# Limits in infinite unions of sets

by clamtrox
Tags: infinite, limits, sets, unions
 P: 930 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?
Math
Emeritus
Thanks
PF Gold
P: 38,456
 Quote by 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 $\bigcup_{n = 0}^{\infty} D_n = [0,1]$.
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?
P: 791
 Quote by 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?
Which D_n contains .10101010...? In fact what you proved is that the set of rationals with terminating binary expansion is countable.

P: 930

## 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
$[0,1] = \lbrace x | x = \sum_{n=0}^{\infty} \frac{a_n}{2^n}, a_n \in \lbrace0,1\rbrace \rbrace.$
Why am I not allowed to equate this with the union of sets
$D_m = \lbrace x |x = \sum_{n=0}^{m} \frac{a_n}{2^n}, a_n \in \lbrace0,1\rbrace$,
$D = \bigcup_{m=0}^{\infty} D_m = \lim_{m\rightarrow \infty} D_m \neq [0,1]$ ?
Emeritus
 Quote by 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 $D_m = \lbrace x |x = \sum_{n=0}^{m} \frac{a_n}{2^n}, a_n \in \lbrace0,1\rbrace$, $D = \bigcup_{m=0}^{\infty} D_m = \lim_{m\rightarrow \infty} D_m \neq [0,1]$ ?
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?
 Quote by micromass 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?