Let $(X,\tau)$ be a topological space. With $T\_2(\tau)$ we denote the collection of $T\_2$-topologies on $X$ that contain $\tau$. 
Is there an example of a topology $\tau$ such that the partially ordered set $(T\_2(\tau),\subseteq)$ contains no minimal members?



Topologies with no minimal $T_2$ topologies above them
*There exists a topology $\tau$ with two non-isolated points on a countable set $X$ such that the poset $T\_2(\tau)$ does not have minimal elements.*


To construct such topology $\tau$, 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 $\tau$ be the topology on the countable set $X=\omega\cup\{-\infty,+\infty\}$ consisting of sets $U\subset X$ such that


$\bullet$ if $-\infty\in U$, then there exists $\alpha\in\omega\_1$ such that $B\_\alpha\subset^\* U$;


$\bullet$ if $+\infty \in U$, then there exists $\alpha\in\omega\_1$ such that $\omega\setminus A\_\alpha\subset^\* U$.


It is clear that $-\infty$ and $+\infty$ are unique non-isolated points of the topological space $(X,\tau)$.


We claim that no topology $\sigma$ in the poset $T\_2(\tau)$ is minimal.


Since $\sigma$ is Hausdorff, the points $-\infty,+\infty$ have disjoint neighborhoods $U\_-,U\_+\in\sigma$.


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.


If the set $A$ is countable, then 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 $\tau\subset\sigma'\subset\sigma$ and $\sigma'\ne\sigma$ as $U\_-\in\sigma\setminus\sigma'$.


The topology $\sigma'$ is Hausdorff since $U\_-\cup A\_\alpha$ and $U\_+\setminus A\_\alpha$ are disjoint neighborhoods of $-\infty$ and $+\infty$, respectively.


If the set $B$ is countable, then 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 $\tau\subset\sigma'\subsetneq\sigma$.


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.


In both cases we have constructed a strictly weaker topology $\sigma'\subset \sigma$ in $T\_2(\tau)$ witnessing that the topology $\sigma$ is not minimal.


