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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)?

Weltschmerz
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    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. – Chris Eagle May 26 '11 at 13:31
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    The one-element topological space also is not uncountable. Perhaps we need normal, Hausdorff, with more than one point. – Qiaochu Yuan May 26 '11 at 13:41
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    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. – Weltschmerz May 26 '11 at 13:45
  • 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. – Syd Mar 30 '12 at 03:22

1 Answers1

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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).

Qiaochu Yuan
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    Counterexamples calls a space with no nonconstant real-valued functions "strongly connected". I don't know if that's at all standard. – Chris Eagle May 26 '11 at 13:38
  • 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. – Chris Eagle May 26 '11 at 13:43
  • @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. – Qiaochu Yuan May 26 '11 at 13:45
  • Normal is a stronger requirement than completely regular, but iirc the spaces that are completely regular but not normal are fairly pathological in nature. – JSchlather May 26 '11 at 15:17
  • @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. – Henno Brandsma May 26 '11 at 18:33
  • @Henno, I recall incorrectly then. I might have been thinking of regular spaces that are not completely regular. – JSchlather May 26 '11 at 18:38
  • @Jacob yes, these are more uncommon. It is even possible to have regular spaces that have the property that all continuous mappings into the reals are constant. But these require some work, and are "artificial" in that sense. Counterexamples in Topology has one of those. – Henno Brandsma May 26 '11 at 18:41