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 continuousreal-valued functions on this space. In this entry, we shall examine the maximalideals 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 $f(p)=0$ forall $f\\in I$ .

Proof.

Assume that, for every point $p\\in X$ , there exists a continuous function $f\\in I$ such that $f(p)\ot=0$ . Then, by continuity, there must exist anopen set $U$ containing $p$ so that $f(q)\ot=0$ for all $q\\in U$ . Thus, we mayassign to each point $p\\in X$ a continuous function $f\\in I$ andan open set $U$ of $X$ such that $f(q)\eq 0$ for all $q\\in U$ . Since thiscollection of open sets covers $X$ , which is compact, there must exists a finitesubcover which also covers $X$ . Call this subcover $U_{1},\\ldots,U_{n}$ and thecorresponding 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 everypoint $p\\in X$ , there exists an integer $i$ between $1$ and $n$ such that $f_{i}(p)\ot=0$ . This implies that $g(p)\eq 0$ . Since $g$ is a continuous function ona 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 ifand 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 offunctions which vanish at a given point. Hence, it only remains to prove thatideals 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 $f(p)\eq 0$ (otherwise, the inclusion would not be proper). Since $f$ is continuous, there will exist an open neighborhood $U$ of $p$ such that $f(x)\eq 0$ when $x\\in U$ . By Urysohn’s theorem, there exists a continuousfunction $h\\colon X\\to\\mathbb{R}$ such that $f(p)=0$ and $f(x)=0$ forall $x\\in X\\setminus U$ . Since $I$ was assumed to contain all functions vanishingat $p$ , 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 apositive number. Thus $g$ has a continuous inverse, so $I=C(X)$ , hence theideal of functions vanishing at $p$ is maximal.∎

单词	CharacterizationOfMaximalIdealsOfTheAlgebraOfContinuousFunctionsOnACompactSet
释义	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 continuousreal-valued functions on this space. In this entry, we shall examine the maximalideals 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 $f(p)=0$ forall $f\\in I$ . Proof. Assume that, for every point $p\\in X$ , there exists a continuous function $f\\in I$ such that $f(p)\ot=0$ . Then, by continuity, there must exist anopen set $U$ containing $p$ so that $f(q)\ot=0$ for all $q\\in U$ . Thus, we mayassign to each point $p\\in X$ a continuous function $f\\in I$ andan open set $U$ of $X$ such that $f(q)\eq 0$ for all $q\\in U$ . Since thiscollection of open sets covers $X$ , which is compact, there must exists a finitesubcover which also covers $X$ . Call this subcover $U_{1},\\ldots,U_{n}$ and thecorresponding 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 everypoint $p\\in X$ , there exists an integer $i$ between $1$ and $n$ such that $f_{i}(p)\ot=0$ . This implies that $g(p)\eq 0$ . Since $g$ is a continuous function ona 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 ifand 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 offunctions which vanish at a given point. Hence, it only remains to prove thatideals 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 $f(p)\eq 0$ (otherwise, the inclusion would not be proper). Since $f$ is continuous, there will exist an open neighborhood $U$ of $p$ such that $f(x)\eq 0$ when $x\\in U$ . By Urysohn’s theorem, there exists a continuousfunction $h\\colon X\\to\\mathbb{R}$ such that $f(p)=0$ and $f(x)=0$ forall $x\\in X\\setminus U$ . Since $I$ was assumed to contain all functions vanishingat $p$ , 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 apositive number. Thus $g$ has a continuous inverse, so $I=C(X)$ , hence theideal of functions vanishing at $p$ is maximal.∎
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