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 $C\mathit{}\mathrm{(}X\mathrm{)}$. Then either $I\mathrm{=}C\mathit{}\mathrm{(}x\mathrm{)}$ or there exists a point $p\mathrm{\in }X$ such that $f\mathit{}\mathrm{(}p\mathrm{)}\mathrm{=}\mathrm{0}$ for all $f\mathrm{\in }I$.

Proof.

Assume that, for every point $p\in X$, there exists a continuous function $f\in I$ such that $f(p)\ne 0$. Then, by continuity, there must exist an open set $U$ containing $p$ so that $f(q)\ne 0$ 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 $f(q)\ne 0$ 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 $U_1,\mathrm{\dots },U_n$ and the corresponding functions $f_1,\mathrm{\dots }{f}_n$. Consider the function $g$ defined as $g(x)={(f_1(x))}^2+\mathrm{\cdots }+{(f_n(x))}^2$. 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 $f_i(p)\ne 0$. This implies that $g(p)\ne 0$. 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 $f(p)\ne 0$ (otherwise, the inclusion would not be proper). Since $f$ is continuous, there will exist an open neighborhood $U$ of $p$ such that $f(x)\ne 0$ when $x\in U$. By Urysohn’s theorem, there exists a continuous function $h:X\to ℝ$ 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. ∎

Title	characterization of maximal ideals of the algebra of continuous functions on a compact set
Canonical name	CharacterizationOfMaximalIdealsOfTheAlgebraOfContinuousFunctionsOnACompactSet
Date of creation	2013-03-22 17:45:08
Last modified on	2013-03-22 17:45:08
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	9
Author	rspuzio (6075)
Entry type	Theorem
Classification	msc 46L05
Classification	msc 46J20
Classification	msc 46J10
Classification	msc 16W80