As an operator algebraist, I think the space of continuous functions to $\mathbb{C}$ which vanish at infinity is my preferred choice. Let me tell you why.
One of the basic ideas of noncommutative topology/geometry (and probably algebraic geometry, but I don't know much about that) is that we can trade the space for algebras of functions on that space. This is afforded by the Gelfand transform. The spectrum of a commutative $C^\ast$-algebra is the space of characters, i.e., $\ast$-algebra homomorphisms to $\mathbb{C}$.
- If $X$ is compact Hausdorff, then the spectrum of $C(X)$ is $X$.
- If $X$ is locally compact Hausdorff, but not compact, the spectrum of the non-unital $C^\ast$-algebra $C_0(X)$ is $X$. The spectrum of the unitalization of $C_0(X)$ ($C_0(X)\oplus \mathbb{C}$) is the one point compactification of $X$. The spectrum of the unital $C^\ast$-algebra $C_b(X)$ is $\beta X$, the Stone-Cech compactification of $X$. One should note that $C_b(X)$ is also the multiplier algebra of $C_0(X)$.
- If $X$ is compact, but not Hausdorff, then $C(X)$ corresponds to some type of "Hausdorffization" of $X$.
Actually $C(X)$ and $C_0(X)$ are the same if $X$ is compact, but you want to denote it $C(X)$ to emphasize the fact that the algebra is already unital. Otherwise, when you add a unit, you take the one point compactification of a compact space which adds an extra point, which is not what you want.
Now let's suppose you have some additional structure, like $X$ is a compact manifold. Then you probably want the $C^\infty$-functions on $X$. However, these can be recovered from $C(X)$ as those operators whose iterated commutator with the Dirac operator is bounded. This inspired the notion of a spectral triple.
EDIT: In my haste to answer this question, I made some mistakes in the earlier answer as pointed out by @Jonas.