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?



Minimal Hausdorff topologies compatible with a bunch of functions
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](https://mathoverflow.net/questions/289740/maximal-elements-in-the-partially-ordered-set-of-image-spaces/292637#292637) to [this MO problem](https://mathoverflow.net/questions/289740/maximal-elements-in-the-partially-ordered-set-of-image-spaces). 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 $T\_2(\mathcal F)$.


