"Pointwise equicontinuity implies uniform" implies compact
Suppose $(M,d)$ is a metric space such that every sequence $f_n:M\to \Bbb R$ which is pointwise equicontinuous is also uniformly equicontinuous. Does this imply that $M$ is a compact metric space?
My thoughts: If true, we could try to show compactness directly or show sequential compactness and totally bounded.
Starting with the first idea, we take any open cover of $M$. One would think that we want to use this to construct some pointwise equicontinuous sequence of functions. No good ideas come to my mind.
Ok, sequential compactness, let ${x_n}$ be a Cauchy sequence in $M$. Maybe we take $f_n(x) = x_n$ the constant function, but that's dumb because it's already uniformly equicontinuous. I can't think of any interesting pointwise equicontinuous functions.
If false we want to construct a counter-example, which means finding a non-compact domain where pointwise equicontinuity implies uniform equicontinuity. My first thought is to try to take the domain to be some subset of the reals which is either not closed or not bounded. Probably start by trying "not bounded", like maybe the domain is $\Bbb N$. Is every pointwise equicontinuous sequence of functions uniformly equicontinuous? It doesn't seem like it should be--having an infinity of points seems like room enough to make a pointwise equicontinuous sequence which is not uniformly so.
I think any domain of reals containing an open interval will not have the property that p.w.e.c implies u.e.c., so maybe I should stop looking for examples in the reals.
What happens with the discrete metric on some set? The only compact sets are the finite sets, so if I can find any infinite domain where p.w.e.c. implies u.e.c. then I'm done. But if the natural numbers didn't work, then this probably can't work either.
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whybecause | (no comment) | Feb 11, 2022 at 21:30 |
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