# Non separable metric space implies an uncountable set with lower bounded distances?

Can this be generalized to arbitrary metric spaces?

That is, if $(X, d)$ is a metric space, does the fact that it is not separable imply that there exists an uncountable set $N \subset X$ and a constant $M > 0$ such that $\forall x, y \in N (x \neq y \Rightarrow d(x, y) > M)$?

For each $r>0$ by $A_r$ we denote a family of all subsets $S\subset X$ with the property $$x\neq y,\; x,y\in S \implies d(x,y)>r$$ Assume all sets in $A_r$ are at most countable for each $r>0$, then $X$ is separable. Indeed, fix $r>0$, then by Zorn's lemma $A_r$ have maximal element $S_r$ which is countable by assumption. From maximality it follows that $d(x,S_r)\leq r$ for all $x\in X$. Now consider countable set $$S_0=\bigcup\limits_{q\in\mathbb{Q}_+} S_q$$ By construction it is dense in $X$, hence $X$ is separable.

As the conseqence, if $X$ is not separable, then there is an uncountable set $S\in A_r$ for some $r>0$.

Similarly one can show that metric space have dense set of cardinality $\kappa$ if there is no set of cardinality $>\kappa$ with pairwise distance between elements bigger than some constant.

Yes.

First, suppose for contradiction that there is no constant $M>0$ such that, for any countable $C \subset X$, there is an $x \in X\setminus C$ so that $d(x,y) > M$ for all $y \in C$. Then taking $C_n$ to be a counterexample for $M = \frac{1}{n}$, we can see that $\bigcup_n C_n$ is a countable dense set. Since $X$ is not separable this is a contradiction and there must be such an $M>0$. Fix such an $M$.

Now proceed by transfinite recursion to construct $N$. Let $N_0$ consist of a single element, and construct $N_\alpha$ for each countable ordinal $\alpha$, so that $N_\alpha$ satisfies $\forall x, y \in N_{\alpha} (d(x,y)>M)$, in the following way: at countable stages $\alpha+1$, $N_{\alpha+1} = N_\alpha \cup \{x\}$, where $x \in X\setminus N_\alpha$ is the element which must exist by the preceding paragraph. Then $N_{\alpha+1}$ has the property that $N_{\alpha+1}$ is countable and $d(x,y) > M$ for all $x, y \in N_{\alpha+1}$. At limit stages, just take unions, so that $\displaystyle N_\alpha = \bigcup_{\beta<\alpha} N_\beta$ for limit ordinals $\alpha$. It's easy to see that $N_\alpha$ has the required property in this case as well. This recursion can proceed up to $\alpha=\omega_1$, the first uncountable ordinal. Take $\displaystyle N = N_{\omega_1} = \bigcup_{\alpha<\omega_1} N_\alpha$. This is our required set $N$.

• Thank you for your answer! Do I understand correctly that your solution is similar to Norbert's, but he's hidden the transfinite induction details in Zorn's lemma? Is the axiom of choice implicitly used in your proof? – kharvd Dec 26 '13 at 18:47
• That is correct. The implicit use of choice arises from the fact that $N_{\alpha+1}$ is not uniquely determined by $N_\alpha$. – universalset Dec 26 '13 at 19:30

Define $x_\alpha\in X$ and $k_\alpha\in\mathbb N$ for $\alpha\lt\omega_1$ by transfinite induction, as follows. Let $\alpha\lt\omega_1$ and suppose that $x_\beta,k_\beta$ have already been defined for all $\beta\lt\alpha$. Since $(X,d)$ is not separable, the countable set $\{x_\beta:\beta\lt\alpha\}$ is not dense in $(X,d)$. Choose a point $x_\alpha\in X\setminus\text{cl}(\{x_\beta:\beta\lt\alpha\})$, and choose $k_\alpha\in\mathbb N$ so that $d(x_\alpha,x_\beta)\gt1/k_\alpha$ for all $\beta\lt\alpha$.

Choose $k\in\mathbb N$ so that the set $\{\alpha\lt\omega_1:k_\alpha=k\}$ is uncountable, and let $M=1/k$. Then the set $N=\{x_\alpha:k_\alpha=k\}$ is an uncountable subset of $X$, and $d(x,y)\gt M$ whenever $x,y\in N,x\ne y$.