Subset of Metric Space is Closed iff contains all Zero Distance Points

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Theorem

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

Let $H \subseteq A$.


Then $H$ is closed in $M$ if and only if:

$\forall x \in A: \map d {x, H} = 0 \implies x \in H$

where $\map d {x, H}$ denotes the distance between $x$ and $H$.


Proof

Necessary Condition

Let $H$ be closed in $M$.

Let $x \in A: \map d {x, H} = 0 \implies x \in H$.

By Existence of Sequence in Subset of Metric Space whose Limit is Infimum there exists a sequence $\sequence {a_n}$ of points of $H$ such that:

$\displaystyle \lim_{n \mathop \to \infty} \map d {x, a_n} = 0$

So every neighborhood of $x$ contains points of $H$.

If some $a_n \in H$ then $x \in H$.

Otherwise each $a_n$ is different from $x$

Then $x$ is the limit of $\sequence {a_n}$ and hence of $H$.

Thus $x \in H$ by definition of closed set.

$\Box$


Sufficient Condition

Suppose $H$ is such that:

$\forall x \in A: \map d {x, H} = 0 \implies x \in H$

Let $x$ be a limit point of $H$.

Then $\map d {x, H} = 0$.


Thus $H$ contains all its limit points and so is by definition a closed set.

$\blacksquare$


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