$C(X)$ separates points? Let $X$ be a compact Hausdorff space and $C(X)$ the space of continuous functions on $X$. In many proofs in abstract harmonic analysis we use the fact $C(X)$ separates points in $X$.
It might seem trivial but the only way I can think of why this is true is by applying Urysohn's lemma.
I wonder whether there is some more elementary way to show that $C(X)$ separates points.
Thanks!
 A: Though I'm not an expert, it seems unlikely to me that you can have something better, more elementary, than Urysohn's lemma for what you want.
Even if you're right that it may seem that Urysohn's lemma is too much, because what it gives is the separation of closed sets and you're asking just for separation of points.
Moreover, at first glance, you can easily prove implications like the following ones. If there are enough continuous functions $f: X \longrightarrow [0,1]$


*

*to separate points (that is, for every pair of points $x,y \in X$ there is a continuous function $f$ such that $f(x)= 0$ and $f(y) = 1$), then $X$ must be a Hausdorff space. Indeed, $f^{-1}([0,1/2))$ and $f^{-1}((1/2, 1])$ are disjoint open sets, cointaining $x$ and $y$, respectively.

*to separate points from closed sets (that is, for every point $x$ and every closed set $A \subset X$, $x \notin A$, there exists a continuous function such that $f(x) = 0$ and $f(a) =1$ for all $a\in A$), then $X$ must be regular. An analogous prove aplies.

*to separate closed sets from closed sets, then $X$ must be normal. Ditto.


In this context, the Urysohn's lemma says that, in fact, the last implication is an equivalence; that is, an space is normal if and only if there are enough continuous functions to separate closed sets.
So, a natural question would be: Why aren't there analogous reciprocal implications in the other two cases? That is, why we don't have Urysohn type lemmas for Hausdorff and regular spaces?
Well, I know at least two reasons:


*

*In the case of regularity, the separation of points from closed sets with continuous functions is an stronger condition than separation with open sets. Together with the requirement of the space being Fréchet, this is called complete regularity. There are examples of regular spaces which are not completely regular ones; that is, spaces where you can separate points from closed sets with open sets, but not with continuous functions.  You can find such an example in the book of Munkress (section 5.2, exercise 6; see also example 1 after theorem 2.1 in the same section).

*I don't know the situation for the Hausdorff condition. [EDIT. See Brian M. Scott's comments: there is also a notion of "completely Hausdorff".] But I can tell you where the standard proof of Urysohn's lemma will break down with Hausdorff (or regular) spaces: you construct your separating function as a certain limit of step functions, which miraculously is continuous. You begin with $f_0$ the characteristic function of your point $y$. Then you find an open set $V$ such that $x\in V \subset \overline{V} \subset X\backslash \{ y\}$ and define $f_1$ as being $0$ on $x$, $1/2$ on $V\backslash \{x\}$ and $1$ on $X\backslash V$. In the following steps, you introduce new open sets at each inclusion $\{x\}\subset V  $ and $\overline{V} \subset X \backslash \{ y\}$ and define your next step function with $1/2^n$ height steps. The Hausdorff condition gives you that open set $V$ here, but the restriction about its clousure seems really necessary in order to assure the continuity of the limit of these step functions. This is unfortunate, because, for the next steps, you are going to need in general that,  given a closed set $A$ (that $\overline{V}$ that has already appeared in the first step) and an open set $U$ such that $A \subset U$, there exists an open set $V$ such that $A \subset V \subset \overline{V} \subset U$. And you can do this for every such $A$ and $U$ if and only if the space $X$ is normal.

