Given any poset $(P,\leq)$ and $x, y\in P$ we set $[x,y] = \{p\in P: x\leq p \leq y\}$. For any set $X$, let $\text{Top}(X)$ denote the set of topologies on $X$. The set $\text{Top}(X)$ is a complete lattice with respect to $\subseteq$.


Is there a set $X$ and $\tau \in \text{Top}(X)$ such that


* $\tau \neq \{\emptyset, X\}$, and
* for all $\sigma \in \text{Top}(X)$ with $\sigma\subseteq \tau$ and $\sigma\neq\tau$ we have $[\sigma,\tau] \neq \{\sigma,\tau\}$? (In other words, $\tau$ has no direct lower neighbor.)




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**EDIT**. As Will Brian pointed out the remark below is erroneous.


Every non-discrete topology $\tau$ has a direct upper neighbor: Pick $x\in X$ such that $\{x\}\notin \tau$, then the topology generated by $\tau\cup\{x\}$ is a direct upper neighbor.



Topology with no direct lower neighbor
An affirmative answer to the question appears in Claim 1 of
[this answer](https://mathoverflow.net/q/237249).


The claim that the topology generated by $\tau\cup\{\{x\}\}$ is a cover of $\tau$ in the lattice
of topologies seems not to be true. For example, if $\tau$ is the usual topology 
on the real line and $x=0$, then the topology generated by $\tau\cup\{[0,\infty)\}$
lies strictly between $\tau$ and the topology generated by $\tau\cup\{\{0\}\}$.


