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Quick question: I am aware of Cantor's intersection theorem for closed sets, but I was wondering if the following statement was true:

Let $A_{n}$ be any decreasing sequence of sets (in the sense that $A_{n+1} \subset A_{n}$ for all $n$). Suppose that there is also a non-empty set $A$ such that $A \subset A_{n}$ for all $n$. Is it true that $A=\cap A_{n}$? Thanks!

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  • $\begingroup$ It is a requirement that $A \subset \cap_k A_k$ but there is no reason it can not be a proper subset. Indeed if $ \cap_k A_k$ isn't empty every proper subset will had the condition you desire without being the intersection itself. $\endgroup$ – fleablood Jun 9 '17 at 15:49
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Not necessarily. Note that any subset of $A$ would have the same property, so you certainly can't prove any equality in general. All you can say is that $A\subset \bigcap_n A_n$ (by definition).

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The answer is no. Take $A_n=[0,1+\frac{1}{n}], A=[0,\frac{1}{2}]$ is a counter example.

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No. $A \subset \cap_{n\in \mathbb N} A_n$ but it it does not follow that $A = \cap_{n\in \mathbb N} A_n$ if it is a proper subset set.

Let $A_k = (-\frac 1k, 5 + \frac 1k)$. $\cap_{n} A_n = [0, 5]$ (left as an exercise to the reader ... Ha! I always wanted to say that .... oh, wait, no.... I never wanted to say that .... oh, well.)

But let $A = (2.3, 4)$. Clearly $(2.3, 4) \subset (-\frac 1k, 5 + \frac 1k)$ and $(2.3, 4) \subset [0,5]$ but obviously $(2.3, 4) \ne [0,5]$.

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