# Lower neighbors in the lattice of topologies

Given any poset $(P,\leq)$ and $x, y\in P$ we set $[x,y) = \{p\in P: x\leq p < y\}$, and $(x,y]$ is defined in an analogous manner. For any set $X$, let $\text{Top}(X)$ denote the set of topologies on $X$. It is well-known that $\text{Top}(X)$ is a complete lattice with respect to $\subseteq$.

Does there exist an infinite set $X$ and topologies $\sigma \subseteq \tau$ on $X$ with $\sigma\neq \tau$ such that $[\sigma,\tau)$ does not contain a maximal element?

EDIT. The following remark, which I thought to be trivially true, is false as pointed out by Ramiro de la Vega in the comments below.

The dual question has an easy positive answer: if $\sigma \subseteq \tau$ with $\sigma\neq \tau$ are members of $\text{Top}(X)$, pick $U_0\in\tau\setminus\sigma$ and then the topology generated by $\sigma\cup\{U_0\}$ is a minimal element of $(\sigma,\tau]$.

• The statement "...then the topology generated by $\sigma \cup \{U_0\}$ is a minimal element of $(\sigma, \tau]$..." doesn´t seem to be true. – Ramiro de la Vega Apr 22 '16 at 18:51
• Oh I'm sorry if I got it wrong! Can you give an example? – Dominic van der Zypen Apr 22 '16 at 19:26
• $X=4$, $\tau=\mathcal{P}(X)$, $\sigma=\{\emptyset,\{0,1\},\{2,3\},X\}$, $U_0=\{0,2\}$. – Ramiro de la Vega Apr 22 '16 at 20:58

The answer to the question Does there exist $\dots$'' is Yes.

Let $X = [0,\infty)$ be the nonnegative part of the real line. Let $\tau$ be the topology on $X$ consisting of all sets $[0,r)$ for $r\in X$. Let $\sigma$ be the indiscrete topology on $X$ and let $\rho$ be the restriction of the usual topology of the real line to $X$.

Claim 1. $[\sigma,\tau)$ has no maximal element.

Claim 2. $(\tau,\rho]$ has no minimal element.

Proof of Claim 1: Given a topology $\tau'$ contained in $\tau$, let $U(\tau') = \{r\in X\;|\; [0,r)\in\tau'\}$. That is, $U(\tau')$ is the set of upper limits'' of open sets in $\tau'$. Observe that if $\tau'$ is contained in $\tau$, then $U(\tau')$ is a subset of $X=[0,\infty)$ that contains $0$ and is closed under the formation of suprema (since $\tau'$ is closed under the formation of unions).

Now choose and fix some $\tau'$ contained in $\tau$.

Case 1: $U(\tau')$ is dense in $X$.

In this case, since $U(\tau')$ is closed under sup's, $U(\tau')=X$, so $\tau'=\tau$. Such a $\tau'$ is not in $[\sigma,\tau)$.

Case 2: $U(\tau')$ is not dense in $X$.

There is a small interval $(a-\epsilon,a+\epsilon)$ contained in $X$ and disjoint from $U(\tau')$. The topology generated by $\tau'\cup\{[0,a)\}$ strictly extends $\tau'$ yet is still contained in $\tau$. This shows that $\tau'$ is not maximal in this case.

This completes the proof of Claim 1.

Remember that Claim 2 asserts $(\tau,\rho]$ has no minimal element''.

Proof of Claim 2: Let's reuse the notation $\tau'$, now for a proposed minimal element of $(\tau,\rho]$. Necessarily $\tau'$ can be generated by $\tau$ and one set $A\in\rho\setminus\tau$. Since $A$ is not in $\tau$, it is either (i) not connected in $\langle X;\rho\rangle$ or else (ii) does not contain the element $0$. Whichever is the case, it is possible to find $r\in X$ such that $B:=[0,r)\cap A\;(\in\rho)$ is a proper subset of $A$ that is still not in $\tau$. To finish, we argue that the topology generated by $\tau\cup\{B\}$ does not contain $A$, hence is strictly smaller than $\tau'$.

The topology generated by $\tau\cup\{B\}$ consists of (i) sets from $\tau$, (ii) sets of the form $V\cap B$ where $V\in\tau$, or (iii) any union of a set from (i) with a set from (ii). $A$ is not of this form. By construction, $\sup(B)<\sup(A)$. Any set $W$ of any of the three types which does not belong to $\tau$ must have $\sup(W)\leq\sup(B)<\sup(A)$, hence cannot equal $A$.