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So, I'm trying to make sense of this question:

"Suppose we have an $A_n$ for each n $\in$ N (i.e. the Naturals). Fill in the blanks with two words to get a true statement. Justify your answer.

x $\in$ $\bigcap_{k=1}^{infinity}\bigcup_{n=k}^{infinity}$ $A_n$ iff x $\in$ $A_n$ for [blank] [blank] n $\in$ N"

I'm very new unfortunately to set theory and families of sets so I'm very stuck, but here's what I've got so far:

When k=1, $\bigcup_{n=1}^{infinity}A_n$ = {1,2,3...}

When k=2, $\bigcup_{n=2}^{infinity}A_n$ = {2,3,4...}

So on, so forth. Hence the total function is $\bigcap_{k=1}^{infinity}$ { {1,2,3...}, {2,3,4...}, {3,4,5...},....}

Let { {1,2,3...}, {2,3,4...}, {3,4,5...},....} = I

This is where I become unsure: I don't know much about infinity, but it seems like this total intersection of I has no fixed list of elements? The intersection of the first element to the n'th is the n'th element. But I has an infinite number of sets as its elements, so the intersection of all its elements is its infinite'th element. And the infinite'th element of I itself has an infinite number of elements.

Hence, $\bigcap_{k=1}^{infinity}$ { {1,2,3...}, {2,3,4...}, {3,4,5...},....} = { $\infty$, $\infty$+1, $\infty$+2,$\infty$+3...., $\infty$}

I'm new to this type of maths, but I know that $\infty$ isn't a number, so I doubt I've understood at all what the $\bigcap_{k=1}^{infinity} I$ equals? Let alone how to advance from this understanding to answering the question.

My only idea goes as follows: instead of thinking about $\bigcap_{k=1}^{infinity} I$ in total, and then thinking about an element from that completed intersection, you can only consider individual elements from the intersection as you 'calculate' it whilst iterating through k?

Hence, as you iterate through k=1 to k=infinity, x $\in$ $\bigcap_{k=1}^{infinity}\bigcup_{n=k}^{infinity}$ $A_n$ iff x $\in$ $A_n$ for x > n $\in$ N

Apologies for likely explaining this poorly. Has anyone got any insight on the solution, and how to understand the concept of the question?

Many thanks, indeed.

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  • $\begingroup$ What is $A_n$? An arbitrary set? $\endgroup$ – blat Nov 27 '17 at 19:56
  • $\begingroup$ Yeah, an arbitrary set. It confused me at first because I thought it meant an a set for every integer in N, that each contains only that integer. But indeed it is referring to arbitrary sets instead. $\endgroup$ – SuperDeliciousCake Nov 29 '17 at 16:51
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For fixed values of $k$, you have $$\bigcup_{n=k}^{\infty} A_n = A_k \cup A_{k+1} \cup A_{k+2} \cup \cdots$$ So $x \in \bigcup\limits_{n = k}^{\infty} A_n$ means that $x \in A_n$ for some $n \ge k$.

Now to say that $x \in \bigcap\limits_{k=1}^{\infty} \bigcup\limits_{n=k}^{\infty} A_k$ is precisely to say that for all $k \ge 1$, there is some $n \ge k$ such that $x \in A_n$.

In plainer words: no matter how large a value of $k$, you give me, I can give you a value of $n$ larger than that $k$ such that $x \in A_n$.

As such, the words you seek are: infinitely many.

You should verify that this is equivalent to the claim that $x \in \bigcap\limits_{k=1}^{\infty} \bigcup\limits_{n=k}^{\infty} A_k$.

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  • $\begingroup$ Thanks very much! That helped clear it up very much, though I had to work hard to appreciate the more abstract and much more streamlined approach to thinking about it that you had. I've got a similar question about the union of the intersection of the An sets that I'm stuck on too, I'm hoping I can apply lessons learned here to that. $\endgroup$ – SuperDeliciousCake Nov 28 '17 at 0:25
  • $\begingroup$ I've actually gotten a bit stuck on this again going over your answer (which I'm sure is correct, but I'm a bit lost on it). If x is a member of the intersection of all those unions, why must there be infinitly many different An sets that x must belong to, rather than just one An that is infinitely far into the Naturals that, as the intersection is taken over more an more union's unto infinity, this An is still in all sets? $\endgroup$ – SuperDeliciousCake Nov 29 '17 at 18:30
  • $\begingroup$ @SuperDeliciousCake: The reason is that it's not possible to 'infinitely far into the naturals'. Although there are infinitely many natural numbers, every natural number is finite, and what it means for $x$ to be an element of $\bigcup\limits_{n=k}^{\infty} A_n$ is that $x \in A_n$ for some natural number $n \ge k$. (In particular, it appears at some finite stage.) $\endgroup$ – Clive Newstead Nov 29 '17 at 19:02
  • $\begingroup$ (P.S. On a tangentially related note, I kind of don't like the $\bigcup\limits_{n=k}^{\infty}$ notation because the $\infty$ is kind of misleading and it confuses people who are new to it, such as yourself.) $\endgroup$ – Clive Newstead Nov 29 '17 at 19:03
  • $\begingroup$ Ah ok, I get you now. I thought it might be something like that, but don't have enough experience in analysis and dealing with infinity to have been sure. Cheers, I really appreciate it. $\endgroup$ – SuperDeliciousCake Nov 30 '17 at 10:18

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