Basic properties of closed sets

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[edit] Theorem

Let $X$ be a topological space and $A$ a subset of $X$ . Then

$A\subseteq \overline A$
If $(x n)$ is a convergent sequence of points in $A$ , its limit is an element of $\overline A$ .
If $X$ satifies the first countability axiom and $x\in \overline A$ , then there exists a convergent sequence $(x n)$ of points in $A$ such that $x_n\to x$ .

[edit] Proof

1. The definition of adherent point implies immediately that all points of $A$ are adherent points for $A$ .

2. Let $(x n)$ be a convergent sequence of points in $A$ and $x$ its limit, and let $V\in \mathcal V(x)$ . By definition of limit there exists $m\in \mathbb N$ such that for all $n > m$ , $x_n\in V$ . But, by hypothesis, for all $n > m$ , $x_n\in A$ , hence $V\cap A\not =\emptyset$ , meaning that $x$ is an adherent point for $A$ .

3. Let $x\in \overline A$ and consider a countable fundamental system of neighbourhoods of $x$ , $\{B_n^'\}_{n\in \mathbb N}$ (which exists by definition). Construct a chain (ordered by inclusion) $\{B_n\}_{n\in \mathbb N}$ by setting

$B_n=\bigcap_{m\leq n} B_m^'$

for all $n\in \mathbb N$ . Note that this chain is still a fundamental system. Now for each $n$ , let $x_n\in B_n$ . Let us prove that $x_n\to x$ . Suppose $V\in \mathcal V(x)$ . Then (by definition) there exists $m\in \mathbb N$ such that $B_m\subseteq V$ . But since $\{B_n\}_{n\in \mathbb N}$ is a chain, we have