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 $\subseteq$.

Given $\tau\in\text{Top}(X)$ and $E\subseteq X$ we set $\tau_E$ to be the topology generated by $\tau\cup\{E\}$.

If $\tau\subseteq\tau'$ are members of $\text{Top}(X)$ with $\tau\neq\tau'$, is there $E\in\tau'\setminus\tau$ such that $[\tau,\tau_E]=\{\tau,\tau_E\}$? (In other words, there is no topology strictly between $\tau$ and $\tau_E$.)

(This question was inspired by a comment that Ramiro de la Vega made on this question.)


1 Answer 1


The answer is negative.

For $X=\mathbb R$, let $\tau=\{(a,\infty)\subseteq\mathbb R\mid a\in[-\infty,\infty]\}$ and let $\tau'$ be the Euclidean topology. Choose any set $E\in\tau'\setminus\tau$. We can express $E$ as a countable disjoint union of open intervals $$E=\bigcup_{n=1}^{\infty}(a_n,b_n),\qquad-\infty\leq a_n\leq b_n\leq\infty.$$ Since $E\notin\tau$, we may additionally assume that $a_1<b_1<\infty$. Now choose any $c\in(a_1,b_1)$ and let $E'=E\cap(c,\infty)$. Then $\tau_{E'}$ lies strictly between $\tau$ and $\tau_E$.


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