Limits in infinite unions of sets

[clamtrox]

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

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

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

 

[HallsofIvy]

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

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

Now for the question: is this equal to \bigcup_{n = 0}^{\infty} D_n? Clearly we have D_1 \subset D_2 \subset ... \subset D_n so I am tempted to think of this as \lim_{n \rightarrow \infty} D_n. If I am allowed to take the limit, then it would seem that \bigcup_{n = 0}^{\infty} D_n = [0,1].

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]

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

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

Now for the question: is this equal to \bigcup_{n = 0}^{\infty} D_n? Clearly we have D_1 \subset D_2 \subset ... \subset D_n so I am tempted to think of this as \lim_{n \rightarrow \infty} D_n. If I am allowed to take the limit, then it would seem that \bigcup_{n = 0}^{\infty} D_n = [0,1]. 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]

Let me rephrase this slightly: I can write any real number between 0 and 1 in binary expansion, and therefore
[0,1] = \lbrace x | x = \sum_{n=0}^{\infty} \frac{a_n}{2^n}, a_n \in \lbrace0,1\rbrace \rbrace.
Why am I not allowed to equate this with the union of sets
D_m = \lbrace x |x = \sum_{n=0}^{m} \frac{a_n}{2^n}, a_n \in \lbrace0,1\rbrace,
D = \bigcup_{m=0}^{\infty} D_m = \lim_{m\rightarrow \infty} D_m \neq [0,1] ?

 

[micromass]

Let me rephrase this slightly: I can write any real number between 0 and 1 in binary expansion, and therefore
[0,1] = \lbrace x | x = \sum_{n=0}^{\infty} \frac{a_n}{2^n}, a_n \in \lbrace0,1\rbrace \rbrace.
Why am I not allowed to equate this with the union of sets
D_m = \lbrace x |x = \sum_{n=0}^{m} \frac{a_n}{2^n}, a_n \in \lbrace0,1\rbrace,
D = \bigcup_{m=0}^{\infty} D_m = \lim_{m\rightarrow \infty} D_m \neq [0,1] ?

Again, which D_m contains x=0.1010101010101010101... ?

Remember that x being in the union means that it is an element of one of the sets. So if x\in \bigcup D_n, then x\in D_n for an n. Does there exist such an n?

 

[clamtrox]

Again, which D_m contains x=0.1010101010101010101... ?

Remember that x being in the union means that it is an element of one of the sets. So if x\in \bigcup D_n, then x\in D_n for an n. Does there exist such an n?

Thanks, got it! :)