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Limits in infinite unions of sets

  1. Mar 2, 2012 #1
    Suppose I define sets [itex]D_n = \lbrace x \in [0,1] | [/itex] x has an n-digit long binary expansion [itex]\rbrace [/itex].

    Now consider [itex]\bigcup_{n \in \mathbb{N}} D_n[/itex]. This is just the set of Dyadic rationals and therefore countable for sure.

    Now for the question: is this equal to [itex]\bigcup_{n = 0}^{\infty} D_n[/itex]? Clearly we have [itex] D_1 \subset D_2 \subset ... \subset D_n [/itex] so I am tempted to think of this as [itex] \lim_{n \rightarrow \infty} D_n [/itex]. If I am allowed to take the limit, then it would seem that [itex]\bigcup_{n = 0}^{\infty} D_n = [0,1][/itex]. Where am I doing a naughty physicist mistake?
  2. jcsd
  3. Mar 2, 2012 #2


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    How do you conclude this? It looks to me like this would be the set of all numbers that have terminating decimal expansions which is a subset of the rational numbers in [0, 1].

  4. Mar 2, 2012 #3
    Which D_n contains .10101010...? In fact what you proved is that the set of rationals with terminating binary expansion is countable.
  5. Mar 6, 2012 #4
    Let me rephrase this slightly: I can write any real number between 0 and 1 in binary expansion, and therefore
    [itex] [0,1] = \lbrace x | x = \sum_{n=0}^{\infty} \frac{a_n}{2^n}, a_n \in \lbrace0,1\rbrace \rbrace. [/itex]
    Why am I not allowed to equate this with the union of sets
    [itex] D_m = \lbrace x |x = \sum_{n=0}^{m} \frac{a_n}{2^n}, a_n \in \lbrace0,1\rbrace [/itex],
    [itex] D = \bigcup_{m=0}^{\infty} D_m = \lim_{m\rightarrow \infty} D_m \neq [0,1] [/itex] ?
  6. Mar 6, 2012 #5
    Again, which [itex]D_m[/itex] contains x=0.1010101010101010101... ???

    Remember that x being in the union means that it is an element of one of the sets. So if [itex]x\in \bigcup D_n[/itex], then [itex]x\in D_n[/itex] for an n. Does there exist such an n?
  7. Mar 7, 2012 #6
    Thanks, got it! :)
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