Let $(X,d)$ be a metric space and $M\subset X$ is uncountable. There is $\alpha>0$ such that $d(x,y)=\alpha$. Prove that $X$ is not separable.

Question

Let $(X,d)$ be a metric space and $M\subset X$. Suppose $M$ is uncountable and there is $\alpha>0$ such that $d(x,y)=\alpha$ for every $x,y\in M$ with $x\not=y$. Prove that $X$ is not separable.

Any help would be nice.

Showing that $X$ is separable if and only if $M$ is countable would imply that if $X$ is NOT separable if $M$ is uncountable. Would that be enough?

Answer 1

Let $X$ be separable; so it would be $M$. Now take $x \in M$ and $(x_n)_{n \in \mathbb{N}} \subset M\ \colon\ x_n \longrightarrow x$. Standing to the hypothesis you would have $ d(x_n, x) = \alpha\ \forall n$: that's a contradiction.

Answer 2

Hint: If $A$ is a dense subset, then for every $x\in M$ there is $a(x)\in A$ such that $d(x,a(x))\leq \dfrac{\alpha}{4}$. It is clear that if $x,x'\in M$ with $x\neq x'$ then $a(x)\neq a(x')$, hence $A$ is uncountable set.

Answer 3

Hint $M$ is an uncountable discrete subset of $X$. This means that for $x\in M$; $\{x\}$ is open.

Claim Suppose $(X,d)$ is a separable metric space. Then every collection of open disjoints sets in $X$ is at most countable.

Proof Let $\{x_1,\ldots\}$ be a dense countable subset. By definition, each open set $O$ in the collection $O$ contains some $x_i$, in particular we may choose $x_i$ of least index $i$. Since the sets are disjoint, we have a well-defined injective mapping from the collection of this open sets to $\{x_1,\ldots\}$, and the claim follows.