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

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Theorem

Let $M = \struct {A, d}$ be a metric space.

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

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


Let $a \in A$.

Then $a \in H^-$ if and only if there exists a sequence $\sequence {x_n}$ of points of $H$ which converges to the limit $a$.


Proof

From definition of closure, $H^- = H' \cup H^i$.

Suppose that $a \in H^-$.


If $a \in H^i$, then $a \in H$ and so $\sequence {a, a, \ldots}$ is a sequence in $H$ that converges to $a$.

If $a \in H'$, then by Limit Point is Limit of Convergent Sequence there exist a sequence in $H$ that converges to $a$.

$\Box$


For the converse, let $\sequence {x_n}$ be a sequence in $H$ that converges to a point $a \in A$.

So let $U$ be an open set that contains $a$.


By the definition of an open set, it follows that there exists an open ball centered at $a$ such that $\map {B_\epsilon} a \subseteq U$.

But then because $\sequence {x_n}$ converges to $a$ there exists $m \in \N_{\gt 0}$ such that:

\(\displaystyle \map d {x_m, a}\) \(=\) \(\displaystyle <\) \epsilon
\(\displaystyle \leadsto \ \ \) \(\displaystyle H \cap \map {B_\epsilon} a\) \(\ne\) \(\displaystyle \O\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle H \cap U\) \(\ne\) \(\displaystyle \O\)


Because from Set Intersection Preserves Subsets it is seen that $H \cap \map {B_\epsilon} a \subseteq H \cap U$.


It has been shown that every open set that contains $a$ also contains a point in $H$.

So by Condition for Point being in Closure, it is seen that $a \in H^-$.

$\blacksquare$


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