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?
How to make a topological subspace Hausdorff
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1http://mathoverflow.net/questions/78175/largest-hausdorff-quotient – 2012-10-03
1 Answers
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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0@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? – 2012-10-06