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:
- There exists a sequence $\sequence {x_n}_{n \mathop \in \N} \subset S$ such that $\ds x = \lim_{n \mathop \to \infty} x_n$
Proof
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 $x \in \map \cl S$.
$\Box$
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 Axiom of Archimedes, 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.