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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.

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  • 2
    $\begingroup$ 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 $\endgroup$ – Jakobian Jan 8 at 18:41
  • $\begingroup$ A good first step would be for you to show that Jakobian's comment is true $\endgroup$ – Omnomnomnom Jan 8 at 18:49
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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.

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  • $\begingroup$ . . . nice . . . $\endgroup$ – janmarqz Jan 8 at 19:29
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Hint :consider a sequence which has infinitely many zeros and a sequence which has finitely many zeros.

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