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 $\Bbb R^N$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.