# User:Dfeuer/Compact Separable Perfect Hausdorff Space Cardinality

## Theorem

Let $(X, \tau)$ be a compact separable perfect Hausdorff space.

Then $|X| \ge |2^{\aleph_1}|$.

## Proof

Since $(X, \tau)$ is separable, it has a countable dense subset $D$.

$D$ is perfect when considered as a subspace:

Let $x \in D$, and let $U$ be a $\tau$-open neighborhood of $x$. Then $U$ contains an element $y \in X$, $y ≠ x$. Then $U \setminus \{x\}$ is a non-empty open set, so it must contain an element of $D$.

By Point in Finite Hausdorff Space is Isolated, $D$ is infinite.

Thus $D$ is countably infinite.

Then $D \times D$ is also countably infinite.

Let $A: D \times D \to \mathcal P(\tau \times \tau)$ be defined by letting $A(x,y)$ be the set of all ordered pairs $(A,B)$ of disjoint open sets such that $x \in A$ and $y \in B$.

By the Axiom of Countable Choice, there is a mapping $A': D \times D \to \tau \times \tau$ such that $a(x, y) \in A(x,y)$ for each $(x,y) \in D \times D$.

Let $U: D \times D \to \tau$ be $\operatorname{fst} \circ A'$.

## Axiom of Countable Choice

This theorem depends on the Axiom of Countable Choice.

Although not as strong as the Axiom of Choice, the Axiom of Countable Choice is similarly independent of the Zermelo-Fraenkel axioms.

As such, mathematicians are generally convinced of its truth and believe that it should be generally accepted.

## Credit

The outline of this proof is due to Asaf Karagila on StackExchange: Karagila's outline