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 $A$. $\blacksquare$
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