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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 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?

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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.

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    $\begingroup$ One can make a similar argument by considering a nontrivial open set $U$ that is the union of two non-open sets. $\endgroup$ – Joel David Hamkins Mar 29 '16 at 13:14

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