# How to make a topological subspace Hausdorff

If $X$ is a compact topological space then $C(X)$ is a C* algebra. I'm not going to attempt to discuss the locally compact case with $C_0(X)$ because usually the definition of that space requires the underlying $X$ to be Hausdorff. $C(X)$ has a Gel'fand representation, as some $C(Y)$ with Y compact and Hausdorff. How does Y relate to X? Is it via some standard process by which one can take a nonHausdorff space and maybe design some strange equivalence relation for which the fact set is $Y$, and Hausdorff?

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Isn't it just the Stone–Čech compactification of $X$ (which isn't actually a compactification in this case)? –  Chris Eagle Oct 3 '12 at 8:24

Let $\delta_t$ be the evaluation at the point $t\in X$. It is easy to check that $s\sim t\iff\delta_s=\delta_t$ is an equivalence relation $\sim$ on $X$. Consider $X/\sim$, with the quotient topology (i.e. the smallest topology that makes the quotient map continuous).

Now consider the map $\phi:C(X/\sim)\to C(X)$ given by $\phi(f)(t)=f([t])$, where $[t]$ is the class of $t$ in the quotient. This map is obviously a $*$-homomorphism. It is one-to-one by construction, because if $\phi(f)=0$, then $f([t])=0$ for all $[t]$, i.e. $f=0$. And it is onto: if $g\in C(X)$, then we define a function $g'\in C(X/\sim)$ by $g'([t])=g(t)$; this of course implies that $\phi(g')=g$. All we need to check is that such $g'$ can be defined; this is exactly to say that $g(t)$ does not depend on the value of $g$, and this holds, since $s\sim t$ if and only if $\delta_s=\delta_t$.

The topology on $X/\sim$ is Hausdorff: given $[s]\ne[t]\in X/\sim$, we have that $\delta_s\ne\delta_t$, so there exists $f\in C(X)$ such that $f(s)\ne f(t)$. The construction of the quotient topology guarantees that $f'$, as in the previous paragraph, is continuous; we get that $f'([s])\ne f'([t])$, and so $[s]$ and $[t]$ can be separated.

So, as C$^*$-algebras, $C(Y)\simeq C(X)\simeq C(X/\sim)$. Since $Y$ and $X/\sim$ are Hausdorff, we deduce that $Y$ is homeomorphic to $X/\sim$.

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What exactly is the equivalence relation "non-separable by open sets"? Do you mean $s \sim t$ iff $\delta_s = \delta_t$? Consider the three point space $\{a,b,c\}$ with the topology $\{\emptyset, \{a\}, \{b\}, \{a,b\}, \{a,b,c\}\}$. Then neither $a$ nor $b$ can be separated from $c$ by open sets, but $a$ and $b$ can be separated by open sets. –  commenter Oct 3 '12 at 19:43
You are right, I'm editing the answer accordingly. It is interesting that, in your example, $C(X)$ consists of the constant functions only. –  Martin Argerami Oct 3 '12 at 20:04
Nice. If one wants, one could split your construction up into two steps 1) Take the coarsest topology $\sigma$ on $(X,\tau)$ that makes elements of $C(X)$ continuous. Then $(X,\sigma)$ is completely regular and $(X,\tau) \to (X,\sigma)$ is continuous. 2) Take the Kolmogorov quotient of $(X,\sigma)$ and this will be a completely regular $T_0$ space, and in particular it will be Hausdorff. Step 2) is similar in spirit to what you wrote before: two points are equivalent iff they are topologically indistinguishable. –  commenter Oct 3 '12 at 20:30
@MartinArgerami I have just noticed that it is still safe to define the continuous functions vanishing at infinity on a locally compact nonHausdorf space, but that your construction no longer works in that case. Also, the one point compactification becomes much messier, and the space is no longer the closure of the compactly supported functions on $X$. Is all of this a reflection that the analogy between compact and locally compact breaks without the presence of the Hausdorf assumption? –  Jeff Oct 6 '12 at 1:07