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]

  • 3
    $\begingroup$ 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. $\endgroup$ – KP Hart Nov 7 '18 at 11:31
  • $\begingroup$ Right, thanks @KPHart, will modify this! $\endgroup$ – 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.

  • $\begingroup$ Will ask separately for countable spaces $\endgroup$ – Dominic van der Zypen Nov 9 '18 at 9:37

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