TITLE: Why do we require $X$ to be Hausdorff when defining $C(X)$?
QUESTION [5 upvotes]: In Efton Park's "Complex Topological K-theory", he begins a section by defining the Banach algebra $C(X)$ (continuous functions $f: X \rightarrow \Bbb C$), presupposing that $X$ is Hausdorff and compact. It's obvious why he requires compact: otherwise, the "norm" wouldn't necessarily be a norm, as there could be unbounded functions on $X$. But whence the Hausdorff condition?
I know that we have an anti-equivalence of categories (compact Hausdorff spaces) ~ (commutative unital C* algebras); this fails if we remove the Hausdorff requirement; consider for instance the Sierpinski 2-point space (call it $S$) with topology $\{\varnothing, \{a\}, \{a,b\}\}$. It's easy to see that $C(S) \cong \mathbb C$. Of course, $S$ is not homeomorphic to the one-point space, even though both are compact; so the anti-equivalence fails here. This is a good enough reason for me to not care so much about $X$ non-Hausdorff.
But where, exactly, does $X$ non-Hausdorff cause trouble for us? I'd prefer an answer on a more basic level than the above; how far can we work with $C(X)$ with $X$ non-Hausdorff before it causes serious trouble for us?

REPLY [9 votes]: One point is that compact Hausdorff guarantees normality, so Urysohn's lemma gives us lots of continuous functions (i.e. for every $x \neq y$ we have a continuous function that sends $x$ to $0$ and $y$ to $1$, e.g.) If $X$ is indiscrete, or cofinite e.g. (both are compact) we have no non-constant functions into $\mathbb{C}$. So even for compact $T_1$ spaces, we can have that $C(X) = \mathbb{C}$. 
In general, for topological spaces $X$, we want $X$ to be Tychonoff when considering rings of continuous functions (into the reals or the complex numbers). Otherwise, we just quotient  out the points that we cannot distinguish by continuous functions, to get an isomorphic ring on a better behaved space.
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