# Inclusion Mapping on Metric Space is Continuous

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## 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 $H$.

$\blacksquare$

## Sources

- 1962: Bert Mendelson:
*Introduction to Topology*... (previous) ... (next): $\S 2.7$: Subspaces and Equivalence of Metric Spaces: Theorem $7.2$