# Let $A_{n}=\left\{ f \in \left\{ 0,1\right\}^{\mathbb N}: f(n)=0 \right\}$

Let $$A_{n}=\left\{ f \in \left\{ 0,1\right\}^{\mathbb N}: f(n)=0 \right\}$$ Find
(a) $$| \bigcap_{m \in \mathbb N}^{} \bigcup_{n \ge m}^{} A_{n}|$$
(b) $$|\left\{ 0,1\right\}^{\mathbb N} \setminus \bigcap_{m \in \mathbb N}^{} \bigcup_{n \ge m}^{} A_{n}|$$ Firstly I have a problem with calculation $$\bigcap_{m \in \mathbb N}^{} \bigcup_{n \ge m}^{} A_{n}$$ because I do not understand how I can do this for function which natural numbers converts to $$0$$ or $$1$$ and how use the fact that $$f(n)=0$$. That is why I also cannot find a cardinality of this sets and I need some tips.

• set in (a) is made up of functions on $\{0, 1\}^\mathbb{N}$, that attain $0$ infinitely many times, you can think of it as set of infinite sequences made up from $0$'s and $1$'s with infinitely many $0$'s – Jakobian Jan 8 at 18:41
• A good first step would be for you to show that Jakobian's comment is true – Omnomnomnom Jan 8 at 18:49

For a fixed $$n \in \mathbb{N}$$, the set $$A_n$$ is the set of functions $$f : \mathbb{N} \to \{0,1\}$$ such that $$f(n) = 0$$.

This means that, for fixed $$m \in \mathbb{N}$$, the set $$\bigcup\limits_{n \ge m} A_n$$ is the set of functions $$f : \mathbb{N} \to \{0,1\}$$ such that $$f(n) = 0$$ for some $$n \ge m$$.

To say that $$f \in \bigcap\limits_{m \in \mathbb{N}} \bigcup\limits_{n \ge m} A_n$$ is thus to say that $$f : \mathbb{N} \to \{0,1\}$$ and, for all $$m \in \mathbb{N}$$, there is some $$n \ge m$$ such that $$f(n) = 0$$. That is, no matter how big you make the natural number $$m$$, there will always be some $$n \ge m$$ that $$f$$ sends to zero.

Thus $$\bigcap\limits_{m \in \mathbb{N}} \bigcup\limits_{n \ge m} A_n$$ is the set of all functions $$f : \mathbb{N} \to \{0,1\}$$ that attain the value $$0$$ infinitely many times.

By identifying a function $$f : \mathbb{N} \to \{0,1\}$$ with the subset $$f^{-1}[\{0\}] = \{ n \in \mathbb{N} \mid f(n) = 0 \}$$, there is thus a bijection between $$\bigcap\limits_{m \in \mathbb{N}} \bigcup\limits_{n \ge m} A_n$$ and the set of all infinite subsets of $$\mathbb{N}$$.

It then follows that $$\{0,1\}^{\mathbb{N}} \setminus \left( \bigcap\limits_{m \in \mathbb{N}} \bigcup\limits_{n \ge m} A_n \right)$$ is the set of all functions $$f : \mathbb{N} \to \{0,1\}$$ attaining the value $$0$$ only finitely many times, and the identification $$f \mapsto f^{-1}[\{0\}]$$ defines a bijection from this set to the set of all finite subsets of $$\mathbb{N}$$.

This is now more than enough information to figure out the cardinalities of both sets.

• . . . nice . . . – janmarqz Jan 8 at 19:29

Hint :consider a sequence which has infinitely many zeros and a sequence which has finitely many zeros.