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Let $X\neq \emptyset$ be a set and let ${\cal J} \subseteq {\cal P}(X)\setminus\{\emptyset\}$ be a collection of non-empty subsets of $X$. We say that a topology $\tau$ on $X$ is ${\cal J}$-compatible if for every $J\in {\cal J}$ there is a continuous surjective map $f:X\to J$, where $J$ carries the subset topology inherited from $(X,\tau)$.

Let $\text{Cptb}(X, {\cal J})$ denote the collection of ${\cal J}$-compatible topologies, ordered by $\subseteq$.

Note that ${\cal P}(X)$ is always the greatest element of $\text{Cptb}(X,{\cal J})$, and the indiscrete topology $\{\emptyset, X\}$ is always the smallest element.

If $\tau\in \text{Cptb}(X, {\cal J})$ and $\tau\neq{\cal P}(X)$, is there a minimal element in $\text{Cptb}(X, {\cal J})$ properly containing $\tau$ (= "above" $\tau$, in the poset we are considering)?

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  • $\begingroup$ No, if $X$ is finite and the topology $\tau$ itself is minimal in $Cptb(X,\mathcal J)$. $\endgroup$
    – Taras Banakh
    Commented Sep 1, 2016 at 14:53
  • $\begingroup$ Thanks @TarasBanakh for your remark. I forgot to exclude $\tau = {\cal P}(X)$, the discrete topology. In the case that $X$ is finite, $\text{Cptb}(X,{\cal J})$ is finite, so if $\tau \neq {\cal P}(X)$, then $\tau$ is not the greatest element of the poset $\text{Cptb}(X,{\cal J})$, therefore there is a minimal element in $\text{Cptb}(X,{\cal J})$ above $\tau$. $\endgroup$
    – Dominic van der Zypen
    Commented Sep 2, 2016 at 13:48
  • $\begingroup$ Setting $\mathcal J = \{X\}$ makes $\mathrm{Cptb}(X,\mathcal J)$ just the lattice of topologies on $X$, and then the answer to your question is known to be negative. Do you want to put further restrictions on $\mathcal J$? $\endgroup$
    – Will Brian
    Commented Sep 2, 2016 at 14:01
  • $\begingroup$ Thanks Will! That's already sufficient for me to see that there are no minimal elements as in the question in general. If you can copy your remark into an answer I'll accept it. - What I'm really keen to see is whether $\text{Cptb}(X,{\cal J})$ is a lattice, see my more recent question on that. $\endgroup$
    – Dominic van der Zypen
    Commented Sep 3, 2016 at 7:56
  • $\begingroup$ OK @DominicvanderZypen, I've done that now. $\endgroup$
    – Will Brian
    Commented Sep 6, 2016 at 14:47

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If we put $\mathcal J = \{X\}$, then $\mathrm{Cptb}(X,\mathcal J)$ is just the lattice of topologies on $X$. In this case, the answer to your question is known: if $X$ is infinite, then not every topology on $X$ has a "minimal" strict refinement.

For example, let $p \in X$ and let $\tau$ be the one-point compactification of $X - \{p\}$: that is, neighborhoods of $p$ are sets of the form $X - F$, where $F$ is a finite subset of $X$ not containing $p$, and every other point is isolated. I claim that $\tau$ has no minimal proper refinement. To see this, suppose $\tau'$ properly refines $\tau$. Then there is some infinite closed $Y \subseteq X - \{p\}$. Split $Y$ into two disjoint infinite sets $Y_0$ and $Y_1$, and define $\sigma$ to be the refinement of $\tau$ obtained by declaring $Y_0$ closed (in other words, $\sigma$ is the topology with subbasis $\tau \cup \{X - Y_0\}$). It is not hard to see that $\sigma$ is strictly between $\tau$ and $\tau'$.

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