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I am asked to state whether the following is true or if false to give a counterexample:

If $A_1 \supseteq A_2 \supseteq A_3 \supseteq \ldots $ are all sets containing an infinite number of elements, then the intersection $$\bigcap_{k=1}^\infty A_k$$ is infinite as well.

I believe this statement to be false but I am not sure if the counterexample I have thought up makes sense. I said:

Let $A_n = \{m \in \mathbb{Z} | m> n\}$ for $n \in \mathbb{N}$. Would this be okay?

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    $\begingroup$ Your counterexample is correct. You should probably state explicitly that the intersection is empty (and therefore not infinite). $\endgroup$ Sep 7, 2017 at 0:49
  • $\begingroup$ @AndreasBlass Yes, I did mention it is empty and therefore not infinite. $\endgroup$
    – Sam
    Sep 8, 2017 at 0:51

2 Answers 2

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Hint: Consider $A_n = \left[-\dfrac1n, \dfrac1n \right] \subseteq \mathbb R$.

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Not quite, because then you have a finite intersection, i.e. each $A_n$ has a finite amount of elements. How about trying $A_n = [n,\infty) \cap \mathbb{N}$? What is the intersection then?

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  • $\begingroup$ I am not too sure what you mean. I want to show that the intersection is not infinite. With the set I showed that the intersection would be the empty set. I also just change the > sign. Sorry, it was a typo. $\endgroup$
    – Sam
    Sep 6, 2017 at 23:11
  • $\begingroup$ @Sam intersection is empty, not infinite, exactly as needed to show $\endgroup$
    – gt6989b
    Sep 7, 2017 at 14:03

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