# An analysis proposition, true or false?

Suppose $$(X,d)$$ is a metric topological space with $$|X|>=\aleph$$ (Topological spaces such as $$\mathbb{Q}$$, $$\mathbb{Z}$$ should not be considered).

Prove or disprove: If for every uncountable subset $$S \subseteq X$$, $$\mathop{\inf}\limits_{x \neq y,{x,y\in S}}d(x,y)=0$$, then $$X$$ satisfies $$C_2$$ axiom (or $$X$$ is separable since such two conceptions are equivalent when $$X$$ is a metric space).(*)

Below is the remark:

I want to verify it because its converse proposition is true. If an uncountable subset $$S \subseteq X$$ such that $$\mathop{\inf}\limits_{x \neq y,{x,y\in S}}d(x,y)>0$$, then $$X$$ must not be $$C_2$$. A typical example is $$l_{\infty}$$, which is formed by bounded series. The subset $$Binary$$ which is formed by infinite $$0-1$$ strings is uncountable, and $$\mathop{\inf}\limits_{x \neq y,{x,y\in Binary}}d(x,y)=1$$ (here distance is induced by $$l_{\infty}$$ norm), and $$l_{\infty}$$ is not separable and not a $$C_2$$ space.

To prove its converse proposition, we notice that： $$X$$ is a metric topological space $$\rightarrow$$ ($$X$$ is $$C_2$$ $$\leftrightarrow X$$ is separable) is always true. So we only need to prove $$X$$ is not seperable. Suppose $$S$$ is an uncountable set and $$m=\mathop{\inf}\limits_{x \neq y,{x,y\in S}}d(x,y)>0$$. We can pick neighborhoods $$N=\{x \in S|B(x,\frac{m}{3})\}$$, those neighborhoods in $$N$$ do not intersect with each other so $$N$$ is uncountable. If $$A$$ is dense in $$X$$, we assert that every neighborhood in $$N$$ contains at least one point in $$A$$, so $$|A| \geq |N|$$ and $$A$$ can not be countable.

I really appreciate your help if you prove or disprove (*). If you can prove it in some special cases, I will also thank you for your help.

Let $(X,d)$ be a metric space satisfying the condition: for every uncountable set $S\subseteq X$ we have $\inf\{d(x,y):x,y\in S,\ x\ne y\}=0.$ I will show that $X$ is separable. (No assumption on the cardinality of $X$ is needed.)
For each $n\in\mathbb N$ let $S_n$ be a maximal subset of $X$ with the property that $x,y\in S,\ x\ne y\implies d(x,y)\ge\frac1n.$ (This requires the axiom of choice, e.g., in the form of Zorn's lemma.) It follows directly from our hypothesis on $(X,d)$ that each $S_n$ is countable; hence the set $S=\bigcup_{n\in\mathbb N}S_n$ is countable. I claim that $S$ is dense in $X.$
Consider any point $x\in X.$ For each $n\in\mathbb N,$ it follows from the maximality of $S_n$ that we can choose a point $x_n\in S_n$ with $d(x_n,x)\lt\frac1n.$ Thus $x_1,x_2,x_3,\dots,x_n,\dots$ is a sequence in $S$ converging to $x.$