Closure of Subset of Closed Set of Metric Space is Subset

Theorem

Let $M = \struct {A, d}$ 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 1

From Metric Induces Topology, the topology $\tau$ induced by the metric $d$ is a topology on $M$.

From Metric Closure and Topological Closure of Subset are Equivalent, it is sufficient to show that the topological closure of $H$ is contained in $F$.

From Set is Closed in Metric Space iff Closed in Induced Topological Space:

$F$ is closed in the topological space $\struct{A, \tau}$ induced by the metric $d$.

We have:

\(\ds H\)

\(\subseteq\)

\(\ds F\)

\(\ds \leadsto \ \ \)

\(\ds H^-\)

\(\subseteq\)

\(\ds F^-\)

Topological Closure of Subset is Subset of Topological Closure

\(\ds \)

\(=\)

\(\ds F\)

Set is Closed iff Equals Topological Closure

$\blacksquare$

Proof 2

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:

$H^- \subseteq F$

$\blacksquare$