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Here's a solution (to the exercise "prove that if $X$ is a space with more than one element, and is normal and connected then $X$ is uncountable"):

by Urysohn's lemma, given $A$ and $B$, closed and disjoint in $X$, there exists a continuous function from $X$ into $[0,1]$ such that $f(A)={0}, f(B)={1}$.

If $X$ were countable, so it would be $f(X)\subset [0,1]$; choose $r\in (0,1)\setminus f(X)$, then $X=f^{-1}([0,r))\cup f^{-1}((r,1])$, so $X$ is disconnected.

But the full power of the lemma wasn't used, all that was used is that there exists a continuous function $X\longrightarrow [0,1]$, so the condition of being normal seems too much.

So my question is, what are the weakest conditions on $X$ for the existence of such continuous functions into $[0,1]$ (or $\mathbb{R}$, for that matter)?

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    $\begingroup$ You're tacitly assuming that $X$ has two closed nonempty disjoint subsets. This need not be the case: for example, any set with the trivial topology is normal and connected, but need not be uncountable. $\endgroup$
    – Chris Eagle
    May 26, 2011 at 13:31
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    $\begingroup$ The one-element topological space also is not uncountable. Perhaps we need normal, Hausdorff, with more than one point. $\endgroup$
    – Qiaochu Yuan
    May 26, 2011 at 13:41
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    $\begingroup$ I forgot to mention that it has more than one point. Thank you. On the other hand, according to Munkres' textbook, the definition of normal requires the space to be $T_1$, so the trivial topology can't be considered. $\endgroup$
    – Weltschmerz
    May 26, 2011 at 13:45
  • $\begingroup$ A space for which any two points can be separated by a continuous real-valued function is said to be functionally Hausdorff and if it's connected and has more than two points, it's uncountable. $\endgroup$
    – Syd
    Mar 30, 2012 at 3:22

1 Answer 1

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What was used is that there is a nonconstant continuous function into $[0, 1]$. I don't know a nice name for this condition; it is so weak that I can't think of a space that satisfies this condition that doesn't satisfy the stronger property that points can be separated by continuous functions into $\mathbb{R}$. In addition, all of the spaces I can think of with this property are completely regular, and all of the completely regular spaces I can think of are built from normal spaces in some way (and the proof that they're completely regular in general is via Urysohn's lemma).

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    $\begingroup$ Counterexamples calls a space with no nonconstant real-valued functions "strongly connected". I don't know if that's at all standard. $\endgroup$
    – Chris Eagle
    May 26, 2011 at 13:38
  • $\begingroup$ Simple examples of spaces which have nonconstant functions to $\mathbb{R}$ but not enough to separate points are the product or coproduct of $\mathbb{R}$ with a two-point trivial space. $\endgroup$
    – Chris Eagle
    May 26, 2011 at 13:43
  • $\begingroup$ @Chris: sure. I guess the point I'm trying to make is that in practice a lot of spaces are normal, so it's okay to use it as a hypothesis. $\endgroup$
    – Qiaochu Yuan
    May 26, 2011 at 13:45
  • $\begingroup$ Normal is a stronger requirement than completely regular, but iirc the spaces that are completely regular but not normal are fairly pathological in nature. $\endgroup$
    – JSchlather
    May 26, 2011 at 15:17
  • $\begingroup$ @Jacob they include spaces like an uncountable product of copies of the real numbers, which is a commonly considered topological vector space. So it's not that obscure. $\endgroup$
    – Henno Brandsma
    May 26, 2011 at 18:33

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