${\Omega \choose T}\neq {\Omega \choose \Gamma}$

TL;DR

This paper demonstrates that the selection principles ${ m oldsymbol{igOmega}race T}$ and ${ m oldsymbol{igOmega}race oldsymbol{igGamma}}$ are not equivalent by constructing a specific topological space that satisfies one but not the other, answering a previously open question.

Contribution

It provides a concrete example of a topological space distinguishing between the principles ${ m oldsymbol{igOmega}race T}$ and ${ m oldsymbol{igOmega}race oldsymbol{igGamma}}$, resolving an open problem.

Findings

Constructed a topological space satisfying ${ m oldsymbol{igOmega}race T}$ but not ${ m oldsymbol{igOmega}race oldsymbol{igGamma}}$

Answered an open question from arXiv:math/0301011

Showed the principles are not equivalent in general

Abstract

We show that the selection principles ${\Omega\choose T}$ and ${\Omega\choose\Gamma}$ are not equal constructing a topological space $(X,\tau)$ that satisfies ${\Omega \choose T}$, but not ${\Omega \choose \Gamma}$. This answers a question from arXiv:math/0301011 .

Equations

$$U^{\ast }\notin \{U_1,\dots ,U_n\},$$

$$\beta =\bigcup S\in {\aleph }_1.$$

Full text

1. Introduction and definitions

Throughout this note, let $(X,\tau)$ be a topological space. 111The definitions in this note also make sense in the broader context of hypergraphs $H=(V,E)$, where $V$ is any set and $E\subseteq\mathcal{P}(V)$.

Definition 1.1**.**

We say that ${\mathcal{U}}\subseteq\tau$ is an open cover, or cover for short, if

1. (1)

$X\notin{\mathcal{U}}$, 2. (2)

$\bigcup{\mathcal{U}}=X$.

For $x\in X$ we set ${\mathcal{U}}_{x}=\{U\in{\mathcal{U}}:x\in U\}$ and call ${\mathcal{U}}_{x}$ the star of $x$ (with respect to ${\mathcal{U}}$).

1.1. Thick covers

We call a cover ${\mathcal{U}}\subseteq\tau$

1. (1)

large if ${\mathcal{U}}_{x}$ is infinite for every $x\in X$, 2. (2)

an $\omega$-cover if every finite subset of $X$ is contained in some member of ${\mathcal{U}}$, 3. (3)

a $t$-cover if ${\mathcal{U}}$ is large, and for all $x,y\in X$ at least one of the sets ${\mathcal{U}}_{x}\setminus{\mathcal{U}}_{y}$ and ${\mathcal{U}}_{y}\setminus{\mathcal{U}}_{x}$ is finite, and 4. (4)

a $\gamma$-cover if ${\mathcal{U}}$ is infinite, and for every $x\in X$ the set ${\mathcal{U}}\setminus{\mathcal{U}}_{x}$ is finite.

Let $\Lambda,\Omega,T,\Gamma$ denote the collections of (open) large covers, $\omega$-covers, $t$-covers, and $\gamma$-covers, respectively. An easy argument shows that every $\gamma$-cover is large and therefore a $t$-cover, so $\Gamma\subseteq T$.

1.2. The selection principle

If ${\mathfrak{U}},{\mathfrak{V}}$ are families of covers of $X$, then we define the property ${{\mathfrak{U}}\choose{\mathfrak{V}}}$, read “$\mathfrak{U}$ choose $\mathfrak{V}$”, as follows:

${{\mathfrak{U}}\choose{\mathfrak{V}}}\;:$ For each ${\mathcal{U}}\in{\mathfrak{U}}$ there is ${\mathcal{V}}\subseteq{\mathcal{U}}$ such that ${\mathcal{V}}\in{\mathfrak{V}}$.

2. Construction of the example

We consider the space $(\aleph_{1},\tau)$ where $\aleph_{1}$ is the smallest uncountable cardinal and $\tau$ is the collection of down-sets in the cardinal $\aleph_{1}$, that is $\tau=\aleph_{1}\cup\{\aleph_{1}\}$.

Proposition 2.1**.**

If ${\mathcal{U}}\subseteq(\tau\setminus\{\aleph_{1}\})$ is a cover, then:

1. (1)

${\mathcal{U}}$* is a large cover,* 2. (2)

${\mathcal{U}}$* is an $\omega$-cover,* 3. (3)

${\mathcal{U}}$* is a $t$-cover, but* 4. (4)

${\mathcal{U}}$* is never a $\gamma$-cover.*

Proof.

(1) Let $\alpha\in\aleph_{1}$. Suppose $\alpha$ is only covered by finitely many members $U_{1},\ldots U_{n}\in{\mathcal{U}}$. But then, $\beta:=\bigcup{\mathcal{U}}\in\aleph_{1}$ is not covered by any member of $\{U_{1},\ldots,U_{n}\}$. We have $\beta>\alpha$, and since ${\mathcal{U}}$ is a covering of $\aleph_{1}$, there is $U^{*}\in{\mathcal{U}}$ covering $\beta$ and therefore $\alpha$. Clearly

[TABLE]

contradicting the assumption that the only members of ${\mathcal{U}}$ covering $\alpha$ are $U_{1},\ldots,U_{n}$.

(2) Let $S\subseteq\aleph_{1}$ be finite, and consider

[TABLE]

Then $\beta$ is contained in some $U\in{\mathcal{U}}$, so $S\subseteq U$.

(3) Let $\alpha,\beta\in\aleph_{1}$. We may assume that $\alpha<\beta$. Then ${\mathcal{U}}_{\alpha}\setminus{\mathcal{U}}_{\beta}=\emptyset$, which is finite, so ${\mathcal{U}}$ is a $t$-cover.

(4) Suppose that ${\mathcal{U}}\subseteq\tau\setminus\{\aleph_{1}\}$ is a $\gamma$-cover. So every $\alpha\in\aleph_{1}$ is contained in all but finitely many members of ${\mathcal{U}}$. Note that due to the special nature of cardinals, we have ${\mathcal{U}}\subseteq\aleph_{1}$. So ${\mathcal{U}}$ is a well-ordered set such that all members only finitely many predecessors. This implies that ${\mathcal{U}}$ is either a finite or countable collection of members of $\aleph_{1}$ such that $\bigcup{\mathcal{U}}=\aleph_{1}$. This contradicts the fact that $\aleph_{1}$ is a regular cardinal. $\Box$

By proposition 2.1 (2) and (3), every $\omega$-cover in the space $(\aleph_{1},\tau)$ with $\tau=\aleph_{1}\cup\{\aleph_{1}\}$ is a $t$-cover, so the property ${\Omega\choose T}$ is trivially true. On the other hand, by proposition $\ref{main}(4)$, no cover is a $\gamma$-cover, therefore property ${\Omega\choose\Gamma}$ is false.

So for the space $(\aleph_{1},\aleph_{1}\cup\{\aleph_{1}\})$ we have ${\Omega\choose T}\neq{\Omega\choose\Gamma}$.

As a final remark, the property ${\Omega\choose\Gamma}$ is the celebrated Gerlits-Nagy $\gamma$-property [1]. Since we have seen that $\Gamma\subseteq T$, property ${\Omega\choose\Gamma}$ always implies ${\Omega\choose T}$. But the converse is not true, as Proposition 2.1 shows that there is a space where ${\Omega\choose T}$ is true but ${\Omega\choose\Gamma}$ is false.

Acknowledgement. I am grateful to Boaz Tsaban for a fruitful discussion via e-mail on the subject of this note.