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.
Best Answer
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 $x \in \bigcap\limits_{k=1}^{\infty} \bigcup\limits_{n=k}^{\infty} A_k$.