Sequence Lemma

From ProofWiki
Jump to navigation Jump to search


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.


Assume the sequence of points of $A$ that converges to $x$ is $\sequence {x_i}$.

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 $\family {U_i}_{i \mathop \in \Bbb Z_+}$ 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 $\sequence {x_i}$.

For any open neighbourhood $U$ of $x$, it contains at least one of the set $U_N$, $N \in \Bbb Z_+$ of $\family {U_i}_{i \mathop \in \Bbb Z_+}$.

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 $\sequence {x_i}$ converges to $x$.



  • 2004: James R. Munkres: Topology Chapter $2$ Topological Spaces and Continuous Functions: $\S 21$ The Metric Topology (contiuned): Lemma $21.2$