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Let $X$ be an infinite set, let ${\cal F}$ be a set of functions $f: X\to X$. We say that a topology $\tau$ is compatible with ${\cal F}$ if every $f\in {\cal F}$ is a continuous function $f:(X, \tau)\to (X,\tau)$. We denote the collection of $T_2$-topologies compatible with ${\cal F}$ by $T_2({\cal F})$.

Note that $\big(T_2({\cal F}), \subseteq\big)$ is a poset; it is always non-empty and the discrete topology ${\cal P}(X)$ is its largest element.

Question: Is there an infinite set $X$ and a set ${\cal F}$ of functions $f:X\to X$ such that $\big(T_2({\cal F}\big), \subseteq)$ does not contain minimal elements?

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    $\begingroup$ What would these minimal elements be good for? i.e. Could you provide some motivation for this question? $\endgroup$ – Ramiro de la Vega Nov 18 '15 at 16:04
  • $\begingroup$ Good point, Ramiro. Two answers: 1) I foolishly thought you might find a smallest topology compatible with a collection of function -> that's trivial, because the indiscrete topology is always compatible! So I thought, you might find a smallest $T_2$-topology - but this is almost always wrong. So then I started wondering about minimal such topologies. $\endgroup$ – Dominic van der Zypen Nov 19 '15 at 7:46
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    $\begingroup$ 2) I think questions about "minimal/maximal objects with property X" are often interesting for their own sake. Consider for instance the question "Is every compact topology contained in a maximal compact topology"? It was open for 50 years and only settled a couple years ago by Martin Maria Kovar ( drops.dagstuhl.de/volltexte/2005/118/pdf/… ) $\endgroup$ – Dominic van der Zypen Nov 19 '15 at 7:54
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The answer to this question is affirmative:

Theorem. There exists a countable set $X$ and an uncountable family $\mathcal F$ of self-functions of $X$ such that the poset $T_2(\mathcal F)$ has no minimal elements.
Here $T_2(\mathcal F)$ is the poset of all Hausdorff topologies on $X$ making all functions $f\in\mathcal F$ continuous.

Proof. To construct the space $X$ and the family $\mathcal F$, take any Hausdorff $(\omega_1,\omega_1)$-gap on $\omega$, which is a pair $\big((A_\alpha)_{\alpha\in\omega_1},(B_\alpha)_{\alpha\in\omega_1}\big)$ of families of infinite subsets of $\omega$ satisfying the following two conditions:

(H1) for any $\alpha<\beta<\omega_1$ we have $A_\alpha\subset^* A_\beta\subset^* B_\beta\subset^* B_\alpha$;

(H2) for any set $C\subset\omega$ one of the sets $\{\alpha\in \omega_1:A_\alpha\subset^* C\}$ or $\{\alpha\in\omega_1:C\subset^* B_\alpha\}$ is at most countable.

Here the notation $A\subset^* B$ means that the complement $A\setminus B$ is finite.

It is well-known that Hausdorff $(\omega_1,\omega_1)$-gaps do exist in ZFC.

Let $X=\{-\infty,+\infty\}\cup\omega$ and for every $\alpha\in\omega_1$ consider the functions $f_\alpha,g_\alpha:X\to X$ defined by $$f_\alpha(x)=\begin{cases} x&\mbox{if $x\in \{-\infty\}\cup B_\alpha$};\\ +\infty&\mbox{otherwise}; \end{cases} $$ and $$g_\alpha(x)=\begin{cases} x&\mbox{if $x\in \{+\infty\}\cup (\omega\setminus A_\alpha)$};\\ -\infty&\mbox{otherwise}. \end{cases} $$ For every $a,b\in\omega$ let $p_{a,b}:X\to X$ be the function defined by $$p_{a,b}(x)=\begin{cases} b&\mbox{ if $x=a$},\\ a&\mbox{ if $x=b$},\\ x&\mbox{otherwise}. \end{cases} $$

We claim that the family of functions $\mathcal F=\{f_\alpha\}\cup\{g_\alpha\}_{\alpha\in\omega_1}\cup\{p_{a,b}\}_{a,b\in\omega}$ has the required property: the poset $T_2(\mathcal F)$ has no minimal elements.

Indeed, take any Hausdorff topology $\sigma$ on $X$ in which every map $f\in\mathcal F$ is continuous. The continuity of the maps $p_{a,b}$, $a,b\in\omega$, implies that each point of the set $\omega$ is isolated in the topological space $(X,\sigma)$.

We claim that for every $\alpha\in\omega_1$ the set $W_\alpha^-:=B_\alpha\cup\{-\infty\}$ is a neighborhood of $-\infty$ and $W_\alpha^+:=(X\setminus A_\alpha)\cup\{+\infty\}$ is a neighborhood of $+\infty$ in the topology $\sigma$.

For this choose any two disjoint neighborhoods $U_-,U_+\in\sigma$ of $-\infty$ and $+\infty$, respectively. By the continuity of the maps $f_\alpha$ and $g_\alpha$, there are neighborhoods $V_-,V_+\in\sigma$ of $-\infty$ and $+\infty$ such that $f_\alpha(V_-)\subset U_-$ and $g_\alpha(V_+)\subset U_+$. Then $$V_-\subset f_\alpha^{-1}(U_-)\subset f_\alpha^{-1}(X\setminus\{+\infty\})\subset \{-\infty\}\cup B_\alpha$$ and $$V_+\subset g_\alpha^{-1}(U_+)\subset g_\alpha^{-1}(X\setminus\{-\infty\})\subset\{+\infty\}\cup(\omega\setminus A_\alpha).$$

Therefore, the topology $\sigma$ contains the topology $\tau$ defined in the answer to this MO problem. Repeating the argument from this answer it can be shown that $\sigma$ is not a minimal element of the poset $T_2(\mathcal F)$.

However, for the interested reader let us write more details.

By the condition (H2), one of the sets $A=\{\alpha\in\omega_1:A_\alpha\subset^* U_-\}$ or $B=\{\alpha\in\omega_1:U_-\setminus\{-\infty\}\subset^* B_\alpha\}$ is countable.

First we assume that the set $A$ is countable. In this case we can find a countable ordinal $\alpha$ such that $A_\alpha\not\subset^*U_-$.

Consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions:

$\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\cup A_\alpha\subset^* W$;

$\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\subset^* W$.

It is clear that $\sigma'\subset\sigma$ and $\sigma'\ne\sigma$ (because $U_-\in\sigma\setminus\sigma'$).

The topology $\sigma'$ is Hausdorff since $U_-\cup A_\alpha$ and $U_+\setminus A_\alpha=U_+\cap W^+_\alpha$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively.

To see that $\sigma'\in T_2(\mathcal F)$, observe that for any $a,b\in \omega$ the permutation $p_{a,b}$ is continuous as the points $a,b$ are isolated in the topology $\sigma'$. Next, for every ordinal $\beta\in\omega_1$ the inclusion $A_\alpha\subset^* B_\beta$ implies that the map $f_\beta$ is continuous at $-\infty$. The continuity of the map $f_\beta$ at $+\infty$ in the topology $\sigma$ implies the continuity of this map at $+\infty$ in the topology $\sigma'$.

The continuity of the map $g_\beta$ at $-\infty$ in the topology $\sigma'$ follows from the inclusion $g_\beta(x)\in\{-\infty,x\}$ holding for all $x\in X$. The continuity of $g_\beta$ at $+\infty$ in the topology $\sigma'$ follows from the continuity of $g_\beta$ at $+\infty$ in the topology $\sigma$ and the definition of the topology $\sigma'$.

Now consider the case of countable set $B:=\{\alpha\in\omega_1:U_-\setminus\{-\infty\}\subset^* B_\alpha\}$. In this case we can find a countable ordinal $\alpha$ such that $U_-\setminus\{-\infty\}\not\subset^* B_\alpha$ and hence $\omega\setminus B_\alpha\not\subset^* U_+$.

In this case we can consider the topology $\sigma'$ on $X$ consisting of sets $W\subset X$ satisfying two conditions:

$\bullet$ if $-\infty\in W$, then there exists $U\in\sigma$ such that $-\infty\in U\subset^* W$;

$\bullet$ if $+\infty\in W$, then there exists $U\in\sigma$ such that $+\infty\in U\cup(\omega\setminus B_\alpha)\subset^* W$.

It is clear that $\sigma'\subset\sigma$ and $\sigma'\ne\sigma$ (because $U_+\in\sigma\setminus\sigma'$).

The topology $\sigma'$ is Hausdorff since $U_-\cap B_\alpha$ and $U_+\cup(\omega\setminus B_\alpha)$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively. By analogy with the case of countable set $A$, we can show that each map $f\in\mathcal F$ remains continuous in the topology $\sigma'$.

In both cases we have constructed a strictly weaker topology $\sigma'\subset \sigma$ in $T_2(\mathcal F)$ witnessing that the topology $\sigma$ is not minimal in the poset $T_2(\mathcal F)$.

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  • $\begingroup$ Amazing, thanks for this well-written solution, Taras! $\endgroup$ – Dominic van der Zypen Feb 13 '18 at 13:45

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