TITLE: If $(M,d)$ has an uncountable discrete subset then $M$ is not separable?
QUESTION [5 upvotes]: I was thinking a bit about the proof of the fact that $\ell^{\infty}$ is not separable. And from the proof I saw, which uses a subspace which is discrete and uncountable, I thought I can prove it for any space which has the property.
Let's take $M$ a metric space with a discrete subspace $X$. Let $D$ be a dense subset of $M$. For every $x\in X$, we consider the ball $B_x=B_{\frac{1}{3}}(x)$. Then since $D$ is dense, for every such $x$, $B_x\cap D\neq \varnothing$, so let's take an arbitrary fixed element $a(x)\in B_x\cap D$ for each $x$.
Since $x\neq y\Rightarrow B_x\cap B_y=\varnothing$ (because $d(x,y)=1$), so $(B_x\cap D)\cap (B_y\cap D)=\varnothing\Rightarrow a(x)\neq a(y)$, so $x\mapsto a(x)$ is injective, thus $D$ is uncountable and $M$ is not separable.
Is that right?
Edit: Now I see that this can be solved using this since the subset $D$ would not be separable.

REPLY [2 votes]: All is good, just want to point out the definition of a discrete subset $X$. $X$ is discrete if for every $x$ in $X$ there exists an open subset $U_x$ of $M$ such that $U_x \cap X= \{x\}$. If $M$ is a metric space that means for every $x$ in $X$ there exists an open ball $B_x \colon = B(x, r_x)$ so that $B_x \cap X = \{x\}$. Now, apriori, the balls $B_x$ might intersect, but a minor adjustment will do: the balls $B'_x\colon = B(x, \frac{r_x}{2})$ are pairwise disjoint. Indeed, assume that $B'_x \cap B'_y \ne \emptyset$. Then $d(x,y)< \frac{r_x+ r_y}{2}$. If $r_y \le r_x$ that would imply $d(x,y) < r_x$, contradiction.