Given a set $X\neq \emptyset$ it is well-known that the collection $\text{Top}(X)$ of all topologies on $X$ is a (complete) lattice with respect to $\subseteq$.


Let $0$ denote the smallest element of the lattice - in our case it is the indiscrete topology $\{\emptyset, X\}$. Given a lattice $L$ with a bottom element $0$ and $x\in L$ we say that $x^\*\in L$ is a [pseudocomplement](https://en.wikipedia.org/wiki/Pseudocomplement) if


1. $x\wedge x^\* = 0$, and
2. if $x\wedge y = 0$ then $y\leq x^\*$.


(In other words, $x^\*$ is the largest element such that the infimum with $x$ is $0$.)


Question: Is there a set $X$ and $\tau\in\text{Top}(X)$ such that $\tau$ does not have a pseudocomplement?



Pseudocomplements in the lattice of topologies
Yes, and in fact, most familiar topologies do not have a pseudo-complement.


To see this, notice that that it often happens with a topology
$\tau$ on a set $X$ that there are non-open sets $A$ and $B$ with
$A\cap B$ open but not empty or $X$. For example, in the usual
topology on the reals, construct $A$ and $B$ by adding different
junk to each side of an interval, so that the intersection is the
interval.


I claim that any topology with such sets $A$ and $B$ will not have
a pseudo-complement. To see this, suppose $\tau^\*$ is a
pseudo-complement. Notice that the topologies $\{\emptyset,A,X\}$
and $\{\emptyset,B,X\}$ will both meet with $\tau$ to $0$ in your
sense, and so they would both have to be below $\tau^\*$. So $A$
and $B$ will both be open in $\tau^\*$, and so $A\cap B$ will
also be open with respect to $\tau^\*$, contradicting that $A\cap
B$ is not in the indiscrete topology.


