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I am reading the book "Operator Algebras and Quantum Statistical Mechanics" by O. Bratteli and D. W. Robinson, and Example 2.1.9 says the following:

Let $\mathfrak{A}=C_0(X)$, the commutative $C^*$-algebra(...). If $F$ is a closed subset of $X$, and $\mathfrak{B}$ consists of the elements in $\mathfrak{A}$ which are zero on $F$ then $\mathfrak{B}$ is a closed two-sided ideal of $\mathfrak{A}$(...). Using the Stone-Weierstrass theorem one can show that each closed, two-sided ideal in $\mathfrak{A}$ has this form.

Here $X$ is a locally compact Hausdorff space.

I found a proof which shows this fact for compact $X$ using Stone-Weierstrass theorem (Ideal in $C(X)$), and I tried to show the statement above based on this argument. Unfortunately, in the link we need to use that $X/F$ is compact when $X$ is compact and $F$ is closed, which is not true for locally compact space: i.e. $X/F$ could be not locally compact when $X$ is locally compact and $F$ is closed(see Quotient of a locally compact space). This makes us hard to use the Stone-Weierstrass theoem, since we do not know the ${}^*$-subalgebra of $C_0(X/F)$ which separates points and vanishes nowhere is dense(see https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem for definitions).

Thus, what should I do? Am I missing something, or is Stone-Weierstrass theorem also holds for the quotiented locally compact space? Or should I take a different route?

haru
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1 Answers1

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The best way to understand $C_0(X)$ for locally compact Hausdorff $X$ is to think about the 1-point compactification $X^*=X\cup\{\infty\}$. Note that $C_0(X)$ can be identified with the subset of $C(X^*)$ consisting of functions $f$ such that $f(\infty)=0$, which is a closed ideal in $C(X^*)$. So a closed ideal $I\subseteq C_0(X)$ is just a closed ideal in $C(X^*)$ which is contained in the ideal of functions that are $0$ at $\infty$. Such an ideal must consist of all functions that vanish on some closed subset $F\subseteq X^*$, which must contain $\infty$. Restricting the functions to $X$, this means $I$ is just the functions in $C_0(X)$ that vanish on the closed subset $F\setminus\{\infty\}\subset X$.

Eric Wofsey
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  • Great answer. The reason why we cannot use the Stone-Weierstrass theorem in $C_0(X)\subset C(X^*)$ is because $C_0(X)$ is not unital, right? Thank you for your nice answer. – haru Sep 04 '23 at 15:01
  • That's correct, you can't directly apply Stone-Weierstrass to it. – Eric Wofsey Sep 04 '23 at 16:00
  • Why is $I$ also an ideal of $C(X^*)$? – Sha Vuklia Jan 02 '24 at 16:01
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    @ShaVuklia: Any element of $C(X^*)$ is a constant (the value at $\infty$) plus an element of $C_0(X)$. – Eric Wofsey Jan 02 '24 at 16:23
  • Thank you! A small argument is needed for the claim: If $\alpha\in\mathbb F$ and $f\in I$, then $\alpha f\in I$. Proof: Let $\epsilon>0$ and let $K={x:\vert f(x)\vert\geq\epsilon}$. Since $f\in C_0(X)$, $K$ is compact. By Urysohn's Lemma (for locally compact Hausdorff spaces) there exists $\phi\in C_c(X)$ such that $\phi\vert_K\equiv 1$ and $\vert\phi\vert\leq 1$. Then $\vert \alpha\phi f-\alpha f\vert\leq 2\vert\alpha\vert\epsilon$. Since $\alpha\phi f\in I$ and $I$ is closed, it follows that $\alpha f\in I$. – Sha Vuklia Jan 02 '24 at 16:46
  • @ShaVuklia: In this context, "ideal" is usually taken to mean "ideal of $\mathbb{F}$-algebras" rather than "ideal of rings" which requires being an $\mathbb{F}$-vector subspace. You're right that in this case that is actually automatic, though. – Eric Wofsey Jan 02 '24 at 16:54
  • Thanks for mentioning (the book I use tends to leave out such definitions). – Sha Vuklia Jan 02 '24 at 17:07