Question: I have a question for an assignment that involves this term.

$$x\in \bigcup\_{k=1}^{\infty}\bigcap\_{n=k}^{\infty}A\_{n}.$$

I have a faint idea of what it means, but perhaps someone can tell me if I am wrong.

first im considering the intersection sign by itself, assuming that n=k=1 at first, meaning a family of sets A1...A(inf) all joined by intersection such that its only true (ie. not the null set) if there is some common element in all the sets.

then i do the same thing but starting with n=k=2 so that you have an identical family of sets with the exception of it not including A1. again X must be in all the sets.

then repeat so you have an infinite family of families of sets, all one set smaller than the last. and take the union of all of them. 

I'm having trouble though seeing what the point of the union is, It makes sense if the union and intersection signs were switched because then they both impose a condition, but having them this way, the union doesn't seem to do anything, or rule anything out?

if there is something obvious that I'm missing Id greatly appreciate your help.

Answer: 
This set is called $\lim\inf A\_n$.

To understand what it means, you have to use the definitions of the union and intersection.
Let $$x\in\bigcup\_{k=1}^{\infty}\bigcap\_{n=k}^{\infty}A\_n$$
By definition of the union, there exists an integer $k$ such that
$$x\in\bigcap\_{n=k}^{\infty}A\_n$$
Then by definition of the intersection:
$$\forall n\geq k,\, x\in A\_n$$

So $$x\in\bigcup\_{k=1}^{\infty}\bigcap\_{n=k}^{\infty}A\_n\iff\exists k\in\mathbb{N},\, \forall n\geq k,\, x\in A\_n$$
This means exactly that $x$ belongs to all the $A\_n$ except finitely many.

If you permute the intersection and the union, you get $\lim\sup A\_n$ which is the set of all $x$ that belong to an infinity of $A\_n$.

You can check that $\lim\inf A\_n\subset\lim\sup A\_n$.
Traditionally, $\bigcup\_{k=1}^{\infty}\bigcap\_{n=k}^{\infty}A\_{n}$ is called $\textit{the inf limit}$ of $(A\_n)\_n$ and it is sometimes noted $\liminf\_n A\_n.$ If you want any "interpretration" of this construction, I can give you this $$\bigcup\_{k=1}^{\infty}\bigcap\_{n=k}^{\infty}A\_{n} = \{x : x \in A\_n \,\,\forall n, \text{excepted for a finite number of}\,\, n \}.$$