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Is there a $\text{T}_1$ topology on $\mathbb Q$ with these properties? It should be connected such removing any point disconnects the space.

Forever Mozart
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  • I'm coming up dry on this one. If I had to guess I'd say it's not possible. But I don't have a proof either way. – Gregory Grant Jan 02 '16 at 00:30
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    @GregoryGrant My idea is to take the one-point compactification $\mathbb Q ^=\mathbb Q \cup {\infty}$ of $\mathbb Q$ (which is $T_1$ and connected) and extend each point to be the $\infty$ point of another copy of $\mathbb Q ^$, then extend again, etc, $\omega$-many times. This should be a countable space with the proeprties I wanted. I have to think about exactly how to do this. – Forever Mozart Jan 02 '16 at 00:34
  • You're extending the space, do you want the property on the extended space or on the restriction back to $\Bbb Q$? Because the one-point compactification would induce the finite complement topology on $\Bbb Q$ and that doesn't satisfy your property. – Gregory Grant Jan 02 '16 at 00:37
  • $\mathbb Q$ with the order topology is a subspace of $\mathbb Q ^$. So if I remove $\infty$, I disconnect $\mathbb Q ^$. – Forever Mozart Jan 02 '16 at 00:39
  • Ok right, but there's still the same issue, if you restrict your topology back to $\Bbb Q$, it can't satisfy your criterion because $\Bbb Q$ will not be connected. Perhaps you need to start with a different topology than the order topology? Or maybe I'm completely missing the point. – Gregory Grant Jan 02 '16 at 00:47
  • A $T_0$ space satisfying this isn't hard - e.g. take a poset with a maximal element, where each element covers two more and is covered by exactly one - so a binary tree - and let open sets be the downwards closed ones. I can't seem to find any similar construction that gives a $T_1$ space though. – Milo Brandt Jan 02 '16 at 05:21
  • @MiloBrandt Instead of downward closed sets, what about cofinite downward closed sets (downward closed sets minus finite number of points)? – Forever Mozart Jan 02 '16 at 05:26
  • @ForeverMozart I thought of that too, but that condition isn't closed under unions; there are countably many disjoint downwards closed sets. Take a point out of each of them and you have a set which is not cofinite downwards closed. – Milo Brandt Jan 02 '16 at 05:28
  • @ForeverMozart Oh wait, my original example doesn't even work - oops! Taking off the top element leaves it disconnected, but taking off lower elements does not, since the only open set containing the top element is the whole set. (This is true enough to be a problem in the generated topology as well) – Milo Brandt Jan 02 '16 at 05:35

1 Answers1

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You should definitely doublecheck, since I’m not entirely awake, but I think that this does what you want; it’s essentially your idea, but with the topology simplified a bit. Let $X$ be the set of all finite sequence of rationals, including the empty sequence $\varnothing$. If $n\in\omega\setminus 1$, $p=\langle q_1,\ldots,q_n\rangle\in X$, $U$ is an open nbhd of $q_n$ in $\Bbb Q$, and $K$ is a compact set in $\Bbb Q$, define

$$\begin{align*} &B(p,U,K)=\\ &\{\langle q_1,\ldots,q_{n-1},q_n',\ldots,q_m'\rangle:n\le m\in\omega\text{ and }q_n'\in U\text{ and }q_k'\notin K\text{ for }n<k\le m\}\;. \end{align*}$$

Take

$$\mathscr{B}(p)=\{B(p,U,K):U\text{ is an open nbhd of }p\text{ in }\Bbb Q\text{ and }K\subseteq\Bbb Q\text{ is compact}\}$$

as a local base at $p$ in $X$. For $K$ a compact subset of $\Bbb Q$ let

$$B(\varnothing,K)=\{\langle q_1,\ldots,q_n\rangle:n\in\omega\setminus 1\text{ and }q_k\notin K\text{ for }1\le k\le n\}\;,$$

and take

$$\mathscr{B}(\varnothing)=\{B(\varnothing,K):K\subseteq\Bbb Q\text{ is compact}\}$$

as a local base at $\varnothing$.

Brian M. Scott
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  • Yes! That's what I wanted to do but I just couldn't find a way to write it. I still need to check the details because it's a little subtle, but initially it looks good. – Forever Mozart Jan 02 '16 at 21:50
  • @ForeverMozart: Great! I did a quick check, but I didn’t try to write it out carefully, so I could have overlooked something. – Brian M. Scott Jan 02 '16 at 21:52