Topologies with no minimal $T_2$ topologies above them 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?
 A: 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.
