Lattice-theoretic complement of the Euclidean topology

Question

It is known that the class of topologies $Top(X)$ of a given set $X$ is a bounded complete lattice (least element is the indiscrete topology and greatest element is the discrete topology, whilst the meet of any family of topologies is the intersection of such family). In the case $X=\mathbb R$, what can be said about lattice-theoretic complements of the Euclidean topology? Do they even exist? Can one of them be constructed?

Answer 1

Here is one way to obtain such a complement. Partition $\mathbb{R}$ into countable dense (with respect to the Euclidean topology) subsets $(A_i)$ (for instance, the $A_i$ could be the cosets of $\mathbb{Q}$). Enumerate each $A_i$ as $\{a^i_n:n\in\mathbb{N}\}$. Let $T$ be the collection of all sets $U\subseteq\mathbb{R}$ with the property that if $a^i_n\in U$, then $a^i_m\in U$ for all $m\leq n$. It is easy to see that $T$ is a topology.

Suppose $U\in T$ and $U$ is not all of $\mathbb{R}$. Let $x\in\mathbb{R}\setminus U$; then $x=a^i_n$ for some $i$ and $n$. Since $U\in T$, this means $a^i_m\not\in U$ for all $m\geq n$. Since $A_i$ is dense in $\mathbb{R}$ with respect to the Euclidean topology, so is $\{a^i_m:m\geq n\}$. So $U$ has empty interior with respect to the Euclidean topology, and in particular cannot be open in the Euclidean topology unless it is empty. This shows that the intersection of $T$ and the Euclidean topology is the indiscrete topology.

On the other hand, note that for any $x\in \mathbb{R}$, there exists $U\in T$ such that $x\in U$ and $U$ is finite. Namely, if $x=a^i_n$, let $U=\{a^i_m:m\leq n\}$. There then exists an open interval $V$ such that $U\cap V=\{x\}$. So, $\{x\}$ is open in the join of $T$ and the Euclidean topology. Since $x\in\mathbb{R}$ was arbitrary, this shows the join is the discrete topology.

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Since you can vary the $(A_i)$ and their enumerations to get different topologies, this construction in fact gives many $2^{2^{\aleph_0}}$ different complements to the Euclidean topology. In fact, by an additional variation on this construction, you can get $2^{2^{2^{\aleph_0}}}$ different complements to the Euclidean topology.

Let $(A_i)_{i\in I}$ be as above and choose their enumerations $A_i=\{a^i_n:n\in\mathbb{N}\}$ such that the set $B=\{a^i_0:i\in I\}$ has intersection of size $2^{\aleph_0}$ with every interval and $0\in B$. Let $F$ be any ultrafilter on $B$ which contains the cobounded filter. Let $T_F$ be the set of elements $U\in T$ (as defined above) with the additional property that if $0\in U$, then $U\cap B\in F$. Then $T_F$ is a topology and $F$ can be recovered from $T_F$ as the set of subsets $S\subseteq B$ such that $\{0\}\cup S\in T_F$ (this works since every subset of $B$ is in $T$). Since there are $2^{2^{2^{\aleph_0}}}$ different choices of $F$, this gives $2^{2^{2^{\aleph_0}}}$ different topologies $T_F$, which I claim are all complements to the Euclidean topology.

Since $T_F\subset T$, it is clear that the intersection of $T_F$ and the Euclidean topology is the indiscrete topology. If $x$ is not in the $A_i$ that contains $0$, then $x$ has the same neighborhoods with respect to $T_F$ as with respect to $T$, and so $\{x\}$ is open in the join of $T_F$ and the Euclidean topology. Finally, suppose $x$ is in the $A_i$ that contains $0$; say $0=a^i_0$ and $x=a^i_n$. Let $C=\{a^i_0,\dots,a^i_n\}$. Since $F$ contains the cobounded filter, there is some $S\in F$ which is disjoint from an open interval $V$ which contains $C$. The set $C\cup S$ is then in $T_F$, and $(C\cup S)\cap V=C$. Since $C$ is finite, we can intersect it with another open interval to find that $\{x\}$ is open in the join of $T_F$ and the Euclidean topology.