Countable connected space where removing $1$ point destroys connectedness

Is there a countable connected space $$(X,\tau)$$ such that for all $$x\in X$$ the space $$X\setminus\{x\}$$ is not connected any more with the induced subspace topology?

• Can’t resist: The one point space satisfies this. Dec 25 '18 at 19:24
• @Thomas: But the empty space is connected! Dec 25 '18 at 20:15
• @FredRohrer This is a convention, but usually, I would say no. For the same reason that 1 isn't prime. Otherwise you cannot say that a space uniquely decomposes into a disjoint union of connected spaces... Dec 25 '18 at 21:32
• @FredRohrer: It depends on the convention indeed. Dec 25 '18 at 21:35
• @FredRohrer Your statement is also true if $\varnothing$ is not connected. I think there are many reasons to want it to not be connected: unique decomposition in connected components; $\hom(X,-)$ preserves coproducts if $X$ is connected; for a (path-)connected space, $\pi_0(X) = *$; a product is connected iff both factors are connected. (This isn't off the top of my head, I'm reading this.) Anyway, this is a bit tangential, and as you say we're certainly rehashing old arguments... Dec 26 '18 at 8:54

3 Answers

Let $$\mathbb{R}$$ be with its usual topology, and let $$f:\mathbb{R}\to \mathbb{Z}$$ defined by: $$f(x)=\left\{\begin{matrix} 2k & x=2k,\text{ where } k\in\mathbb{Z}\\ 2k+1 & 2k Let $$X$$ be $$\mathbb{Z}$$ with the quotient topology induced by $$f$$.

• In other words it is the Alexandroff topology for the partial order$$\cdots<a_{-3}>a_{-2}<a_{-1}>a_0<a_1>a_2<a_3>\cdots,$$right? Dec 25 '18 at 18:47
• @მამუკაჯიბლაძე Indeed Dec 27 '18 at 9:55
• @მამუკაჯიბლაძე what endpoint? Dec 30 '18 at 17:28
• I was responding to a comment that is no longer there. In any proper connected subspace of your example there are points $k$ such that either $k+1$ or $k-1$ does not belong to the subspace. It is natural (I think) to call these endpoints. If one removes such point from the subspace, what remains will be connected. Dec 30 '18 at 17:32

The other answer describes the "Khalimsky line". It is not $$T_1$$, but it is possible to obtain Hausdorff examples by starting with a countable connected Hausdorff space $$X$$, blowing up its points into more copies of $$X$$, and continuing this process infinitely many times. This ever-branching countable "tree" of $$X$$'s can be topologized so that it is connected, Hausdorff, and removing any point disconnects the space.

• Could you provide more detail? I don't think it is easy to see. I certainly don't see it at all. Dec 25 '18 at 18:52

To be a bit more explicit: in Countable connected spaces Proc. Amer. Math. Soc. 26 (1970) 355-360 G. G. Miller describes a countable connected Urysohn space with a dispersion point.

• That doesn't answer the question - it asks for a space where removal of any point disconnects the space. Dec 29 '18 at 20:14
• Indeed, I reacted more to the title than to the content of the question. Still, it indicates that a construction may be possible, though difficult. Dec 30 '18 at 22:02
• The answer by @ForeverMozart seems to contain such construction, but I could not see the details, only the devil in them Dec 31 '18 at 16:16
• The operative word being `seems' Jan 3 '19 at 9:16