Show that a metric space has uncountably many disjoint dense subsets if every ball is uncountable.

Question

I was wondering how to show that a metric space $X$ has uncountably many dense subsets $Y_{\alpha}$ such that $Y_{\alpha_1} \cap Y_{\alpha_2} = \emptyset$ if $\alpha_1 \neq \alpha_2$, under the assumption that every ball in $X$ is uncountable. When $X = (a,b)$ is a non-empty interval, this is fairly trivial because we can just choose an (uncountable) Hamel basis $H \subset \mathbb{R}$ when ${\mathbb R}$ is a vector space over $\mathbb{Q}$, and then let $Y_\alpha = \alpha {\mathbb Q} \cap X$ for $\alpha \in H$. But what about an arbitrary metric space $X$ where every ball is uncountable?

Answer 1

Claim 1: Let $\kappa$ be an uncountable cardinal and $X$ a metric space in which each open ball has size $\kappa$. Let us call such a space $\kappa$-homogeneous. Then $X$ can be divided into $\kappa$ many dense subsets.

Proof: Clearly $|X| = \kappa$ and hence $X$ has a basis of size $\kappa$ (use rational radii balls), say $\{B_{\alpha} : \alpha < \kappa\}$. Inductively construct $\{D_{\alpha} : \alpha < \kappa\}$ by putting, at stage $\alpha$, one point from each $B_{\beta}$, $\beta < \alpha$ into every $D_{\beta}$, $\beta < \alpha$.

Claim 2: Let $X$ be a metric space in which every open ball is uncountable. Then there is a family of pairwise open balls $U$ of $X$, such that the union of the family is dense in $X$ and each $U$ is a $\kappa$-homogeneous metric space for some uncountable cardinal $\kappa$.

Proof: Every open ball in $X$ contains a homogeneous ball as there is no infinite decreasing sequence of cardinals. So take a maximal family of pairwise disjoint homogeneous open balls in $X$.

Claims 1 + 2 give your result.

The fact that $X$ is a metric space was important here. There are examples of Hausdorff spaces in which every ball is uncountable yet there is no dense codense subset. Such spaces are called irresolvable.

Answer 2

I have a proof that almost works for arbitrary $X$, and definitely works for complete $X$. For any point $x \in X$, there must be infinitely many "shells" $S(x,1/2^k) = B_{1 / 2^k}(x) - B_{1 / 2^{k+1}}(x)$ around $x$ ($k \geq 0$ an integer) that contain uncountably many points, because if there were only finitely many than we would find some ball $B_{1 / 2^k}$ which has only countably many points. So proceed as follows. By Zorn's lemma, define $X_0$ to be a maximal set of points in $X$ which are pairwise distance at least 1 apart. Then similarly for $n \geq 1$ define $X_n \supset X_{n-1}$ to be a maximal set of points in $X$ which contains $X_{n-1}$ and such that all points have pairwise distance at least $1/2^n$ apart. Note that by maximality, any point in $X$ must be within distance $1/2^n$ of some point in $X_n$. For each $x \in \cup_n X_n$, let $(k_n(x))_{n \in {\mathbb Z}_{\geq 0}}$ be an infinite sequence of increasing integers $ > 1$ such that the shells $S(x,1/2^{k_n(x)})$ all have uncountably many points. Note the shells $S(x,1/2^{k_n(x)})$ around a given point $x$ are disjoint for different values of $n$. Then define dense subsets $Y_\alpha \subset X$ for the uncountably many $\alpha \in [0,1]$ as follows. First, for every $x \in X_0$, assign a distinct point $y_\alpha(x)$ to each $Y_\alpha$, where each $y_\alpha(x)$ is chosen from the uncountable shell $S(x,1/2^{k_0(x)})$. Note this is possible for all $x \in X_0$ because the shells $S(x,1/2^{k_0(x)})$ are disjoint because $1/2^{k_0(x)} \leq 1/4$ for all $x$, and all points $x \in X_0$ are distance at least 1 apart. Then similarly, for each $x \in X_1$, assign a distinct point $y_\alpha(x)$ to each $Y_\alpha$, where each $y_\alpha(x)$ is chosen from the uncountable shell $S(x,1/2^{k_1(x)})$. Again all shells will be disjoint because $1/2^{k_1(x)} \leq 1/8$ for all $x$ and all points $x \in X_1$ are distance at least $1/2$ apart. Note all $Y_\alpha$ are disjoint by construction. Furthermore every $Y_\alpha$ will contain a point inside the ball $B_{1/2^n}(x)$ for every $x \in X_n$, which guarantees $Y_\alpha$ is dense.

There is just one complication we need to consider. Namely, there could be a point $x_n \in X_n$ which is contained in the chosen shell for a previous point $x_k \in X_k$ where $k < n$. To handle this case, we modify the chosen points from the shell around $x_k$ as follows. We note that for $x_k \in X_k$, there must be some $n > k$ such that there are two or more points in $X_n$ that are in some common uncountable shell around $x_k$. Choose such $n$ and choose two such points in $X_n$. Then note that for the two chosen points in $X_n$, they must have disjoint uncountable shells contained in the shell around $x_k$ such that each disjoint shell around the two chosen points in $X_n$ each contains at least two points $x_m \in X_m$ where $m > n$. We thus get an infinite binary tree of shells ordered by inclusion, where all shells at each fixed level of the tree are disjoint. Since $X$ is complete, choosing any infinite path in the tree starting at the root will give a distinct limit point in $X$. So starting at the root of the tree, create a mapping from $[0,1]$ to points in the tree by mapping $0. a_1 a_2 a_3 \ldots$ in binary to the path in the tree where at level $j$ in the tree, we traverse left if $a_j = 0$ and traverse right if $a_j = 1$. Then note that the Cantor set will map to an uncountable set, such that for each node in the tree, the number of points left in the corresponding shell that are not chosen is uncountable. This is how we choose uncountably many points from a given shell around a point $x_k \in X_k$, while guaranteeing that all subsequent shells around $x_n \in X_n$ for $n > k$ will still have an uncountable number of points left to choose from, if they had an uncountable number of points to begin with. Note in particular this works because for each $x_n \in X_n$, appearing in a tree, the possible parent $x_k \in X_k$ of $x_n$ is uniquely defined for each $k < n$.

Answer 3

We can define our sets by transfinite recursion. Let's take the set of countable ordinals as our indexing set, and proceed as follows:

Let us take the set of all open balls in $X$, and using the axiom of choice, take one point from each of the open balls, and define that set of points to be $Y_1$.

For any countable ordinal $\alpha$, let us take the set of all open balls in $X$, and using the axiom of choice, define $Y_\alpha$ by taking one point from each of the open balls, but not choosing a point that's already been chosen for any of the previous sets $Y_\beta$ for $\beta < \alpha$ (we can do this because each open ball is uncountable, and we've only taken out countably many points from each open ball for our previous sets, since there are only countably many ordinals less than $\alpha$).

Since there are uncountably many countable ordinals, we're done.

EDIT: As @user2566092 points out, you could accidentally pick all the points in some open ball for $Y_1$, so subsequent sets $Y_\alpha$ wouldn't be able to choose any points for that open ball. This can be remedied as follows: use the axiom of choice to well-order the set of all open balls in $X$, and let the order type of that well-ordered set be $\gamma$. Let us define $Y_1$ be transfinite recursion. Choose an arbitrary point $y_1$ in $B_1$. For any $\alpha < \gamma$, first try choosing a point $y_\alpha$ in $B_\alpha$ that has already been chosen as one of the previous points $y_\beta$ for $\beta < \alpha$. If no such point exists, then choose any point $y_\alpha$ in $B_\alpha$. And then define $Y_1$ to be the set of all $y_\alpha$ for $\alpha < \gamma$. (The fact that $Y_1$ exists relies on the axiom of choice, of course.)

We can similarly choose points for any $Y_\alpha$ (where $\alpha$ is a countable ordinal) in the same manner, maintaining the condition from before that the points you choose for $Y_\alpha$ can't be the same as any of the points that have been chosen for the sets $Y_\beta$ for $\beta < \alpha$.

EDIT 2: Here is a proof that my construction works, under the assumption that $X$ is a separable metric space and the continuum hypothesis holds. (I'm not sure whether one or both assumptions can be dropped.)

Assuming the continuum hypothesis, a separable metric space must have cardinality less than or equal to aleph_1, so $X$ has cardinality at most aleph_1, and thus so does the set $B$ of open balls on $X$. Using the axiom of choice, well-order the set $B$, and let $\gamma$ be the order type of the well-ordering. Note that $\gamma$ is less than or $\omega_1$.

Since $X$ is separable, it's second countable, so there's a countable subset $C$ of $B$ which is a basis for the topology on $X$, i.e. if $U$ is an open set and $x$ is a point in $U$, there's an element of $C$ which contains $x$ and is a subset of $U$. Let $\delta$ be the smallest ordinal $\alpha$ such that the set of all $B_\beta$ for $\beta < \alpha$ contains C. Then $\delta < \gamma$, so $\delta$ is a countable ordinal.

Let $y_1$ be any element of $B_1$. For any ordinal $\alpha < \gamma$, if any of the previous elements $y_\beta$ for $\beta < \alpha$, then let $y_\alpha$ be the $y_\beta$ in $B_\alpha$ with the least index. If no such point exists, let $y_\alpha$ be any element of $B_\alpha$. Then the set of all $y_\alpha$ for $\alpha < \delta$ is the same as the set of all $y_\alpha$ for $\alpha < \gamma$, because $C$ is a basis for the topology on $X$. Thus this set $Y_1$ is a countable dense subset of $X$. (It's countable because $\delta$ is countable.) Since it's countable, it doesn't contain any open balls, since every open ball is uncountable.

We can do a similar construction for each $Y_\alpha$ (where $\alpha$ is a countable ordinal), maintaining the condition from before that the points you choose for $Y_\alpha$ can't be the same as any of the points that have been chosen for the sets $Y_\beta$ for $\beta < \alpha$. Then each $Y_\alpha$ will be countable, and there are only countably many ordinals less than a given countable ordinal, so we'll never cover all the points in an open ball in the stages before we get to any $Y_\alpha$.