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Why do we need $M$ to be compact in the proof of the completeness of $C(M)$ with usual uniform norm? I have seen the proofs for $C[a,b]$ and for $C(M)$ and although $M$ compact is supposed to be more general than $[a,b]$ I couldn't notice the slightest difference nor the use of the compactness property. Why doesn't this imply that the $C(M)$ is complete for $M$ any set?

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    $\begingroup$ How can you define the norm if $M$ is not compact? $\endgroup$ Jan 14 at 18:37
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    $\begingroup$ Please use MathJax $\endgroup$ Jan 14 at 18:38
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The uniform norm is defined by $\|f\|=\sup_{x\in M}|f(x)|$. If $M$ is not compact, then this supremum doesn't exist for most functions.

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    $\begingroup$ Why can't we allow the norm to be infinite (if that is what you mean by not existing)?The 4 properties that define a norm seem to be true anyway (en.wikipedia.org/wiki/Norm_(mathematics)) $\endgroup$
    – J.C.VegaO
    Jan 14 at 18:45
  • $\begingroup$ @J.C.VegaO Wikipedia's definition says $\| \cdot \| : X \to \mathbb{R}$, but $\infty \not\in \mathbb{R}$ $\endgroup$ Jan 14 at 18:47
  • $\begingroup$ Because, by definition, a norm is a function which takes real values. And infinity is not a number. $\endgroup$ Jan 14 at 18:48
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    $\begingroup$ For instance, what would $\|v-w\|\geqslant\bigl|\|v\|-\|w\|\bigr|$ mean then? $\endgroup$ Jan 14 at 18:53
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    $\begingroup$ Note that if we work on the space of continuous bounded functions, that everything works fine. $\endgroup$ Jan 14 at 18:54
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lFor more general spaces (generaly assumed to be completely regular $T_1$ to keep it interesting), we need the consider $C^\ast(M)$, the set of bounded real functions and we're still OK (i.e. we have a Banach space). Otherwise the norm need not be defined for all elements. There is a nice theory for locally compact Hausdorff $M$ (duality, Riesz etc) and for compact Hausdorff this also applies plus some nicer algebraic properties too.

So the theory is most convenient and elegant for $M$ compact Hausdorff. (See Semadeni's classic book *Banach spaces of continuous functions * and for non-Banach theory Gilman and Jerrison, Rings of continuous functions).

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You get a complete space for arbitrary $M$ if you don't insist on having a norm but only a metric. Equip $C(M)$ with the metric $d_\infty(f,g)=\max\{\|f-g\|_\infty,1\}$. This metric induces the topology of uniform convergence. If $M$ is compact, cutting off at $1$ does not make a difference for completeness because all that matters are small values of the metric.

With this metric, $C(M)$ is always complete: Let $(f_n)$ be a Cauchy sequence. By definition, $f_n(x)$ is a Cauchy sequence for every $x\in X$, so that there exists $f\colon M\to\mathbb C$ such that $f_n\to f$ pointwise. In fact, this convergence is uniform: $$ |f(x)-f_n(x)|=\lim_{m\to\infty}|f_m(x)-f_n(x)|\leq \liminf_{m\to\infty}d_\infty(f_m,f_n), $$ which is independent of $x\in M$ and goes to zero as $n\to\infty$. By standard arguments, $f$ is continuous and $d_\infty(f,f_n)\to 0$.

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