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?
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asked Dec 25, 2018 at 8:35
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1$\begingroup$ Can’t resist: The one point space satisfies this. $\endgroup$– Thomas RotCommented Dec 25, 2018 at 19:24
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1$\begingroup$ @Thomas: But the empty space is connected! $\endgroup$– Fred RohrerCommented Dec 25, 2018 at 20:15
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1$\begingroup$ @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... $\endgroup$– Najib IdrissiCommented Dec 25, 2018 at 21:32
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$\begingroup$ @FredRohrer: It depends on the convention indeed. $\endgroup$– Thomas RotCommented Dec 25, 2018 at 21:35
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1$\begingroup$ @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... $\endgroup$– Najib IdrissiCommented Dec 26, 2018 at 8:54
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3 Answers
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Let $\mathbb{R}$ be with its usual topology, and let $f:\mathbb{R}\to \mathbb{Z}$ defined by: $$ f(x)=\left\{\begin{matrix} 2k & x=2k,\text{ where } k\in\mathbb{Z}\\ 2k+1 & 2k<x<2k+2,\text{ where } k\in\mathbb{Z} \end{matrix}\right. $$ Let $X$ be $\mathbb{Z}$ with the quotient topology induced by $f$.
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2$\begingroup$ 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? $\endgroup$ Commented Dec 25, 2018 at 18:47
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$\begingroup$ 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. $\endgroup$ Commented Dec 30, 2018 at 17:32
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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.
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1$\begingroup$ Could you provide more detail? I don't think it is easy to see. I certainly don't see it at all. $\endgroup$ Commented Dec 25, 2018 at 18:52
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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.
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1$\begingroup$ That doesn't answer the question - it asks for a space where removal of any point disconnects the space. $\endgroup$– WojowuCommented Dec 29, 2018 at 20:14
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$\begingroup$ 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. $\endgroup$– KP HartCommented Dec 30, 2018 at 22:02
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$\begingroup$ The answer by @ForeverMozart seems to contain such construction, but I could not see the details, only the devil in them $\endgroup$ Commented Dec 31, 2018 at 16:16
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