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?
Best Answer
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.