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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?

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 $(x_n)_nin X$ 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 $$C_0(mathbb{N})={(a_n)_nin mathbb{R}^mathbb{N} mid lim a_n=0}$$
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 $pi:Xrightarrow Y$ 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.