This is little Rudin's proof 7.25: If K is compact, f_n complex and continous functions and if {f_n} is pointwise bounded and equicontinuous on K, then {f_n} contains a uniformly convergent subsequence.(adsbygoogle = window.adsbygoogle || []).push({});

The proof requires a countable dense subset of K, called E. Then by compactness (and density???), there exists finitely many x_i elements of E s.t K is covered by U V(x_i, delta). Then the usual triangle inequality applies.

I don't see why a countable dense subset of K is needed. Why can we not directly invoke compactness and choose the x_i's elements of K, instead? U V(x, delta), x elements of K, is an infinite cover of K, so why can we not directly choose the x_i's? The only requirement in the proof is that d(x,x_i)<delta be valid for some i from 1 to m and for all x elements of K.

What am I completely missing?

Thanks in advance!

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# Rudin's proof of Arzela-Ascoli teorem

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