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Sep 6, 2016 at 18:40 vote accept Dominic van der Zypen
Sep 6, 2016 at 14:47 comment added Will Brian OK @DominicvanderZypen, I've done that now.
Sep 6, 2016 at 14:46 answer added Will Brian timeline score: 1
Sep 3, 2016 at 7:56 comment added Dominic van der Zypen 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.
Sep 2, 2016 at 14:01 comment added Will Brian 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 $\mathcal J$?
Sep 2, 2016 at 13:55 history edited Dominic van der Zypen CC BY-SA 3.0
excluded a trivial case
Sep 2, 2016 at 13:48 comment added Dominic van der Zypen 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 $\tau$.
Sep 1, 2016 at 14:53 comment added Taras Banakh No, if $X$ is finite and the topology $\tau$ itself is minimal in $Cptb(X,\mathcal J)$.
Aug 28, 2016 at 9:32 history asked Dominic van der Zypen CC BY-SA 3.0