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)?
Best Answer
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).