Let $X$ be a compact space, $C(X)$ be the metric space of continuous functions from $X$ to $\mathbb{R}$ with sup norm.
Is $C(X)$ complete?
I can prove it when $X$ is a metric space.
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$C(X)$ is never compact (with respet to the topology of uniform convergence or any other reasonnable topology). – Jochen May 04 '24 at 09:58
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You change the question from completeness to compactness. The constant functions $1,2,3...$ are in $C(X)$. Do they form a bounded set? – Kavi Rama Murthy May 04 '24 at 12:27
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I was confusing compact with complete. I will correct it. – user46190 May 04 '24 at 12:32
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It is a consequence of the uniform limit theorem.
user46190
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While that theorem does address the issue of determining that a uniformly convergent sequence of continuous functions can only converge to a continuous function, it does not address the fundamental issue of proving that a sequence of functions that is Cauchy in the uniform metric is uniformly convergent. – Sassatelli Giulio May 04 '24 at 12:58
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It is sufficient nonetheless, because being a Cauchy sequence in the uniform metric implies pointwise convergence, therefore the limit function exists and is continuous by the theorem indeed. – Desteny May 04 '24 at 20:09
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@destinysfather If you need a separate lemma to do it, the it is by definition not sufficient. To address your portion of doing half the theorem and forgetting important details, you need uniform convergence, not just pointwise convergence. – Sassatelli Giulio May 04 '24 at 21:27
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If you start with uniform convergent sequence the pointwise convergence follows directly (and I really mean directly from the definition not by a lemma). So your sequence converges uniformly AND pointwise. – Desteny May 05 '24 at 08:59