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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 $\subseteq$.

Is there a set $X$ and $\tau \in \text{Top}(X)$ such that

  • $\tau \neq \{\emptyset, X\}$, and
  • for all $\sigma \in \text{Top}(X)$ with $\sigma\subseteq \tau$ and $\sigma\neq\tau$ we have $[\sigma,\tau] \neq \{\sigma,\tau\}$? (In other words, $\tau$ has no direct lower neighbor.)

EDIT. As Will Brian pointed out the remark below is erroneous.

Every non-discrete topology $\tau$ has a direct upper neighbor: Pick $x\in X$ such that $\{x\}\notin \tau$, then the topology generated by $\tau\cup\{x\}$ is a direct upper neighbor.

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    $\begingroup$ Your comment about direct upper neighbors is incorrect. For example, let $\tau$ be the usual topology on $\omega+1$ (making it a convergent sequence). The refinement of $\tau$ that you're describing is the discrete topology. But this is not a direct upper neighbor of $\tau$. In fact, it's a good exercise to show that $\tau$ does not have any direct upper neighbors. (Proof sketch: Every non-discrete refinement of $\tau$ corresponds to a non-principal filter on $\omega$, with $\tau$ itself corresponding to the Frechet filter. The Frechet filter has no "neighbor" in the lattice of filters.) $\endgroup$ – Will Brian Apr 28 '16 at 14:17
  • $\begingroup$ Oh wow .. my intuition was false again. Thanks for pointing this out, @Will. Seems like $\text{Top}(X)$ is treacherous territory... $\endgroup$ – Dominic van der Zypen May 1 '16 at 19:14
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An affirmative answer to the question appears in Claim 1 of this answer.

The claim that the topology generated by $\tau\cup\{\{x\}\}$ is a cover of $\tau$ in the lattice of topologies seems not to be true. For example, if $\tau$ is the usual topology on the real line and $x=0$, then the topology generated by $\tau\cup\{[0,\infty)\}$ lies strictly between $\tau$ and the topology generated by $\tau\cup\{\{0\}\}$.

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