User:TheDoctor/Library

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

Let $S \subset A$.

Then the following are equivalent:


 * $x \in \operatorname{cl}\left({S}\right)$


 * There exists a sequence $\left \langle {x_n} \right \rangle_{n \in \N} \subset S$ such that $\displaystyle x = \lim_{n \to \infty} x_n$

Necessary condition
Let $ \left \langle {x_n} \right \rangle_{n \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 $B_{\varepsilon}\left({x}\right)$ is contained in $U$.

By the definition of convergence in metric spaces there is a $n \in \N$, such that $d\left({x, x_n}\right) < \varepsilon$.

Thus $x_n \in B_{\varepsilon}\left({x}\right)$ by the definition of open ball.

Case 1: $d\left({x_n, x}\right) = 0$, thus $x_n = x$.

Then $x \in S$, because $ \left \langle {x_n} \right \rangle_{n \in \N} \subset S$.

Hence $x \in \operatorname{cl}\left({S}\right)$ by Set is Subset of its Topological Closure.

Case 2: $d\left({x_n, x}\right) > 0$, thus $x_n \ne x$.

We have $x_n \in S \cap \left({ B_{\varepsilon} \left({x}\right) \setminus \left\{ {x} \right\} }\right) \subset S \cap \left({ U \setminus \left\{ {x} \right\} }\right)$.

If case 1 applies to any open neighborhood $U$ of $x$, then we have $x \in \operatorname{cl}\left({S}\right)$.

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 $x \in \operatorname{cl}\left({S}\right)$.

Sufficient condition
Let $x \in \operatorname{cl}\left({S}\right)$.

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 $\left \langle {x} \right \rangle_{n \in \N} \subset S$.

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

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

Let $B_{\delta}\left({x}\right)$ be the open $\delta$-ball of $x$.

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

By the definition of limit point, $S \cap \left({ B_{\delta}\left({x}\right) \setminus \left\{ {x} \right\} }\right)$ is not empty.

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

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

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 $d\left({x, x_N}\right) \le \frac 1 {N+1} < \varepsilon$ holds by construction.

This implies $\displaystyle \lim_{n \to \infty} x_n = x$ 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&mdash;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)