Show that a metric space has uncountably many disjoint dense subsets if every ball is uncountable. 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?
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 $\{B_{\alpha} : \alpha &lt; \kappa\}$. Inductively construct $\{D_{\alpha} : \alpha &lt; \kappa\}$ by putting, at stage $\alpha$, one point from each $B_{\beta}$, $\beta &lt; \alpha$ into every $D_{\beta}$, $\beta &lt; \alpha$.

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.
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 $\beta &lt; \alpha$ (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 $\alpha &lt; \gamma$, first try choosing a point $y_\alpha$ in $B_\alpha$ that has already been chosen as one of the previous points $y_\beta$ for $\beta &lt; \alpha$.  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 $\alpha &lt; \gamma$.  (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 $\beta &lt; \alpha$.

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 $\beta &lt; \alpha$ contains C.  Then $\delta &lt; \gamma$, so $\delta$ is a countable ordinal.

Let $y_1$ be any element of $B_1$.  For any ordinal $\alpha &lt; \gamma$, if any of the previous elements $y_\beta$ for $\beta &lt; \alpha$, 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 $\alpha &lt; \delta$ is the same as the set of all $y_\alpha$ for $\alpha &lt; \gamma$, 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 $\beta &lt; \alpha$.  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 $Y_\alpha$.