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    I want to prove the set of reals in [0,1] whose decimal expansion does not contain 3 is not dense.

    Here's what I thought: suppose it is dense. Then there is a point of the set in between an open interval where \displaystyle \frac{33}{100} < x = \sum^{\infty}_{k=1} a_k 10^{-k} < \frac{34}{100}. Then I claim we must have a_1 = 3. Suppose not: 4 \leq a_1 \leq 9 or 0 \leq a_1 \leq 2.

    The former interval seems to lead to a contradiction without much effort by bounding the positive series above by a negative number. But I can't do it with the latter. Any help?

    Got another set, too: prove the set of rational numbers whose denominators is a power of 2 is dense. I have no idea how to begin with this one. Any hints for making it slightly more tractable?

    I made an observation (albeit simple and I'm not quite sure if it's true) that we may find a natural number n where 1/2^n < 1/n < ɛ when n > 0 as a simple consequence of the Archimedean property. Perhaps this will be useful? However, I guess I'll have to convert my "every open interval" definition of dense in to terms of ɛ to use it.

    Thanks!
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    For the first, note that there is no number in (0.3, 0.399999] whose decimal expansion doesn't include a 3.

    For the 2nd, you can easily use your observation to show that for any x in [0,1] we can find a sequence a_n s.t. |a_n / 2^n - x| < 1/n. Then a_n/2^n -> x.
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    You've lost me on the first one I'm afraid, isn't that what I'm trying to do?
 
 
 
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