# How far is being star compact from being countably compact？

How far is star-compactness from countably compactness？

A topological space $X$ is said to be star compact if whenever $\mathscr{U}$ is an open cover of $X$, there is a compact subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$.

I've found that every regular $T_1$ star compactness space $X$ is countably compact, which leads a paradox. However, I don't know where I'm wrong.

Prove: In every regular $T_1$ space $X$, countably compactness is equiavalent to this condition: every point finite countable open covering of $X$ has a finite subcover (see the theorem 2 of Pseudocompact and Countably Compact Spaces). For any point finite countable open covering $\mathscr{U}$ of $X$, there is a compact subspace $K$ of $X$ such that $St(K,\mathscr{U})=X.$ Suppose $X$ is not coutably compact, then $\mathscr{U}$ has not finit subcover. For any $x_1 \in K$, there exists a finit subcollection $\mathscr{U_1}$ of $\mathscr{U}$, such that $x_1$ is in the every element of $\mathscr{U_1}$ and $\cup \mathscr{U_1}$ can't cover $X$. So there exists $x_2 \in K\setminus \cup \mathscr{U_1}$ and $\mathscr{U_2}$, such that $x_2$ in the every element of $\mathscr{U_2}$ and $\cup \{\mathscr{U_i}, i= 1,2\}$ can't cover $X$. We can work on it until $\omega$ times! Therefore there exists a subset $\{x_i:i \in \omega\}$ of $K$ which is a countable closed discrete set of the compact $K$. Contradiction!

• By star-compact you mean compact and star-countable? Dec 26, 2011 at 10:28
• No. A topological space $X$ is said to be star compact if whenever $\mathscr{U}$ is an open cover of $X$, there is a compact subspace $K$ of $X$ such that $X = \operatorname{St}(K,\mathscr{U})$. I have modified the definition of star-compactness
– Paul
Dec 26, 2011 at 11:04
• @Patrick: More generally, for any property $\mathscr{P}$ of spaces, $X$ is $\text{star-}\mathscr{P}$ iff for every open cover $\mathscr{U}$ of $X$ there is an $A\subseteq X$ with property $\mathscr{P}$ such that $\operatorname{st}(A,\mathscr{U})=X$. Warning: Some authors have used the term starcompact for what is star finite by this definition, and John’s (and my) star compactness has gone by at least two other names. Dec 26, 2011 at 12:11
• Thanks Brian! Very appreciated. Dec 26, 2011 at 13:17

I’m not very familiar with star-$\mathscr{P}$ properties, but a bit of digging turned up the following relevant results.

Theorems 2.1.4 and 2.1.5 of DRRT show every countably compact space is star finite and that the implication can be reversed for Hausdorff spaces. Thus, every countably compact space is star compact (as you already knew). Now suppose that $X$ is star compact, and let $\mathscr{U}$ be an open cover of $X$. There is a compact $K\subseteq X$ such that $\operatorname{st}(K,\mathscr{U})=X$. Let $\mathscr{V}\;$ be a finite subset of $\mathscr{U}$ covering $K$; clearly $\operatorname{st}(\cup\mathscr{V},\mathscr{U})=X$. Thus, $X$ is what is called $1$-starcompact in DRRT, and Theorem 2.1.6 of DRRT implies that $X$ is pseudocompact. Every $T_4$ pseudocompact space is countably compact, so in $T_4$-spaces the following are equivalent:

1. star compactness
2. pseudocompactness
3. countable compactness
4. star finiteness.

However, Song shows in this paper that the space

$$\big(\beta\omega\times(\omega_1+1)\big)\setminus \big(\beta\omega\setminus\omega \times\{\omega_1\}\big)$$

is a Tikhonov space that is star compact but not countably compact. On the other hand, Lemma 2.2.10 of DRRT implies that every second countable, star compact $T_3$-space is countably compact and that it’s consistent that $\mathfrak{c}>\omega_1$ and every first countable, star compact $T_3$-space of weight less than $\mathfrak{c}$ is countably compact.

Added: The error in the argument that you added to the question comes when you say

So there exists $x_2 \in K\setminus \cup \mathscr{U_1}$ and $\mathscr{U}_2$, such that $x_2$ in the every element of $\mathscr{U}_2$ and $\cup \{\mathscr{U_i}, i= 1,2\}$ can’t cover $X$.

You know that $\mathscr{U}_1$ doesn’t cover $X$, but that just means that there is a point $x_2\in X\setminus \cup\mathscr{U}_1$; the point $x_2$ need not be in $K$. Indeed, after some finite number of steps you must reach a point at which $\cup \{\mathscr{U_i}:1\le i\le n\}$ covers $K$ but not $X$, and then $x_{n+1}$ must be in $X\setminus K$.