Subset of Metric Space is Closed iff contains all Zero Distance Points
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:
- $\ds \lim_{n \mathop \to \infty} \map d {x, a_n} = 0$
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:
- $\forall x \in A: \map d {x, H} = 0 \implies x \in H$
From Point at Zero Distance from Subset of Metric Space is Limit Point or Element:
- $\map d {x, H} = 0$
if and only if either:
- $x$ is a limit point of $H$
- $x \in H$
Let $x$ be limit point of $H$.
Then by hypothesis $x \in H$.
That is:
- $H$ contains all its limit points
So by definition $H$ is a closed set of $M$.
$\blacksquare$
Also see
- Point at Distance Zero from Closed Set is Element, which presents this result as:
- If $H$ is closed in $M$, then $\map d {x, H} = 0$ if and only if $x \in H$.
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.9$