Subspace of Metric Space is Metric Space

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Theorem

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

Let $H \subseteq A$.

Let $d_H: H \times H \to \R$ be the restriction $d \restriction_{H \times H}$ of $d$ to $H$.

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


Then $d_H$ is a metric on $H$.


Proof

By definition of restriction:

$\forall x, y \in H: d_H \left({x, y}\right) = d \left({x, y}\right)$

As $d$ is a metric, the metric space axioms are all fulfilled by all $x, y \in A$ under $d$.

As $H \subseteq A$, by definition of subset, all $x, y \in H$ are also elements of $A$.

Therefore the metric space axioms are all fulfilled by all $x, y \in H$ under $d_H$.

$\blacksquare$


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