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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.

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1 Answer 1

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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.

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