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Limits in infinite unions of sets

by clamtrox
Tags: infinite, limits, sets, unions
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clamtrox
#1
Mar2-12, 08:36 AM
P: 939
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?
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HallsofIvy
#2
Mar2-12, 11:25 AM
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Quote Quote by clamtrox View Post
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].
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].

Where am I doing a naughty physicist mistake?
SteveL27
#3
Mar2-12, 12:25 PM
P: 800
Quote Quote by clamtrox View Post
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?
Which D_n contains .10101010...? In fact what you proved is that the set of rationals with terminating binary expansion is countable.

clamtrox
#4
Mar6-12, 06:27 AM
P: 939
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] ?
micromass
#5
Mar6-12, 08:24 AM
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Quote Quote by clamtrox View Post
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] ?
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?
clamtrox
#6
Mar7-12, 07:33 AM
P: 939
Quote Quote by micromass View Post
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?
Thanks, got it! :)


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