About continuous functions on a compact Hausdorff space.

Question

Let $X$ be a compact Hausdorff space (if necessary you can even assume that it is connected). Consider the ring $R= C(X, \mathbb{R})$ of real-valued continuous functions on $X$. I'm interested in determining if the following statement is true:

If $|X| > 1$, then there is an element $f \in R$ that is not the zero function and that vanishes somewhere.

Would it be correct to say that this follows from Urisohn's lemma? Since $|X| > 1$, we can pick $x_1 \neq x_2$ and because $X$ is compact Hausdorff we can separate these by disjoint closed sets. Because a compact Hausdorff space is normal, we can find a continuous function $f: X \to \mathbb{R}$ such that $f = 1$ on one of these closed sets and $f= 0$ on the other. Thus there is a function $f \in R$ that is not the zero function and that vanishes somewhere.

Is this correct?

Answer 1

This is correct. Normality is somewhat stronger than what is needed for Urysohn's Lemma, although it certainly suffices. I will disagree with @Math1000's comment and say that Urysohn's Lemma is perhaps the most non-trivial statement one can see in a first course on point-set topology!

Note that connectivity is not necessary. If there is more than one connected component, then the function that is $1$ on one component and $0$ on the other is continuous.

Answer 2

Yes, Urysohn's lemma implies this, we can even apply Tietze to the closed set $C=\{x_1, x_2\}$ and extend the trivially continuous $x_1 \to 0,x_2 \to 1$ function on $C$. Normality or Tychonoffness gives us "plenty of" continuous functions like this, which is why the combination (locally) compact plus Hausdorff is so common (enough functions to get an interesting ring $C(X)$ (or TVS etc.)