
#1
Mar212, 08:36 AM

P: 937

Suppose I define sets [itex]D_n = \lbrace x \in [0,1]  [/itex] x has an ndigit 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
Mar212, 11:25 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904





#3
Mar212, 12:25 PM

P: 799





#4
Mar612, 06:27 AM

P: 937

Limits in infinite unions of sets
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] ? 



#5
Mar612, 08:24 AM

Mentor
P: 16,703

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? 



#6
Mar712, 07:33 AM

P: 937




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