle \epsilon$

So by definition $d_H$ is continuous at $a$.

As $a \in H$ is arbitrary, it follows that $d_H$ is continuous on $H$.

$\blacksquare$