Definition: A family of functions $\mathcal{F}$ on a set $X$ separates points in $X$ if for every distinct pair $x,y\in X$ there exists $f\in\mathcal{F}$ such that $f(x)\neq f(y)$.

Let $K$ be a compact metric space, is there a finite set of continuous functions $K\rightarrow \mathbb{R}$ that separates points in $K$?

Let $(K,\tau)$ be a compact Hausdorff topological space and suppose there exists a countable number of real valued maps $\mathcal{F}=\{f_n : n<\omega\}$ that separate points in $K$ (by considering $f_n/\|f_n\|_\infty$ if necessary we may assume that $\|f_n\|_\infty\leq 1$ for every $n$). Consider metric $$ d(x,y)=\frac{1}{2}|f_0(x)-f_0(y)|+\frac{1}{2^2}|f_1(x)-f_1(y)|+\cdots. $$ It is easily checked that this is a metric on $K$. It is in fact a metric for the weak topology on $K$ generated by $\mathcal{F}$.

Now the identity map $id:(K,\tau)\rightarrow (K,\tau_d)$ is a continuous (by the Universality Property) bijection from a compact to a Hausdorff space and thus an homeomorphism, meaning that $(K,\tau)$ is metrizable with metric $d$.

On the other hand let $(K,d)$ be a compact metric space, $(K,d)$ is separable. Let $D=\{x_n : n<\omega\}$ be a dense subset of $K$. Consider the family of real valued functions $\mathcal{F}=\{d_n : n<\omega\}$ doing $x\mapsto d_n(x)=d(x,x_n)$. By the triangle ineq. these functions are continuous and by density of $D$ they separate points in $K$. We conclude that

A compact space $K$ is metrizable if an only if there exists a countable family of real valued continuous functions that separates points in $K$.

Moving back to my question, I wonder if the countability condition in the above result can be improved to finiteness. Clearly following the above arguments if there is a finite set of continuous functions on $K$ compact that separates points then $K$ is metrizable. However, is the converse true? Can anyone think of a counterexample?

Background Theorem A compact Hausdorff space $K$ is metrizable if and only if the space $C(K)$ of continuous real valued functions on $K$ with the supremum norm is separable.


2 Answers 2


If the $n$ real-valued continuous functions $f_1, \ldots, f_n$ separate points of $K$, then $(f_1, \ldots, f_n)$ is a homeomorphism from $K$ to a compact subset of $\mathbb R^n$. But not every compact metric space is homeomorphic to a compact subset of $\mathbb R^n$. For example, let $K$ be the Hilbert cube. For each $k$, $K$ has a subset $S_k$ homeomorphic to $\mathbb R^k$. Since there is no homeomorphism of $S_k$ to a subset of $\mathbb R^n$ if $n < k$, such an $F$ can't exist.


No. Say $K$ is the "Hilbert cube", that is, the set of all sequences $x=(x_1,\dots)$ with $$0\le x_n\le 1/n$$for all $n$, and the metric $$d(x,y)=\left(\sum(x_n-y_n)^2\right)^{1/2}.$$The question is equivalent to asking whether there exists $N$ and a continuous injective mapping from $K$ to $\Bbb R^N$. And the answer to that is no. Let $$K_N=\{x\in K:x_n=0,n\ge N+1\}.$$ Then $K_N$ is homeomorphic to the cube $[0,1]^{N+1}$, and some not quite trivial topology shows that there is no injective continuous map from $[0,1]^{N+1}$ to $\Bbb R^N$.


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