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# characterization of maximal ideals of the algebra of continuous functions on a compact set

Let $X$ be a compact topological space and let $C(X)$ be the algebra of continuous real-valued functions on this space. In this entry, we shall examine the maximal ideals of this algebra.

###### Theorem 1.

Let $X$ be a compact topological space and $I$ be an ideal of $C(X)$. Then either $I=C(x)$ or there exists a point $p\in X$ such that $f(p)=0$ for all $f\in I$.

###### Proof.

Assume that, for every point $p\in X$, there exists a continuous function $f\in I$ such that $f(p)\not=0$. Then, by continuity, there must exist an open set $U$ containing $p$ so that $f(q)\not=0$ for all $q\in U$. Thus, we may assign to each point $p\in X$ a continuous function $f\in I$ and an open set $U$ of $X$ such that $f(q)\neq 0$ for all $q\in U$. Since this collection of open sets covers $X$, which is compact, there must exists a finite subcover which also covers $X$. Call this subcover $U_{1},\ldots,U_{n}$ and the corresponding functions $f_{1},\ldots f_{n}$. Consider the function $g$ defined as $g(x)=(f_{1}(x))^{2}+\cdots+(f_{n}(x))^{2}$. Since $I$ is an ideal, $g\in I$. For every point $p\in X$, there exists an integer $i$ between $1$ and $n$ such that $f_{i}(p)\not=0$. This implies that $g(p)\neq 0$. Since $g$ is a continuous function on a compact set, it must attain a minimum. By construction of $g$, the value of $g$ at its minimum cannot be negative; by what we just showed, it cannot equal zero either. Hence being bounded from below by a positive number, $g$ has a continuous inverse. But, if an ideal contains an invertible element, it must be the whole algebra. Hence, we conclude that either there exists a point $p\in x$ such that $f(p)=0$ for all $f\in I$ or $I=C(x)$. ∎

###### Theorem 2.

Let $X$ be a compact Hausdorff topological space. Then an ideal is maximal if and only if it is the ideal of all points which go zero at a given point.

###### Proof.

By the previous theorem, every non-trivial ideal must be a subset of an ideal of functions which vanish at a given point. Hence, it only remains to prove that ideals of functions vanishing at a point is maxiamal.

Let $p$ be a point of $X$. Assume that the ideal of functions vanishing at $p$ is properly contained in ideal $I$. Then there must exist a function $f\in I$ such that $f(p)\neq 0$ (otherwise, the inclusion would not be proper). Since $f$ is continuous, there will exist an open neighborhood $U$ of $p$ such that $f(x)\neq 0$ when $x\in U$. By Urysohn’s theorem, there exists a continuous function $h\colon X\to\mathbb{R}$ such that $f(p)=0$ and $f(x)=0$ for all $x\in X\setminus U$. Since $I$ was assumed to contain all functions vanishing at $p$, we must have $f\in I$. Hence, the function $g$ defined by $g(x)=(f(x))^{2}+(h(x))^{2}$ must also lie in $I$. By construction, $g(g)>0$ when $x\in U$ and when $g(x)\in X\setminus U$. Because $X$ is compact, $g$ must attain a minimum somewhere, hence is bounded from below by a positive number. Thus $g$ has a continuous inverse, so $I=C(X)$, hence the ideal of functions vanishing at $p$ is maximal. ∎

## Mathematics Subject Classification

46L05*no label found*46J20

*no label found*46J10

*no label found*16W80

*no label found*

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