If X is an uncountable set, will countably many random draws be dense in X? I'm interested in how large samples would approximate a distribution.
Suppose $X$ is a set of finite dimension of uncountable cardinality. If I take countably many i.i.d. random draws from the set according to a distribution with full support, will those sampled draws be almost surely dense in $X$? 
I would try a proof on $X=[0,1]$ with uniform draws by contradiction. Suppose it isn't true. Then, there is some neighborhood of length $d$ where no sampled points lie. From $n$ draws, this event has probability $(1-d)^{n}$. Send $n\rightarrow\infty$, for $d>0$ this event has zero probability. 
For $[0,1]^{n}$, simply imagine $d$ as a measure and the proof stays the same, even with a new distribution?
And we also maintain probability zero of drawing two identical points? 
Is it even sensible to talk about uncountably many random draws?
Thanks
 A: A set is dense in $X$ if and only if it intersects every element of a basis of topology for $X$. The spaces we are most familiar with, like $[0,1]$, $\mathbb R$, $\mathbb R^d$ and their subsets, have a countable basis of topology. Let $\{U_n\}$ be such a basis. 
If $P(U_n)=0$ for some $n$, then your countable sample will almost surely miss $U_n$, hence will not be dense. 
If $P(U_n)>0$, then the probability of the sample being disjoint from $U_n$ is zero, because $\lim_{k\to\infty } (1-P(U_n))^k = 0$. The union of countably many events of probability zero still has probability zero. Therefore, the countable sample will be dense almost surely. 


And we also maintain probability zero of drawing two identical points?

Yes. If $P(\{x\})=0$ for every $x\in X$, then the probability of drawing a point that was  drawn at a previous step is zero. 


Is it even sensible to talk about uncountably many random draws?

Sensible, but not as easy as in the countable case. See  continuous-time white noise.
