# Injective choice function for non-separable $T_2$-spaces

For any set $$X$$ and cardinal $$\kappa$$ let $$[X]^\kappa$$ be the collection of all subsets of $$X$$ of cardinality $$\kappa$$.

I was looking for $$T_2$$-spaces $$(X,\tau)$$ with the property that

$$(P)$$ there is an injective function $$f:[X]^\omega\to \tau$$ such that for all $$s\in [X]^\omega$$ we have $$s\subseteq f(s)$$.

Question. If $$(X,\tau)$$ is infinite, Hausdorff, and non-separable, is there a function with the properties described in $$(P)$$?

Note. I had the following remark in the original version of this post, but KP Hart made me aware that it is false - thanks for spotting my mistake!

[False] Obviously, if $$(X,\tau)$$ is separable and contains at least two different countable dense subsets $$s$$, $$s'$$, then the only member of $$\tau$$ that contains $$s$$ or $$s'$$ is $$X$$ itself, so there cannot be an injective function $$f$$ as described in $$(P)$$. [/False]

• The `obviously' is not quite true: in $\mathbb{R}$ we have $\mathbb{Q}\subseteq\mathbb{R}\setminus\{\sqrt2\}$ and $\pi+\mathbb{Q}\subseteq\mathbb{R}\setminus\{0\}$. In fact there is an injective function $f:[\mathbb{R}]^\omega\to\mathbb{R}$ such that $f(A)\notin A$ for all $A$, so $\mathbb{R}$ does admit a function as required. – KP Hart Nov 7 '18 at 11:31
• Right, thanks @KPHart, will modify this! – Dominic van der Zypen Nov 7 '18 at 18:33

Here is a partial answer: if $$|X|^{\aleph_0}=|X|$$ then the answer is yes. First take an injective function $$F:[X\times X]^\omega\to X$$ and then take some function $$G:[X\times X]^\omega\to X$$ such that for all $$A\in[X\times X]^\omega$$ the point $$\langle F(A),G(A)\rangle$$ is not in $$A$$. Translate this via a bijection between $$X$$ and $$X\times X$$ to a function $$H:[X]^\omega\to X$$ such that $$H(A)\notin A$$ for all $$A$$.
Then $$f:A\mapsto X\setminus\{H(A)\}$$ is as required.
Note that this uses nothing but the $$T_1$$-property and works even for the co-finite topology.