Let be a compact metric space, a continuous function and . Prove that is compact iff is finite.

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# Compact function spaces

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2 thoughts on “Compact function spaces”

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Let be a compact metric space, a continuous function and . Prove that is compact iff is finite.

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If is finite there is nothing to prove. Suppose on the other hand that is compact. We will show that every point in is isolated and thus finite due to the compactness of .

Suppose that is not isolated, there is therefore a ball of center and radius such that , and we consider the sequence of balls . Let the Urysohn map such that and and consider the sequence of functions . Clearly, and the sequence converges pointwise to (where is Dirac’s delta function). Since is compact, there must be a subsequence of which converges uniformly to , but this is absurd since, taking sufficiently small we would have . And we're done.

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Alternatively, let us prove that is not totally bounded if is infinite.

Take a sequence of pairwise distinct points in ; by compactness we have a subsequence . Consider the subspace , which is closed in .

Let be the minimum value that takes, which exists by compactness. All points in except for are isolated, so the function is continuous and bounded by . Moreover, if . Since is closed, by Tietze’s extension theorem we may obtain a continuous extension of which is also bounded above by , and in particular by . Since we have that if and the functions belong to , we are done.

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