# 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. – Thomas Rot Dec 25 '18 at 19:24
• @Thomas: But the empty space is connected! – Fred Rohrer 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... – Najib Idrissi Dec 25 '18 at 21:32
• @FredRohrer: It depends on the convention indeed. – Thomas Rot 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... – Najib Idrissi 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? – მამუკა ჯიბლაძე Dec 25 '18 at 18:47
• @მამუკაჯიბლაძე Indeed – user49822 Dec 27 '18 at 9:55
• @მამუკაჯიბლაძე what endpoint? – user49822 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. – მამუკა ჯიბლაძე 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.

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. – Wojowu 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. – KP Hart 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 – მამუკა ჯიბლაძე Dec 31 '18 at 16:16
• The operative word being `seems' – KP Hart Jan 3 at 9:16