Given metric space $X$ with no isolated points, and dense $Y \subset X$, find $Z \subset Y$ so that $Z$ and $Y-Z$ are both dense in $X$.

Question

The question is: Given metric space $X$ with no isolated points, and dense $Y \subset X$, find $Z \subset Y$ so that $Z$ and $Y-Z$ are both dense in $X$. Now, if a metric space $X$ with no isolated points has a countable base $B$ for the induced topology, I see how to answer this question because for each $B_i \in B$, you can always choose two distinct points $y_i,z_i \in Y \cap (B_i - (\{y_1,\ldots,y_{i-1}\} \cup \{z_1,\ldots,z_{i-1}\}))$, and get disjoint $Y',Z \subset Y$ which are both dense in $X$, completing the proof. But I don't see how to do the proof if there is no countable base for the topology on $X$. Does the result hold in general for a metric space $X$ with no isolated points, and if so, what is a more general argument? It's fine to just show $Z$ exists, e.g. if axiom of choice is required or there is an even more indirect argument.

Answer 1

I believe I have a proof by Zorn's lemma. Starting with $\epsilon = 1$, by Zorn's lemma there is a maximal set of points $X_1 \subset X$ such that all points in $X_1$ are distance at least $1$ apart. (Because we can order all subsets with points distance at least $1$ apart by inclusion, and for a chain of subsets we can take the union to get an upper bound). For such $X_1$, every point in $X$ must be within distance $< 1$ of some point in $X_1$ (otherwise we could add another point to $X_1$ and get a larger set). So then we take all balls of radius $1/8$ centered at the points in $X_1$, and choose two distinct points $y_B,z_B \in Y$ inside each ball $B$. Then we similarly define $X_{1/2}$ to be a maximal superset of $X_1$ so that all points in $X_{1/2}$ are distance at least $1/2$ apart. And similarly for each ball $B$ of radius $1/16$ centered around a point in $X_{1/2}$, we choose two distinct points $y_B,z_B \in B \cap Y$ that have not been chosen so far. And so on defining $X_{1/4}$ and taking balls of radius $1/32$ etc. In general for $X_{1/2^k}$ we take balls of radius $1/2^{k+3}$. When we consider $X_{1/2^k}$, every ball $B$ of radius $1/2^{k+3}$ centered around a point in $X_{1/2^k}$ is disjoint and has only had at most finitely many points chosen from it so far, because there is at most one $x \in X_{1/2^m}$ which could have points chosen in its ball that are also in $B$, for each $m< k$ (this can be seen by the triangle inequality). Thus because $X$ has no isolated points and $Y$ is dense in $X$ we can always choose the new distinct points $y_B, z_B \in Y$ from each ball $B$. The collections of chosen points $\{y_B\}_B$ and $\{z_B\}_B$ are disjoint subsets of $Y$ which are both dense in $X$, completing the proof.