Second Countable Crit. for the Exist. of an Acc. Point of Uncount. Sets

# Second Countable Criterion for the Existence of an Accumulation Point of Uncountable Sets

Recall from the Second Countable Topological Spaces page that a topological space $(X, \tau)$ is said to be second countable if the topology $\tau$ has a countable base.

Topological spaces that are second countable possess many nice properties. One of such properties is described below.

Theorem 1: Let $(X, \tau)$ be a second countable topological space and let $A \subseteq X$ be uncountable. Then there exists an point $a \in A$ that is an accumulation point of $A$. |

*Theorem 1 assumes that the space $(X, \tau)$ is uncountable to begin with so that such an uncountable subset $A$ of $X$ exists.*

**Proof:**Let $(X, \tau)$ be second countable and let $\mathcal B$ be a countable base for the topology $\tau$. Let $A$ be an uncountable set and suppose instead that every $x \in A$ is NOT an accumulation point of $A$.

- Then for each $x \in A$ there exists an open neighbourhood $U_x$ of $x$ which contains no points of $A$ different from $x$, that is:

\begin{align} \quad U_x \cap A = \{ x \} \end{align}

- Since $\mathcal B$ is a base for $\tau$, for each $U_x$ there exists a $B_x \in \mathcal B$ such that $x \in B_x \subseteq U_x$. But then:

\begin{align} \quad B_x \cap A = \{ x \} \end{align}

- Observe that the set $\{ B_x \cap A : x \in A \}$ is countable since $\mathcal B$ is countable. Thus, the above equality implies $A$ is countable - a contradiction. So the assumption that no $x \in A$ is an accumulation point of $A$ is false.

- Thus, there exists an $a \in A$ that is an accumulation point of $A$. $\blacksquare$