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1) If $X$ is an infinite set and $T_X$ the set of all infinite topologies on $X$ is it in general true that there is no injection $f_T:T_X \to X$?

2) What conditions on $X$ assure an injection (if that´s ever possible)?

By infinite topologies, I mean topologies with an infinite number of sets (subsets of $X$) as its elements.

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    $\begingroup$ For every subset $Y$ of $X$ with $|Y|\le |X|$, there exists a metrizable topology on $X$ for which the set of accumulation points is exactly $Y$. Hence $X$ has $\ge 2^{|X|}$ topologies. This is enough to answer the question. [Each topology being an element of $2^{2^X}$, the upper bound is $2^{2^{|X|}}$, so it would be more interesting to ask whether the number of topologies on $X$ is $2^{2^{|X|}}$. For instance I remember once checking that the number of group topologies on the symmetric group $S_\omega$ is $2^{2^{2^{\aleph_0}}}$.] $\endgroup$
    – YCor
    Commented Mar 24, 2020 at 9:24
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    $\begingroup$ @YCor : the number of ultrafilters on $X$ is $2^{2^{|X|}}$, and there is an injection $\beta X\to T_X$ ( a subset is open iff it is in the ultrafilter) $\endgroup$
    – Maxime Ramzi
    Commented Mar 25, 2020 at 13:32
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    $\begingroup$ @MaximeRamzi thanks! (of course you mean iff it's empty or in the ultrafilter). BTW the number of metrizable topologies is $=2^{|X|}$: the upper bound just consists in counting the number of possible distances. Also if you fix $x_0$ in $X$ and consider, for a nonprincipal ultrafilter $\xi$ on $X-\{x_0\}$, open subsets as finite subsets of $X-\{x_0\}$ and subsets containing $x_0$ and intersecting $X-\{x_0\}$ in an element of $\xi$, one gets $2^{2^{|X|}}$ Hausdorff topologies. $\endgroup$
    – YCor
    Commented Mar 25, 2020 at 13:38

4 Answers 4

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If $X$ is an infinite set, then there are $2^{|X|}$-many ways to well-order $X$. Each such order determines an Alexandroff topology on $X$, which will have infinitely many open sets. The order is recoverable from the topology, so there are too many different infinite topologies on $X$ for there to be an injection $f_T: T_X\to X$.

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  • $\begingroup$ Is much changed if every topology consists of an infinite number of sets such that each of them is also infinite? $\endgroup$
    – user153451
    Commented Mar 24, 2020 at 7:57
  • $\begingroup$ @Ante This will not change much: for each subset $Y$ there is a topology consisting of those sets which contain $Y$ (together with the empty set). These topologies are all different for different $Y$, and for infinite $Y$ they consist of infinite sets only (except for the empty set, which should be excluded by you anyway). $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 24, 2020 at 12:58
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Similar to Keith Kearnes’ answer, but made slightly more concrete:

  • there’s an injection from “partial orders on $X$” to “infinite topologies on $X$”, sending any partial order to its Alexandrov topology, whose open sets are just the down-closed sets of the order. This is always infinite if $X$ is, since the sets $\mathord{\downarrow} \{x \}$ are distinct for each $x \in X$; and it’s an injection since any partial order can be recovered from its Alexandrov topology as the specialisation order.

  • there’s an injection from “proper subsets of $X$” to “partial orders on $X$” sending $S \subseteq X$ to the order where $x <_S y$ just if $x \in S$ and $y \notin S$. (I.e. all of $X \setminus S$ is above all of $S$, and the order is otherwise discrete.) This is an injection since $S$ can be recovered as $\{ x \mathrel{|} \exists y,\ x <_S y \}$.

  • composing these gives an injection from “proper subsets of $X$” to “infinite topologies on $X$”

  • so any injection “infinite topologies on $X$” to $X$ would give an injection from “proper subsets of $X$” to $X$, which is impossible.

I’m assuming throughout that you work in ZFC. Without choice, the earlier parts all work, but the final step breaks, because we don’t know that there’s no injection $\mathcal{P}(X) \to X$.

In a comment, you mention restricting further, to topologies whose (nonempty) open sets are all infinite. That doesn’t essentially change anything, as can be shown by a slight modification of the above construction. Fix a countable and co-infinite set $K \subseteq X$. Now take the injection from “proper subsets of $X \setminus K$” to preorders on $X$, where $\leq_S$ puts $K$ codiscretely at the bottom, $S$ discretely in the middle, and $X \setminus (S \sup K)$ discretely on top. Now send this to its specialisation order. This gives an injection from “proper subsets of $X \setminus K$” to “infinite topologies on $X$ with all nonempty opens infinite”. But since $K$ was countable and co-infinite, $|X \setminus K| = |X|$, so we are done as before.

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Let

$$ \forall_{A\subseteq X}\quad \mathcal B_A \ :=\ \{\{a\} :\ a\in A\}\cup X $$

Then each $\ \mathcal B_A\ $ is a topological base in $\ X\ $ while every two of them generate different topologies in $\ X.\ $ When $\ X\ $ is infinite then

$$ |\,\{\mathcal B_A : \, A\subset X\ \ \text{and}\ \ |A|=\infty \}\,| \,\ =\,\ 2^{|X|} $$

We see that an injection of the set of all topologies in $\ X\ $ into $\ X\ $ is impossible when $\ X\ $ infinite.

Remark 1   There is no such injection also when $\ X\ $ is finite but $\ |X|\ne 1.\ $ Indeed, let

$$ \forall_{A\subseteq X}\quad T_A \ :=\ \{\emptyset\,\ A\,\ X\} $$

These are topologies, and

$$ |\,\{T_A:\,A\subseteq X\}\,|\ =\ 2^{|X|}-1 $$

for every non-empty finite $\, X.\, $ In the empty case we get $\, 0\, $ points and $\, 1\, $ topology, $\ 0<1.$

Remark 2   The situation, cardinality wise, is even more dramatic.

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For an infinite set $X$, there are $2^{2^{|X|}}$ topologies with infinitely many elements on $X$.

The upper bound is obvious, the lower bound is given by my comment (and the slight correction in YCor's): for any ultrafilter $U$ on $X$, define $T_U$ to have as opens the empty set and the elements of $U$. It is clear that this forms a topology, and that you can recover $U$ from it.

Moreover, clearly $T_U$ has infinitely many open sets.

So $U\mapsto T_U$ defines an injection $\beta X\to T_X$, and it's well known that $|\beta X| = 2^{2^{|X|}}$.

So there can be no injection $T_X\to X$ (in fact, no injection $T_X\to P(X)$)

YCor's comment below mine elaborates a bit on what one can ask of these topologies and how many we get (Hausdorff, metrizable)

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