Classification of compact Hausdorff spaces $X$ of cardinality $2^{\aleph_0}>|X|\geq \aleph_1$

cardinalsgeneral-topologyordinalsreference-requestset-theory

On my own time, I've been curious/investigating some properties of compact Hausdorff spaces of cardinality $\aleph_{\alpha}$, for $\alpha \geq 0$ a particular ordinal. I am aware of the following results:

  • A compact Hausdorff space without isolated points is uncountable (various proofs including Munkres and the classic diagonal argument);
  • Compact Hausdorff spaces which are finite or countably infinite are homeomorphic to $\omega^{\beta} \times n + 1$, where $\omega$ is the first infinite ordinal, $n \geq 1$ is a finite positive ordinal, and $\beta \geq 0$ is an arbitrary countable ordinal (Sierpinski-Mazurkiewicz);
  • A compact, uncountable subset of $\mathbb{R}$ has cardinality $2^{\aleph_0}$ (Cantor-Bendixson).

My question is:
What is known in ZFC regarding the shape/structure of uncountable compact Hausdorff spaces of cardinality $\kappa < 2^{\aleph_0}$?
I'm particularly interested in the case of cardinality $\kappa = \aleph_1$.
Is there nothing that can be said to characterize compact spaces $X$ of cardinality $\aleph_1$, in the absence of CH, other than that the successor ordinal spaces of cardinality $\aleph_1$ are compact?
In particular, I'm curious if there exists a known counterexample: a topological space $X$ of cardinality $\aleph_1$ which cannot be embedded in such a compact ordinal space.

Best Answer

Following Alessandro Codenotti's assumption, let $X$ be the one-point compactification of an uncountable, discrete space. Then $X$ cannot be embedded into a linearly ordered space:

Assume on the contrary. Then, by compactness, there is a linear order $\le$ on $X$, such that the order topology is the given topology. For $x \in X$ define $X^{\le x} = \{y \in X: y \le x \}$. Analogously, $X^{< x} = \{y \in X: y < x \}$. Let $\infty$ be the non-isolated point of $X$. W.l.o.g., $X^{\le\infty} $ is uncountable. If $x \in X^{< \infty}$, then $X^{\le x}$ is finite, hence the map $X^{< \infty} \rightarrow \mathbb N, x \mapsto |X^{\le x}|$ is well-defined and easily seen to be strictly monotonically, in particular one-to-one. Contradiction!

Update
Here is a very different example: $X := (\omega_1 +1) \times (\omega +1)$, both factors with the order topology. Since they are compact, $X$ is compact, $|X| = \aleph_1$. Since $X$ contains the non-normal deleted Tychonoff plank, it cannot be embedded into any linearly ordered topological space.

Related Question