Question about finite/infinite intersection 
*

*$A= \bigcap_{i \in \mathbb{N}} (-1+\frac{1}{i}, \frac{1}{i})$


*$B= \bigcap_{i=1}^{n} (-1+\frac{1}{i}, \frac{1}{i})$ with $n<\infty$
In case of $A,$ I think the intersection should be empty.
but I am unsure about $B$
the first few intervals would look like this
$i=1: (0,1)$
$i=2: (-1/2,1/2)$
$i=3: (-2/3, 1/3)$
$\vdots$
$i=n: (\frac{1-n}{n}, \frac{1}{n})$ $n<\infty$
so the value on the left would get smaller and 0 would be included in all intervals, but how about the right side?, the last value $\frac{1}{n}$ would be the smallest of all, could I just say the finite intersection would be $B=(0,\frac{1}{n})$?
 A: As $0 \not \in (-1 +\frac 11, \frac 11)= (0,1)$ we can't have $0$ in any of either the finite or infinite intersections.
If $r > 0$ then there is an $m > \frac 1r$ and $\frac 1m < r$ so $r \not \in (-1 +\frac 1m, \frac 1m)$ and so $r$ will not be in any intersection that includes an index $m > \frac 1r$.  As the infinite intersection includes that index, no $r>0$ will be in the infinite intersect.
As for whether $r$ is in the finite intersection, we'll get to that in a moment.
If $r < 0$.... then $r < 0 = -1 + \frac 11$ so $r \not \in (-1 +\frac 11, \frac 11)=(0,1)$ so $r$ is not in any of the intersections finite or infinite.
So there is no values that are in the infinite intersection.  So $\bigcap_{m\in \mathbb N}(-1+\frac 1m, \frac 1m) =\emptyset$.
And if $\bigcap_{m=1}^n (-1+\frac 1m, \frac 1m) \subset (0,\infty)$.
Okay.... back that that observation that if $m > \frac 1r$ then $r\not\in (-1+\frac 1m, \frac 1m)$.
Well..... let's step back.   If $0< r < \frac 1n$ then $r < \frac 1n < \frac 1{n-1} <....$ and $r < \frac 1m$ for all $\frac 1m$ (and, of course, $r > 0 \ge -1+\frac 1m$) so $r\in (-1+\frac 1m, \frac 1m)$ for all $m \le n$ so $r$ is in the finite intersection.
But if $r\ge \frac 1n$ then $r\not \in (-1+\frac 1n, \frac 1n)$ and $r$ is not in the finite intersection.
So $\bigcap_{m=1}^n (-1+\frac 1m, \frac 1m) = (0, \frac 1n)$.
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Here's a slicker way to view it.
Consider $r \in \cap_{\alpha \in A} (a_\alpha, b_\alpha)$.  Then $r $ is in every interval $(a_\alpha, b_\alpha)$ so $r > \max a_\alpha$ if a maximum element exists and $r \ge \sup \max a_\alpha$ if a maximum element doesn't exist.
And the same reasoning to conclude $r < \min b_\alpha$ if a minimum element exist or $r \le \inf b_\alpha$ if it doesn't.
ANd thus the intersection can be determined by considering whatever interval can be made from $\max,\sup a_\alpha, \min,\inf b_\alpha$.
(If $\max,\min$ exist that side of the interval will be open and not include the $\max, \min$; otherwise it will be closed and will include the $\sup, \inf$; If $\max,\sup  a_\alpha > \min,\inf b_\alpha$ then the intersection will be empty; and if $\sup a_\alpha = \inf b_\alpha$ it will be the single point.
So $\max_{m\in \mathbb N} (-1+\frac 1m)=\max_{m\le n}(-1+\frac 1m) = (-1+\frac 11)=0$.
And $\min_{m< n}\frac 1m = \frac 1n$. So $\cap_{m=1}^n (-1+\frac 1m,\frac 1m) =(0,\frac 1n)$.
And $\min_{m\in \mathbb N}\frac 1m$ does not exist and $\inf_{m\in \mathbb N}\frac 1m = 0$.
So if $r\in \bigcap_{m\in \mathbb N}(-1+\frac 1m,\frac 1m)$ then $r > \max -1 + \frac 1m = 0$ and $r \le \inf \frac 1m = 0$.  So $0 < r \le 0$ which is impossible.  So  $ \bigcap_{m\in \mathbb N}(-1+\frac 1m,\frac 1m)= \emptyset$.
