# 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 $h:X\to \mathbb{R}$ 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. ∎

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 |