What the difference of euclidean metric and sup metric in this proof? And since the theorem is proofed why there would be a special case where the neighborhood theorem does not hold?
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all details are in Theorem 4.6 and its proof. 1. What the difference of euclidean metric and sup metric in this proof?Both of them use triangle inequality in the proof. 2. I think for any given X, there is certain ε is the minimum value of f(x). So, there would be no special case. And what's wrong is the case. Thx very much
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Answer:
The particular metric plays no role in the proof; it would work for either metric. In fact, the theorem immediately generalizes to the following: If [math]X[/math] is a metric space, [math]K\subset X[/math] is compact, and [math]U\subset X[/math] is an open set containing [math]K[/math], then [math]U[/math] contains an [math]\epsilon[/math]-neighborhood of [math]K[/math]. The proof is identical to the one given here. The counterexample following the theorem is showing compactness is necessary; the [math]x[/math]-axis is not a compact subset of [math]\mathbb{R}^2[/math]. The function [math]f(x)[/math] does not attain its minimum in the example; its value can be made as close to 0 as you want.
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