The sequence lemma

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Lemma

Let $A$ be a subset of a topological space $X$. If there is a sequence of points of $A$ converging to $x$, then $x\in\bar{A}$; the converse holds if $X$ is first-countable.

Proof

Assume the sequence of points of $A$ that converges to $x$ is $\{x_i\}$. Then for any open set $U$ of $x$, there exists a positive natural number $N$ such that when $i>N$, $x_i\in U$. Thus $U\cap A$ is nonempty, $x\in \bar{A}$.

Let the topological space $X$ be first-countable, then there is a countable collection of open neighbourhood $\{U_i\}_{i\in \Bbb Z_+}$ of $x$ such that any open neighbourhood $U$ of $x$ contains at least one of the sets $U_i$.

Because $x\in\bar{A}$, $U_1\cap A$ is nonempty, we can select a point $x_1$ in it. In a similar manner, $U_1\cap U_2\cap A$ is nonempty, hence we can select a point $x_2$ in it. The point $x_i$ is selected from

$U_1\cap U_2\cap \cdots \cap U_i\cap A$

We then obtain a sequence $\{x_i\}$. For any open neighbourhood $U$ of $x$, it contains at least one of the set $U_N$,$N\in \Bbb Z_+$ of $\{U_i\}_{i\in \Bbb Z_+}$. Thus it contains the set

$U_1\cap U_2\cap \cdots \cap U_N\cap A$, $U_1\cap U_2\cap \cdots \cap U_N\cap U_{N+1}\cap A$, $U_1\cap U_2\cap \cdots \cap U_N\cap U_{N+1}\cap U_{N+2}\cap A$, ...

Or the set $U_1\cap U_2\cap \cdots \cap U_i\cap A$ with $i>N$, hence the points $x_i$ with $i>N$. The sequence $\{x_i\}$ converges to $x$. $\blacksquare$


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