Let X be a complete metric space in which every closed ball is uncountable. Prove that X has cardinal number greater or equal than the continuum Let X be a complete metric space in which every closed ball is uncountable. Prove that X has cardinal number >= c (continuum) (Can you please prove with properties of Separability of a Metric Space? Thank you^^.)
 A: One way to go about it: start with two disjoint closed balls $D(0)$ and $D(1)$ both with some radius $r_0 \le 1$. Inside $D(0)$ find two disjoint closed balls $D(0,0)$ and $D(0,1)$ and inside $D(1)$ two disjoint closed balls $D(1,0)$ and $D(1,1)$, all with some radius $r_1 \le {r_0 \over 2}$. This can be done as we always have two distinct points, which we separate by disjoint closed balls, and shrink radii as needed. This you continue, so each time we are splitting $D(i_0, i_1,\ldots,i_n)$ into $D(i_0, i_1,\ldots,i_n,0)$ and $D(i_0, i_1,\ldots,i_n,1)$ with all radii at level $n$ at least half the former radius.
Then for each infinite sequence $(i_n)_n$ of zeroes and ones we get a corresponding sequence of decreasing diameter closed balls $D(i_0), D(i_0,i_1), \ldots D(i_0,i_1,\ldots, i_n)$ etc. which has a unique point $f((i_n))$ in its intersection by completeness, and this is a 1-1 map, clearly, by the disjointness of the levels. The set of all infinite sequences of zeroes and ones has size continuum, so that's at least how many points in $X$ we have.
