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Let $C(X)$ be the set of all real-valued continuous functions on a set $X$. X is a compact topological space. Let $d_\infty(f,g):=\sup d(f(x),g(x))$, where $d$ is the standard distance on the real numbers) We wish to prove that the compactness of the closed sphere $B_1(0)$ in the metric space $(C (X), d_\infty)$ is equivalent to $X$ being a finite set.

I tried to prove that $B_1(0)$ is total bounded, but I couldn’t. Would someone tell me how to prove this?

nowhere dense
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sally
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  • One question: what is $X$? If it's just a set, then what does it mean for a function defined on $X$ to be continuous? Presumably it's at least a topological space, but are there more assumptions? – user3482749 Dec 13 '18 at 12:54
  • I missed the important thing. (X,O) is compact topological space. – sally Dec 13 '18 at 12:57
  • I think $X$ should be hausdorff, otherwise take $X$ any infinite set with the indiscrete (aka trivial) topology. – nowhere dense Dec 13 '18 at 13:09

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Is easy to see that, for any compact space, the metric space $C(X)$ with the distance that you defined above is actually a normed vector space for the norm $\|f\|_\infty=\max_x f(x)$ (the uniform norm). So we need to translate the fact that the unity sphere in this space is compact in terms of the topological space $X$.

  • Suppose $X=\{x_1,...x_n\}$ is finite.

If $X$ is Hausdorff then this set is discrete and so $C(X)$ is the vector space of all real functions from $\{x_1,...x_n\}$ to $\mathbb{R}$. This normed vector space is isometric to $\mathbb{R}^n$ with the $\|\cdot\|_\infty$ norm and so its unit sphere is compact.

If $X$ is not Hausdorff then $C(X)$ is just a vector subspace of the vector space of all functions on $\{x_1,...,x_n\}$ above and so the unit sphere is the intersection of the unit sphere above (compact) with this vector subspace (closed) and hence is compact as well.

  • Now suppose the unit sphere of $C(X)$ is compact.

This is equivalent to $C(X)$ being of finite dimension. If $X$ is infinite and Hausdorff is not so hard to construct a sequence of points $(x_n)_n\in X$ with only one limit point. Let ${x}$ be this limit point and consider the closed set $F={x_n}\cup \{x\}$.

Now for each element $(a_n)_n$ of $$C_0(\mathbb{N})=\{(a_n)_n\in \mathbb{R}^\mathbb{N} \mid \lim a_n=0\}$$ we can construct an element $f$ of $C(X)$ by letting $f(x_n)=a_n$, $f(x)=0$ and extending by Tietze's theorem. Is not difficult to prove that in this way linearly independent families goes to linearly independent families and so $C(X)$ has infinite dimension.

Now if $X$ is not Hausdorff it is no longer true that $X$ should be finite as you can take $X$ any infinite vector space with the trivial topology or $X=\mathbb{N}$ with the cofinite topology (this example is T1).

But what is true is that its Largest Hausdorff quotient $Y$ should be finite. For this suppose it is not, Then $C(Y)$ is infinite dimensional by the fact above and if $\pi:X\rightarrow Y$ is the natural projection then $\pi^*:C(Y)\rightarrow C(X)$ give you and injection of an infinite dimensional vector space into $C(X)$. Reciprocally if the Largest Hausdorff quotient is finite then $C(X)$ is finite dimensional.

nowhere dense
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