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 $H$.

$\blacksquare$


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