Closure of Subset of Closed Set of Metric Space is Subset/Proof 2

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Theorem

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

Let $F$ be a closed set of $M$.

Let $H \subseteq F$ be a subset of $F$.

Let $H^-$ denote the closure of $H$.


Then $H^- \subseteq F$.


Proof

Let $x \in H^-$.

From Point in Closure of Subset of Metric Space iff Limit of Sequence

there exists a sequence $\sequence {a_n}$ of points of $H$ which converges to the limit $x$.

By assumption:

$\sequence {a_n}$ is also a sequence of points of $F$

From Subset of Metric Space contains Limits of Sequences iff Closed:

$x \in F$

Thus it has been shown:

$H^- \subseteq F$

$\blacksquare$

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