How to evaluate infinite intersections or unions on any intervals? I'm having trouble understanding the basic principles of evaluating infinite intersections or unions on any intervals. Such as
$$\bigcap_{i=1}^{\infty} [2+i^{-2}, 5-i^{-2}]$$
I've been using the following:
https://math.stackexchange.com/a/1438760/248602
as an example, but I have trouble adapting this to infinite intersections since the arguments should be slightly different.
Can anyone explain the basic rules / common techniques for evaluating infinite intersections or unions for any intervals?
 A: "Prove" is the wrong word.  Maybe you meant "evaluate" or "simplify" or express in some simple form. One proves theorems; one proves identities; one proves propositions, lemmas, corollaries, etc.; one prives things that can be true of false. One does not "prove" intersections of sets.
As it stands now, the expression you wrote is $$\bigcap_{i=1}^\infty \{2+i^{-2}, 5-i^{-2}\}.$$ Maybe you meant $$\bigcap_{i=1}^\infty (2+i^{-2}, 5-i^{-2})$$ or $$\bigcap_{i=1}^\infty [2+i^{-2}, 5-i^{-2}].$$
In conventional notation $\{a,b\}$ means a set with only two members $a$ and $b$, and $(a,b)$ means an open interval from $a$ to $b$, with infinitely many members (except when that notation refers to an ordered pair, which is quite a different thing), and $[a,b]$ means the closed interval from $a$ to $b$.
Let us suppose for the moment that you meant $$\bigcap_{i=1}^\infty (2+i^{-2}, 5-i^{-2}).$$
The main thing you need to know is this: A number (or other object) is a member of the intersection of a specified collection of sets if and only if it is a member of every one of them.
You're looking at
$$
(3,4) \cap (2+\tfrac 1 4, 5-\tfrac 1 4) \cap (2+\tfrac 1 9, 5 - \tfrac 1 9) \cap \cdots.
$$
These intervals are actually getting longer, so the first one is a subset of all of the others.  Therefore the only numbers that are in every one of these intervals are those that are in the first one, $(3,4)$, so that is the intersection.
I am suspicious: I wonder if you saw $\displaystyle \bigcap_{i=1}^\infty (2-i^{-2}, 5 + i^{-2})$ and interchange the plus and minus signs.  In that case, you'd have
$$
(1,6) \cap (2-\tfrac 1 4, 5 + \tfrac 1 4) \cap (2-\tfrac 1 9, 5+\tfrac 1 9) \cap \cdots.
$$
In that case, the lower bound of each interval is less than $2$ and the upper bound is more than $5$.  $2$ and $5$ are both members of every one of those intervals, and so are all numbers between $2$ and $5$. But every number less than $2$ will get excluded by some of the intervals, as will every number more than $5$.  Thus the intersection in that case is $[2,5]$, a closed interval that includes both endpoints.
