Herediterily sigma-compact space means countable. Miklos Schweitzer 2019, problem 1.

The first problem of the last Miklos Schweitzer contest has a very short statement.

Problem 1 (Miklos Schweitzer 2019). Prove that if every subspace of a Hausdorff space X is \sigma – compact, then X is countable.

Seeking for a contradiction, suppose on the contrary X is not countable. Since X is \sigma – countable itself, it can be represented as countable union of compact sets. One of those sets, say X_1 , is also non countable. Clearly, X_1 has the same property as X – it is Hausdorff and every its subspace is \sigma compact. Hereafter, we forget about X and all considerations will be with respect to X_1. The proof is based on the following claims.

Claim 1. Let the compact and Hausdorff topological space X be a disjoint union of two everywhere dense sets Y,Z. Then either Y or Z is not \sigma-compact.

Proof. Suppose on the contrary both Y,Z are \sigma-compact and let Y=\bigcup_{i=1}^{\infty}Y_i, Z=\bigcup_{i=1}^{\infty}Z_i where Y_i,Z_i are compact. Note that Y_i,i\in \mathbb{N} are nowhere dense sets, since they are free of points of Z and Z is dense in X. In the same way, Z_i are also nowhere dense. Hence,

X=\displaystyle \left(\bigcup_{i=1}^{\infty}Y_i\right)\bigcup \left(\bigcup_{i=1}^{\infty}Z_i \right)

where Y_i,Z_i are closed (compact) nowhere dense sets. But X , as being Hausdorff and compact, has Baire property (see [3]) and cannot be represent as a countable union of closed nowhere dense sets, a contradiction. \blacksquare

Definition. Given a topological space X which can be represent as disjoint union of two everywhere dense subspaces, we call X a resolvable topological space.

Claim 2. Every compact Hausdorff topological space X without isolated points is resolvable.

It is a known result. I think due to Edwin Hewitt [1] (not sure), who had introduced the notion of resolvability. It has many generalization, e.g. it’s true for any locally compact Hausdorff space (see [2], Theorem 3.7). \blacksquare

Back to the problem. Let A_0 be the set of all isolated points of X_1. In case A_0 is uncountable, we are done. Indeed each point of A_0 is an open set, and \bigcup_{a\in A_0}\{a\} is a cover of A_0, which doesn’t have a countable subcover. It contradicts with the fact A_0 being \sigma-compact, is a Lindelof space. Thus, A_0 is at most countable.
Let \omega_1 be the first uncountable ordinal. We construct a family of countable sets A_{\alpha}\subset X_1, \alpha<\omega_1 using the transfinite induction as follows. A_0 is as constructed above. Suppose A_{\beta}, \beta<\alpha have been already chosen. Consider X_1':=X_1\setminus \left( \bigcup_{\beta<\alpha} A_{\beta}\right). It is non empty set, since X_1 is uncountable and \bigcup_{\beta<\alpha} A_{\beta} – at most countable. There are two cases:

  1. X_1' has no isolated points. Then the same holds for \overline{X_1'}. Using Claim 2 for \overline{X_1'} we can find two everywhere dense subsets of it and by Claim 1, one of them is not \sigma-compact, a contradiction.
  2. In the other case let A_{\alpha}\neq \emptyset be the set of all isolated points of X_1'. The same argument as above shows A_{\alpha} cannot be uncountable, otherwise it wouldn’t be \sigma-compact.

By transfinite induction, the construction of the family A_{\alpha}, \alpha<\omega_1 is completed. Further, we choose points a_{\alpha}\in A_{\alpha},\forall \alpha<\omega_1. Since a_{\alpha} is an isolated point of X_1\setminus \left( \bigcup_{\beta<\alpha} A_{\beta}\right) , there exists an open set U_{\alpha} (in X_1) such that U_{\alpha} is free of points of \left( X_1\setminus \left( \bigcup_{\beta<\alpha} A_{\beta}\right)\right)\setminus \{a_{\alpha}\}. In particular, a_{\beta} \notin U_{\alpha}, \forall \beta, \alpha<\beta<\omega_1. Let X':=\{a_{\alpha}:\alpha<\omega_1\}. Since X' is Lindelof as being \sigma-compact and \{U_{\alpha}: \alpha<\omega_1\} is a cover of X' , there exists a countable subcover \{U_i : i\in I\subset \omega_1\} . Let \alpha := \sup I+1. Then \alpha<\omega_1 and i< \alpha, \forall i\in I. It means a_{\alpha}\notin U_{i},\forall i\in I, a contradiction.

About motivation.

The interval [0,1] is Hausdorff, compact and uncountable. So, why it is not herediterily \sigma-compact? Because the set of irrational numbers inside [0,1] is not \sigma-compact. The proof of it exploits the Baire category theorem. If the irrationals can be represented as a countable union of compact sets, these sets are nowhere dense, since the rationals are dense. Hence, [0,1] can be represented as countable union of compact, nowhere dense sets – those compact sets plus all the points \{q\}, q\in\mathbb{Q}\cap [0,1]. It contradicts the Baire category theorem.

So, the same approach can be applied for a compact, Hausdorff space, which is separable and does not have isolated points. After some unsuccessful tries, I realized that the existence of a countable dense set can be replaced by existence of two disjoint dense sets, the base space can be partitioned into. That’s how Claim 1 came. I searched the net when a topological space can be partitioned like that and was surprised to find there is a spacial name for such cases. And indeed, the compact Hausdorff spaces without isolated points are resolvable.

But having isolated points is a big problem. For example, any isolated point is not nowhere dense set. So, imagine you have a compact Hausdorff space X with countably many isolated points which are dense in X. Then we cannot apply the Baire theorem like the described case of [0,1] , since any isolated point \{a\} is not nowhere dense.

Thus, we try to remove the isolated points, but the space that remains may also have isolated points. The process of “nibbling” the isolated points was the final obstacle.

References:
[1] A problem of set-theoretic topology by E. Hewitt, Duke Math J., 1943
[2] On resolvability of topological spaces, Oleg Pavlov. Theorem 3.7
[3] Baire category theorem. https://en.wikipedia.org/wiki/Baire_category_theorem

2 thoughts on “Herediterily sigma-compact space means countable. Miklos Schweitzer 2019, problem 1.”

  1. Pingback: Banach – Mazur Game. Miklos Schweitzer 2020, Problem 9. – A Point of View.
  2. Pingback: USEMO 2023, Problem 2. Or Again on Baire Category Theorem. – A Point of View.

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