Random set of rationals topological properties

Question

Flip a coin (probability of heads is p, strictly greater than 0 and strictly less than 1) for every rational number. For each toss, if heads include the number in a set S, if tails exclude it. What is the probability S is dense?

I came up with this question while I was working on my masters and it fascinated me. I have some more time now and was thinking about it again but can't make any headway so I thought I'd reach out for help here.

My gut feeling originally was that the answer is 1, but I can't prove it. Back when I was in school I asked one of my profs and his intuition was that my conclusion was incorrect. Here is what is guiding my intuituon. Given a specified real number r, it is almost surely a limit point of S. (Take any sequence (q_n) of rationals converging to r, there must exist some infinitely many points in the sequence or else there exists some N such that for all n>N the coin toss corresponding to q_n is tails. Then p^n converges to 0. i think one can construct a similar argument for finitely many limit points. It gets murkier with infinitely many. The argument I constructed doesn't really look at S, it just looks at sequences then relates it back to S.

Answer 1

The probability is indeed $1$. For each open interval $U\subset\mathbb{R}$ with rational endpoints, note that $\mathbb{Q}\cap U$ is infinite. So $P(S\cap U=\emptyset)=0$, since if you flip infinitely many coins there is a probability of $0$ that all of them will be tails. Now there are only countably many such $U$, so by countable additivity, the union of all of these events still has probability $0$. Thus $$P(\text{there exists $U$ such that $S\cap U=\emptyset$})=0.$$ But there exists such a $U$ iff $S$ is not dense in $\mathbb{R}$, so $$P(\text{$S$ is dense in $\mathbb{R}$})=1.$$