First, notice that each of the sets is a maximal ideal in . In fact, if 1 denotes the constant function 1 then for any we can write . Since , it follows that (which is trivially an ideal) has codimension 1 in and is therefore maximal.

Now let be a maximal ideal of . If there exists such that each element of vanishes at , then is contained in (and hence by maximality is equal to) , and we are done.

So suppose that there is no such . Then for each there exists such that . The set is an open neighbourhood of t. These neighbourhoods cover the unit interval, so by compactness there is a finite subcover, given by elements of . Then is an element of that is strictly positive throughout the interval and hence invertible. But an ideal containing an invertible element must by the whole ring. So is not a proper ideal.