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I know that a connected $T_{3}$ space with more than one point is necessarily uncountable. The proof which I'm familiar with uses the fact a $T_{3}$ space is in particular a $T_{1}$ space. I'm wondering whether the claim remains true if you demand only that the space is regular but not necessarily that it is $T_{1}$.

Thanks in advance!

Serpahimz
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    What about an indiscrete space with exactly two points? – David Mitra Jun 23 '13 at 18:34
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    By the way, you mean "A connected $T_3$ space with more than one point is uncountable". – David Mitra Jun 23 '13 at 19:22
  • Indeed to both comments, I'll correct the question. The Trivial topology is a bit of a dull (albeit correct) example since it is only vacuously connected and vacuously regular. I would be happy to see another example with a topology which is not Trivial. – Serpahimz Jun 23 '13 at 19:26
  • I only knew that connected Tychonoff spaces are uncountable. Could you provide me with a reference for the stronger result? – Stefan Hamcke Jun 23 '13 at 19:58
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    @StefanH. See here. If I recall correctly, a connected Tychonoff space having at least two points has cardinality at least $c$. – David Mitra Jun 23 '13 at 20:20
  • @Stefan: A countable space is Lindelöf, and a regular Lindelöf space is normal. – Brian M. Scott Jun 23 '13 at 20:26
  • @BrianM.Scott: Thanks, I've already checked the link David posted, and saw that it uses a number of facts all of which I once learned about 1) A countable space is Lindelöf. 2) A regular Lindelöf space is paracompact. 3) A regular paracompact space is normal. – Stefan Hamcke Jun 23 '13 at 20:48
  • @DavidMitra: Thank you. So this says that a countable regular space is also completely regular. – Stefan Hamcke Jun 23 '13 at 20:49
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    @Stefan: You don’t need to introduce the complication of paracompactness, though: the proof that a regular, Lindelöf space is normal is pretty straightforward. – Brian M. Scott Jun 23 '13 at 20:49
  • @BrianM.Scott: Okay, I will give it a try :-) – Stefan Hamcke Jun 23 '13 at 20:50
  • @Stefan: No, you don’t immediately get complete regularity, because that doesn’t follow from normality in the absence of $T_1$. What you get is that disjoint closed sets can be functionally separated. – Brian M. Scott Jun 23 '13 at 20:50
  • @BrianM.Scott: If I remember correctly a regular normal space is completely regular. – Stefan Hamcke Jun 23 '13 at 20:51
  • @Stefan: Not true: just take the indiscrete topology on any set with more than one element. Every $T_4$-space is Tikhonov, but you really need $T_1$. – Brian M. Scott Jun 23 '13 at 20:54
  • @BrianM.Scott: By completely regular I mean Tychonoff without $T_1$ ( point and closed set can be functionally separated). Isn't the indiscrete space ${a,b}$ then completely regular? – Stefan Hamcke Jun 23 '13 at 20:56
  • @Stefan: I know; that’s also how I use the term. My example was nonsense. In fact the argument that I gave in my answer actually shows that you do get complete regularity, though you have to work for it a little, since singletons need not be closed. – Brian M. Scott Jun 23 '13 at 20:59
  • @BrianM.Scott: No problem. But I'm still struggling with the regular Lindelöf implies normal. Maybe I'll get back to you if I don't succeed. – Stefan Hamcke Jun 23 '13 at 21:11
  • @Stefan: If you get stuck, take a look at this question and my answer. – Brian M. Scott Jun 23 '13 at 21:23
  • @BrianM.Scott There was no conclusion to whether a normal space is necessarily completely regular or not. If the claim is true I'd appreciate a reference to a proof :) – Serpahimz Jun 23 '13 at 22:38
  • @Serpahimz: The Sierpiński space is vacuously normal, but it’s not completely regular: there is no continuous real-valued function separating the isolated point $1$ from the closed set ${0}$. – Brian M. Scott Jun 24 '13 at 06:08

1 Answers1

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The indiscrete (= trivial) topology is the only counterexample.

Let $X$ be a connected regular space, and suppose that $X$ is countable. Then $X$ is Lindelöf, and a regular Lindelöf space is normal (though not necessarily $T_1$). If $X$ contains two non-empty disjoint closed sets, Uryson’s lemma ensures the existence of a continuous surjection $f:X\to[0,1]$, showing that $|X|\ge 2^\omega$. Thus, $X$ cannot contain a pair of non-empty disjoint closed sets.

Suppose that $x\in X\setminus F$, where $F$ is a non-empty closed set; by regularity $x$ has an open nbhd $U$ such that $x\in U\subseteq\operatorname{cl}U\subseteq X\setminus F$. But then $\operatorname{cl}U$ and $F$ are non-empty disjoint closed sets, which is impossible. It follows that no non-empty proper subset of $X$ is closed and hence that no non-empty proper subset of $X$ is open. Thus, $X$ must have the indiscrete topology.

Brian M. Scott
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  • One thing I'm unsure of is why does Ursyohn's lemma ensure the existence of a surjection to $\left[0,1\right]$? The formulation of the lemma I'm familiar with only ensures a continuous function separating the sets but nothing is said about it needing to be a surjection. – Serpahimz Jun 23 '13 at 22:46
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    @Serpahimz: It gives you a continuous $f:X\to[0,1]$ such that $f(y)=0$ if $y\in\operatorname{cl}U$ and $f(y)=1$ if $y\in F$. Continuous maps preserve connectedness, so $f[X]$ must be a connected subset of $[0,1]$ containing both $0$ and $1$. There is only one such subset, $[0,1]$ itself, so $f$ must be a surjection. – Brian M. Scott Jun 24 '13 at 04:40
  • Ah very nice. Thanks! – Serpahimz Jun 24 '13 at 06:50
  • @Serpahimz: You’re welcome! – Brian M. Scott Jun 24 '13 at 06:57