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Let $S_n$ be the set of all binary strings of length $2n$ with equal number of zeros and ones. Is it correct to say $\lim_{n\to\infty} S_n$ is countable? I wanted to use it to solve this problem. My argument is that each of $S_n$s is countable (in fact finite) thus their union would also be countable. Then $\lim_{n\to\infty}S_n$ should also be countable as it is contained in the union.

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S.B.
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1 Answers1

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The collection of all finite strings of $0$'s and $1$'s is countably infinite. The subcollection of all strings that have equal numbers of $0$'s and $1$'s is therefore countably infinite.

I would advise not using the limit notation to denote that collection. The usual notation for this kind of union is $\displaystyle\bigcup_n S_n$.

André Nicolas
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  • Looking at the context given by the problem the OP is trying to solve, it seems no infinite strings are involved, but rather a count of strings of length 2n with equal numbers of 0's and 1's. – hardmath Jul 03 '13 at 02:31
  • That's why the first part of the answer deals with that. – André Nicolas Jul 03 '13 at 02:35
  • @AndréNicolas Thank you. As I said I had an idea to solve the problem I mentioned above. I interpreted $\binom{2n}{n}/4^n$ as the probability of a number whose binary representation is $(\overline{0.a_1\cdots a_{2n}})_2$ has equal zeros and ones among $a_i$s. Then, the limit of this fraction would be the probability that a uniformly distributed r.v. on $[0,1]$ hits an element of the limit of $S_n$ (suppose that the strings are mapped to the numbers). If this limit is countable then the probability will be zero. Is this reasoning correct? – S.B. Jul 03 '13 at 02:41
  • Basically, I wanted to say as $n\to\infty$ the desired probability tends to the measure of dyadic numbers corresponding to $S_n$. If I can say $S_n$ remains countable in the limit the probability of interest would be zero. – S.B. Jul 03 '13 at 03:12
  • The approximation says that the probability that the continuous uniform is $\le x$ is well approximated by the probability that a finite uniform with large support is $\le x$. I do not see how to make this into an argument for what you want. The limit of a probability need not be the probability of a limit. The system gets upset with long comment strings. Will remove most of mine, suggest you do the same. – André Nicolas Jul 03 '13 at 03:36