# Some set theory proof

While reading "Understanding Analysis" by Stephen Abbott one of the exercises was to check whether the following claim was correct, and then if so, to prove it.

1. If $$A_1 \supseteq A_2 \supseteq A_3 \supseteq A_4 · · ·$$ are all sets containing an infinite number of elements, then the intersection $$\bigcap\limits_{n \in \mathbb{N}} A_n$$ is infinite as well.

It would be a great help if anyone could read through my proof and tell me whether there are any mistakes. Also I want to know if it is coherent and whether there is too little/too much detail.

Lemma: $$\bigcap\limits_{n=1}^{k} A_n=\bigcap\limits_{n=1}^{k-1} A_n \cap A_k$$

Let $$x\in \bigcap\limits_{n=1}^{k} A_n$$, then by definition of intersection, $$x\in A_1, x \in A_2...x\in A_{k}$$. Hence $$x\in A_k$$, and $$x\in A_1, x\in A_2...x\in A_{k-1}$$.It follows that $$\bigcap\limits_{n=1}^{k} A_n \subseteq\bigcap\limits_{n=1}^{k-1} A_n \cap A_k$$. Let $$x\in\bigcap\limits_{n=1}^{k-1} A_n \cap A_k$$, then by definition of intersection, $$x\in A_k$$, and $$x\in A_1,x\in A_2...x\in A_{k-1}$$. Hence $$x\in A_1,...,x\in A_k$$. Thus $$x\in \bigcap\limits_{n=1}^{k} A_n$$. Thus $$\bigcap\limits_{n=1}^{k} A_n \supseteq\bigcap\limits_{n=1}^{k-1} A_n \cap A_k$$. We can conclude that $$\bigcap\limits_{n=1}^{k} A_n = \bigcap\limits_{n=1}^{k-1} A_n \cap A_k$$.

To prove 1) we use mathematical induction. Let $$S$$ be a subset of $$\mathbb{N}$$ so that if $$k \in S$$, if $$A_1 \supseteq A_2 · · ·\supseteq A_k$$ are all sets containing an infinite number of elements, then the intersection $$\bigcap\limits_{n=1}^{k} A_n$$ is infinite as well. It is trivial that this is the case when $$k=1$$. Let the property be fulfilled for some $$k \in \mathbb{N}$$, then we shall prove that $$k+1$$ also fulfills that property(and thus is also in $$S$$). Let $$A_1 \supseteq A_2 · · ·\supseteq A_{k+1}$$ be all sets containing an infinite number of elements. From the lemma, we know that $$\bigcap\limits_{n=1}^{k+1} A_n=\bigcap\limits_{n=1}^{k} A_n \cap A_{k+1}$$. Furthermore, $$\bigcap\limits_{n=1}^{k} A_n \cap A_{k+1}=A_{k+1}$$, since $$A_1 \supseteq A_2 · · ·\supseteq A_{k+1}$$. Hence $$\bigcap\limits_{n=1}^{k+1} A_n=A_{k+1}$$, and since $$A_{k+1}$$ is a set with an infinite number of elements, so is $$\bigcap\limits_{n=1}^{k+1} A_n$$. Hence $$k+1 \in S$$ . By induction, $$S = \mathbb{N}$$. Therefore, if $$A_1 \supseteq A_2 \supseteq A_3 \supseteq A_4 · · ·$$ are all sets containing an infinite number of elements, then the intersection $$\bigcap\limits_{n \in \mathbb{N}} A_n$$ is infinite as well.

If, for each $$n\in\Bbb N$$, $$A_n=[n,\infty)$$, then $$\bigcap_{n\in\Bbb N}=\emptyset$$, in spite of the fact that each $$A_n$$ is an infinite set. You proved correctly that, or each $$N\in\Bbb N$$, $$\bigcap_{k=1}^nA_k$$ is an infinite set. But induction does not allow you to deduce from this fact that $$\bigcap_{k\in\Bbb N}A_k$$ is an infinite set.