A question about locally compact Hausdorff space

Question

If $X$ is a locally compact Hausdorff space, $C_{0}(X)$ denotes the set of continuous functions from $X$ to $\mathbb{C}$ vanishes at infinity. This is a basic example in C*algebra. My question is Why we need to consider the continuous functions vanishes at infinity in locally compact Hausdorff space? I mean, if we only consider the set of continuous functions from $X$ to $\mathbb{C}$, what kind of error may happen?

Answer 1

In topology, one can consider the pointwise convergence topology on $C(X)$, this is often denoted $C_p(X)$, and this needs no assumptions on $X$, except $X$ being Tychonoff (completely regular and $T_1$) to guarantee enough continuous functions on $X$.

But in analysis one often wants to make $C(X)$ into a normed space (which is then complete, so a Banach space), but then we need some form of boundedness. If $X$ is compact (or really, pseudocompact) we know all real-valued (or complex-valued) functions on $X$ are bounded and defining $||f|| = \sup \{ |f(x): x \in X\}$ gives us a norm with nice properties.

Now, if $f$ "vanishes at infinity" (which means, as was pointed out in the comments, that for every $\epsilon>0$ there is some compact $K \subseteq X$ such that $|f(x)| < \epsilon$ outside $K$) we again have that continuous such functions are bounded (the part on the compact $K$ is bounded, and outside is bounded by $\epsilon = 1$ say) and we can define a norm. The local compactness makes the space nicer behaved: we have a one-point compactification $\alpha(X) = X \cup \{\infty\}$ and functions that vanish at infinity can be extended continuously from $X$ to $\alpha(X)$ by setting $f(\infty) = 0$, so that $C_0(X)$ becomes a closed subset (subalgebra even) $\{f \in C(\alpha(X)): f(\infty) = 0 \}$ of $C(\alpha(X))$ which allows us to transfer a lot of the theory of the compact case over to the locally compact case.

When $X$ is not locally compact, we cannot do this one-point compactification, and we also don't have a nice description (I think) of the dual as a space of Radon measures, etc. Life is a lot more complicated then. But it's probably studied in that case as well, I'm no expert.

Answer 2

There's no "error", but such $C(X)$ is not a $C^*$-algebra, in fact not a normed space. It is a Fréchet algebra, under the compact-open topology.