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For a set $S\subset\mathbb R$ let $C(S)$ denote the continuous real-valued functions on $S$. Describe the maximal ideals in $C((0,1))$. For $C([0,1])$ we know that maximal ideals are points in $[0, 1]$ and these are the only maximal ideals, all whose elements vanish at a single common point. The proof relies on the compactness of $[0,1]$. I want to know what happens when the compact interval is replaced by the open interval $(0, 1)$. Thanks for your help.

kimchi lover
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Lawrence Mano
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  • To clarify, you are interested in all continuous functions on $(0,1)$, even the unbounded ones? Just the bounded ones? Just those that vanish at the endpoints? – Mike F Feb 14 '20 at 00:33
  • For what it's worth the inclusion $i: (0,1) \rightarrow [0,1]$ induces a surjection $\bar{i}: C([0,1]) \rightarrow C_0(0,1)$ and it's clear that an ideal $I \subset C_0(0,1)$ is maximal iff $\bar{i}^{-1}(I)$ is a maximal ideal, which implies that the only maximlas ideals in $C_0(0,1)$ again correspond to functions vanishing at a single point. – Jonathan Hole Feb 14 '20 at 01:50
  • @JonathanHole please define your notation $C_0$. – KCd Feb 14 '20 at 01:58
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    The answer by egreg on the page https://math.stackexchange.com/questions/451223/the-stone-%C4%8Cech-compactification-of-a-space-by-the-maximal-ideals-of-the-ring-of shows, taking $X = (0,1)$, that the space of maximal ideals in the ring of continuous functions $(0,1) \rightarrow \mathbf R$, when given the Zariski topology, is homeomorphic to the Stone-Cech compactification of $(0,1)$. That is huge. It's the same as the Stone-Cech compactification of $\mathbf R$; a discussion of that is at https://math.stackexchange.com/questions/1790222/stone-%C4%8Cech-compactification-of-real-line/1790467 – KCd Feb 14 '20 at 02:00
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    The "points" of the Stone-Cech compactification $\beta(0,1)$ of $(0,1)$ that are not points in $(0,1)$ itself are basically impossible to describe in a concrete way. To get a sense of how much bigger the Stone-Cech compatification of $(0,1)$ is than $(0,1)$ itself, the continuous function $\sin(1/x)$ from $(0,1)$ to $[-1,1]$ does not extend by continuity to a continuous function $[0,1] \rightarrow [-1,1]$, but $\sin(1/x)$ on $(0,1)$ does extend to a continuous function $\beta(0,1) \rightarrow [-1,1]$. – KCd Feb 14 '20 at 02:07
  • Very nice answer @KCd – Hugo Feb 14 '20 at 02:14

1 Answers1

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The space $C(0,1)$ has the trivial maximal ideals $$m_a = \{ f \in C(0,1): f(a)=0 \}.$$

Nevertheless, there are other maximal ideals: Consider $$I=\{f \in C(0,1): f \text{ has compact support}\},$$ where we say $f$ has compact support if there are $0<a<b<1$ such that $f$ is zero outside of the interval $[a,b]$. This set $I$ is a proper ideal of $C(0,1)$ and, therefore, it is contained in maximal ideal $m$. Notice this ideal is not an ideal $m_a$, because we can always find a function $f\in C(0,1)$ such that $f$ has compact support and $f(a)\neq 0$. So, this ideal $m$ is a non-trivial maximal ideal.

I'm not sure if there is a general classification of maximal ideals in this case. So, this is a partial answer.

Hugo
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    This is what I was looking for!! Thank you! A complete classification using compactification given by KCd is very deep and involving and interesting also. Thank you very much. – Lawrence Mano Feb 14 '20 at 14:49