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 $\operatorname{fst} \circ A'$.

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