Does $\forall n\in \mathbb N: |A_n| \geq 2^n =|\{0,1\}^n|$ imply that $\underset {n\to \infty} \lim |A_i| \geq |\{0,1\}^\mathbb N|$

Does $$\forall n\in \mathbb N: |A_n| \geq 2^n =|\{0,1\}^n|$$ imply that $$\underset {n\to \infty} \lim |A_i| \geq |\{0,1\}^\mathbb N|$$

This is a lemma I would need to get around some confusing Nested Intervals in a proof of Cantor-Staub, but I am not sure if it is correct. (I know I should just learn nested intervals, but I'm told good mathematicians are also lazy so...)

I went over this again, and I realised, there is some context missing:

$$A_0 := [0, 1]$$

$$A_1 := [0, 1/3] \cap[2/3, 1]$$

$$A_2 := [0, 1/9] \cap[2/9, 3/9] \cap [6/9, 7/9] \cap[8/9, 1]$$

And so forth, each step cutting out a third of the largest continuous sub-intervals.

Now trying to show that $$\bigcap_{i =0}^\infty A_i = |\{0,1\}^\mathbb N|$$.

• Now as $$\mathbb R$$ is compact we know that $$|[a,b]| \geq 1$$ for $$a\leq b$$ call it $$(i)$$
• $$\bigcap_{i=1}^n A_i = A_n$$, as $$\forall n \in \mathbb N:A_{n+1} \subset A_{n}$$, so $$\bigcap_{i=0}^\infty A_i = \underset {n\to \infty} \lim \bigcap_{i=0}^n A_i = \underset {n\to \infty} \lim A_n$$
• Per $$(i)$$ it holds that $$A_n \geq 2^n =|\{0,1\}^n|$$, the number of biggest continuous intervals in $$A_i$$ doubles in every step ($$i\to i+1$$) .
• Now the question is, does this "invariant" persist in the limit, such that $$\underset {n\to \infty} \lim A_n \geq \underset {n\to \infty} \lim|\{0,1\}^n|$$?

Intuitively I would say yes, but this is very dangerous territory, as $$\underset {n\to \infty} \lim 2^n$$ is not properly defined...

• What if for some $k$ for all $n\geq k$ we have $|A_n|=|\mathbb N|$? Commented May 15 at 14:19
• For example, if $A_n=\mathbb N$, then $|A_n|\geq 2^n$, but $\lim|A_n|=|\mathbb N|<|\{0,1\}^{\mathbb N}|$. Commented May 15 at 14:21
• Thanks @amrsa You just answered my question. Thanks. (I feel a little ashamed that I didn't figure it out myself...). Is there some way to equate the growth of a set to the growth of a binary string? Say if $|A_n| = \Omega(2^n)$, i.e. $|A_n|$ grows at least as fast as $2^n$. I can see how this is not helpful notation, as this kind of calculus with infinity doesn't really work... Commented May 15 at 14:51
• Your set theory notation looks off to me. You write for example $$A_0 := \{[0,1]\}$$ This means that $A_0$ is a set of cardinality $1$, having a unique element (namely the set $[0,1]$). Similarly $A_1$ is a set of cardinality $1$, having a unique element (namely the set $[0,1/3] \cup [2/3,1]$). I suspect this is not what you intended, I suspect instead that you intended $A_0 = [0,1]$, $A_1 = [0,1/3] \cup [2/3,1]$, and so on. As currently written, $A_0$ and $A_1$ have no object in common, so $A_0 \cap A_1 = \emptyset$, hence $\bigcap_{i=0}^\infty A_i = \emptyset$. Commented May 15 at 14:51
• Thanks for noticing @LeeMosher, you are of course right. I adjusted it Commented May 15 at 14:54

Thanks to @amrsa's comment I have the answer before I could even type out my question properly ^^ For $$\forall n:A_n = \mathbb N$$ we have $$|A_n| \geq 2^n$$ but $$\lim |A_n| = |\mathbb N| < |\{0,1\}^{\mathbb N}|$$

The invariant is not the size of the set, but the existence of an injective function from $$\{0,1\}^n$$ to $$A_n$$, and therefore it stays greater than or equal. Now I am not sure if this is allowed in the limit.
So what is allowed is to use the binary representation of a real number in $$[0,1]$$ (we already know that $$[0,1] \simeq \mathbb R$$:
$$b := \sum_{i=1}^\infty b_i 2^{-i} \in [0,1], \forall i : b_i \in \{0,1\}$$ And then we show that there is a unique point in $$A_\infty$$ that corresponds to it. This is the classical nested intervals method: We divide an interval into 3, so we have a unique way to distribute 2 possible values. So e.g. 0.000110... could correspond to lllrrl... etc.