Let $X$ be an infinite set and $\mathcal{B} \subseteq \mathcal{T}_{cof}$ be an open base then...

Question

I was given the following proposition in Topology and wasn't told if it's actually true or not.

Let $X$ be an infinite set and $\mathcal{B} \subseteq \mathcal{T}_{cof}$ be an open base then

There exists an infinite set $S \subseteq \mathbb{N} $ such that $\mathcal{B}(S)\subseteq \mathcal{B}$

Where $\mathcal{T}_{cof} := \{A \subseteq X : X\backslash A$ is finite$\}$ is the cofinite topology

and $\mathcal{B}(S) := \{A \subseteq X : |X\backslash A|\in S \}$

I still don't know if it's true, I've tried using:

$S := \mathbb{N}$

$S:=\{n\in \mathbb{N}:n=|X\backslash A|$ for some $A \in \mathcal{B} \}$

But I fail at even proving that $\mathcal{B}(S)\subseteq \mathcal{B}$. And I can't come up with a counterexample.

Could anyone give me any hint?

Thanks

Answer 1

The statement is false. Counter-example:

Consider $X=\mathbb{N}$ and $\tau$ it's co-finite topology, and define close sets family $$\mathcal{C}:=\lbrace\lbrace1\rbrace,\lbrace2,3\rbrace,\lbrace4,5,6\rbrace,...\rbrace$$ formally $$\mathcal{C}=\Bigg\lbrace\Big\lbrace i\Big\rbrace_{i=\frac{k(k-1)}{2}}^{\frac{k(k+1)}{2}}\Bigg\rbrace_{k\in\mathbb{N}^*}$$ and define $\mathcal{B}:=\lbrace U\in\tau|X\setminus U\notin\mathcal{C}\rbrace$. Is easy to see that $\forall n\in\mathbb{N}^*,\exists U\subset X:|X\setminus U|=n\land U\notin \mathcal{B}$, so there can't be some $S\subset\mathbb{N}:\mathcal{B}(S)\subset\mathcal{B}$. All we need is to probe that $\mathcal{B}$ is a base. Let $x\in X$ and $U\in\tau_x$. If $U\in\mathcal{B}$ we win so consider $U\notin\mathcal{B}$, so $C=X\setminus U\in\mathcal{C}$. As $C$ is finite, consider $y\in U\setminus\lbrace x\rbrace$. By definition $F=C\cup\lbrace y\rbrace\notin\mathcal{C}$ so $V=X\setminus F\in\mathcal{B}$ and since $x\neq y$, $x\in V\subset U$.