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Let $X$ be a compact Hausdorff topological set, and $Y$ be its closed subset. Is the ideal of functions vanishing on $Y$ $$ I=\{f\in C(X):\ \forall y\in Y\ f(y)=0\} $$ complementable (as a closed subspace) in $C(X)$ (as a Banach space)?

This is true in the case when $X\subseteq {\mathbb R}^n$ (this follows from: Stein. Singular integrals... VI 2.2), but what about general case?

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Counterexamples are already presented below, but on the positive side, it is true whenever $Y$ is compact metrizable. In such a case, there is a unital positive linear operator $T\colon C(Y)\to C(X)$ such that $T(f)|_Y=f$ for all $f\in C(Y)$. – Narutaka OZAWA Aug 12 '14 at 23:11
@NarutakaOZAWA, could you, please, give the reference? – Sergei Akbarov Aug 13 '14 at 18:02
The result Taka mentioned is due to K. Borsuk, Bull. Internat. Acad. Polon. Sci. Sér. A No. 113 (1933), 1–10. – Bill Johnson Aug 13 '14 at 22:04
Borsuk assumed that $X$ is separable; Dugundji proved that it is enough to assume that only $Y$ is separable: J. Dugundji, An extension of Tietze's theorem. Pacific Journal of Mathematics 1 (1951), no. 3, 353--367. There are two nice papers, one due to Pełczyński and the second one due to Haydon, about spaces which satisfy the above-mentioned theorem. Google for the term: "Dugundji space". – Tomek Kania Aug 14 '14 at 13:42
up vote 7 down vote accepted

Not in general.

It's well-known in Banach space theory that the ideal $c_0$ in $\ell^\infty$ is not complemented (see e.g. Albiac & Kalton).

By the Gelfand representation, $\ell^\infty \simeq C(\beta \mathbb{N})$ as a $C^\ast$-algebra. This maps $c_0$ to the ideal of functions on $\beta \mathbb{N}$ that vanish on $\beta \mathbb{N} \setminus \mathbb{N}$.

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But $\beta{\mathbb N}\setminus{\mathbb N}$ is not closed in $\beta{\mathbb N}$. – Sergei Akbarov Aug 12 '14 at 18:39
@SergeiAkbarov: Actually, it is. $\mathbb{N} \subset \beta \mathbb{N}$ consists of isolated points, so it's open. And in any case, your ideal $I$ only depends on the closure of the set. – Alexander Shamov Aug 12 '14 at 18:47
Yes, excuse me and thank you! – Sergei Akbarov Aug 13 '14 at 6:04

Alexander Shamov answered your question with a classical example, but you might be interested in a modern example constructed by Piotr Koszmider. There is an infinite connected compact Hausdorff space $K$ s.t. $C(K)$ has no complemented subspaces that are both infinite dimensional and infinite codimensional. In particular, if $L$ is a closed subset of $K$ s.t. the ideal of functions vanishing on $L$ is complemented, then $L$ is finite.

Koszmider, Piotr(BR-SPL) Banach spaces of continuous functions with few operators. (English summary) Math. Ann. 330 (2004), no. 1, 151–183.

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Yes, that's unexpected. Thank you, Bill. – Sergei Akbarov Aug 13 '14 at 18:01

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