Why do we need $M$ to be compact in the proof of the completeness of $C(M)$ with usual uniform norm?

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?

• How can you define the norm if $M$ is not compact? Jan 14 at 18:37
• Please use MathJax Jan 14 at 18:38

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.

• 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)) Jan 14 at 18:45
• @J.C.VegaO Wikipedia's definition says $\| \cdot \| : X \to \mathbb{R}$, but $\infty \not\in \mathbb{R}$ Jan 14 at 18:47
• Because, by definition, a norm is a function which takes real values. And infinity is not a number. Jan 14 at 18:48
• For instance, what would $\|v-w\|\geqslant\bigl|\|v\|-\|w\|\bigr|$ mean then? Jan 14 at 18:53
• Note that if we work on the space of continuous bounded functions, that everything works fine. Jan 14 at 18:54

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).

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 $$d_\infty(f,f_n)\to 0$$.