Must uncountable compact Hausdorff spaces have large discrete subsets?

The situation is this. I have a space $X$ which is second countable, compact, and Hausdorff (it's a modified form of a type space, though I don't think that matters here). It has size continuum. It may or may not have isolated points.

Must $X$ have a discrete subset of size continuum?

The obvious inductive constructions yield only countable discrete sets, and there isn't obviously a Zorn's Lemma argument, since the ascending union of discrete sets is not necessarily discrete. But this is not my specialty; is this known one way or the other?

The purpose of this is to produce a many-models theorem, and seemingly has nothing to do with topology.

• Addressing the question in your title (hence ignoring the term "perfect kernel" which is unfamiliar to me): I can't see how to get an infinite discrete subset of the circle with its usual topology... – Yemon Choi Sep 9 '15 at 13:38

Without even using compactness (or metrisability), a subspace of a second-countable space is second-countable. There is no discrete, uncountable, second-countable space.

A second countable, compact Hausdorff space is metrisable (and separable). Subspaces of separable metric spaces are separable too. Uncountable discrete spaces are not separable.

No. A second countable compact Hausdorff space is metrizable and therefore separable. Every separable metric space is the union of a countable set and a set that is dense in itself. These facts are in many books that contain an introduction to metric spaces (baby Rudin comes to mind).

• Just didn't occur to me. I'm so used to thinking the subsets of $\mathbb R$ are maximally weird that I forgot there are some nice properties. – Richard Rast Sep 9 '15 at 15:17
• Baby Rudin doesn't have the Urysohn metrization theorem; it deals only with metric spaces and doesn't discuss general topological spaces at all. – Nate Eldredge Sep 9 '15 at 16:12

A second countable space never contains an discrete set of any uncountable cardinality. Compact Hausdorff is not needed. And you don't need the Urysohn metrization theorem, either.

Let's prove the contrapositive. Suppose $A$ is an uncountable discrete set in $X$. Let $\mathcal{B}$ be any base for the topology of $X$. Since $A$ is discrete, for each $x \in A$ there is an open set $U_x$ with $U_x \cap A = \{x\}$. Since $\mathcal{B}$ is a base, there is a basic open set $V_x \in \mathcal{B}$ with $x \in V_x \subset U_x$. So $V_x \cap A = \{x\}$ as well. In particular no two of the $V_x$ are the same. So we have an injection from $A$ to $\mathcal{B}$, thus $\mathcal{B}$ is uncountable.

It is worth noting that for any cardinal $\kappa$ there is a compact Hausdorff space (not generally second countable) with a discrete set of cardinality $\kappa$: simply equip $\kappa$ with the discrete topology and take its one-point compactification.

• In words, you say that any base of a discrete space must contain all singleton subsets? – Incnis Mrsi Sep 9 '15 at 19:48

A countable discrete subset of the circle can be obtained by taking $\{e^{2\pi i /n} \mid n=1,2,\dots\}$ whose points are obviously isolated from each other. Of course, a closed discrete subset of a $D$ compact Hausdorff space must be finite (otherwise $U_n = (X \setminus D) \cup \{x_1,\dots,x_n\}$ ) is an open cover without finite subcover.