User:TheDoctor/Library

Point in Closure of Metric Subspace iff Limit of a Sequence
Let $M = \struct {A, d}$ be a metric space and $x \in A$.

Let $S \subset A$.

Then the following are equivalent:


 * $x \in \map \cl S$


 * There exists a sequence $\sequence {x_n}_{n \mathop \in \N} \subset S$ such that $\ds x = \lim_{n \mathop \to \infty} x_n$

Necessary Condition
Let $\sequence {x_n}_{n \mathop \in \N} \subset S$ be a sequence with $x_n \to x \in A$.

Let $U \subseteq A$ be a open neighborhood of $x$.

By the definition of open set in metric spaces there is a $\varepsilon > 0$, such that the open ball $\map {B_\varepsilon} x$ is contained in $U$.

By the definition of convergence in metric spaces there is a $n \in \N$, such that $\map d {x, x_N} < \varepsilon$.

Thus $x_n \in \map {B_\varepsilon} x$ by the definition of open ball.


 * Case 1: $\map d {x_n, x} = 0$, thus $x_n = x$.

Then $x \in S$, because $\sequence {x_n}_{n \mathop \in \N} \subset S$.

Hence $x \in \map \cl S$ by Set is Subset of its Topological Closure.


 * Case 2: $\map d {x_n, x} > 0$, thus $x_n \ne x$.

We have $x_n \in S \cap \paren {\map {B_\varepsilon} x \setminus \set x} \subset S \cap \paren {U \setminus \set x}$.

If case 1 applies to any open neighborhood $U$ of $x$, then we have $x \in \map \cl S$.

Otherwise, case 2 applies to every open neighborhood of $x$, thus every open neighborhood of $x$ contains a point different from $x$.

Hence $x$ is a limit point by the definition of limit point and $x \in \map \cl S$.

Sufficient condition
Let $x \in \map \cl S$.

By the definition of closure, $x$ is either an element of $S$ or a limit point or $S$.


 * Case 1: $x \in S$.

$x$ is limit of the constant sequence $\sequence x_{n \mathop \in \N} \subset S$.


 * Case 2: $x$ is a limit point of $S$.

Let $n \in \N$ and $\delta = \frac 1 {n+1} > 0$.

Let $\map {B_\delta} x$ be the open $\delta$-ball of $x$.

$\map {B_\delta} x$ is a open neighborhood of $x$, because open balls are open sets and $x \in \map {B_\delta} x$.

By the definition of limit point, $S \cap \paren {\map {B_\delta} x \setminus \set x}$ is not empty.

Thus, by the definition of intersection and set difference, there is a $y \in S$ such that $y \in \map {B_\delta} x$ and $y \ne x$.

By the definition of open ball, we further have $\map d {x, y} < \delta$.

Define $x_n := y$.

Now let $\varepsilon > 0$.

By the Archimedean Principle, there is a natural number $N \in \N$ such that $N > \frac 1 \varepsilon$, thus $\frac 1 N < \varepsilon$.

Now $\map d {x, x_N} \le \frac 1 {N + 1} < \varepsilon$ holds by construction.

This implies $\ds \lim_{n \mathop \to \infty} x_n = x$ by the definition of convergence in metric spaces.


 * Note: if we don't have this theorem up and you want to add it, please do not use that name -- the term "subspace" is not appropriate there. "Subset of Metric Space" makes more sense than "Metric subspace" in this context. --Dfeuer (talk) 16:36, 11 June 2013 (UTC)