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](https://mathoverflow.net/questions/237002/lower-neighbors-in-the-lattice-of-topologies) question.)



"Discrete jumps" in the collection of all topologies on a set $X$
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 $\tau\_E$.


