Let $A_i = (-\infty, y + i^{-1}]$ for all $y \in \mathbb{R}$. Show $\bigcap_{i=1}^{\infty}A_i = (-\infty, y]$. Let $A_i = (-\infty, y + i^{-1}]$ for all $y \in \mathbb{R}$. Show $\bigcap_{i=1}^{\infty}A_i = (-\infty, y]$.
I am not satisfied with the solution I was given. I do agree that it is quite obvious that $(-\infty, y] \subset A_i$ for all $i$, hence also applying to the intersection, so we're good there. But the other direction is proven like the following:
$$\bigcap_{i=1}^{\infty}A_i = \{x \in \mathbb{R}: x \leq y + \dfrac{1}{n} \text{ for all } n \geq 1 \} \subset \{x \in \mathbb{R}: x \leq y  \text{ for all } n \geq 1 \} = (-\infty, y]\text{.}$$
I don't agree with this solution, for it is possible for an element $x \in \left[y, y + \dfrac{1}{n}\right]$ (and thus the subsetting fails).
 A: The entire proof consists of proving that inclusion, so I agree that the "proof" you presented here is quite unsatisfactory if not unclear. 
Hopefully you'll find the following argument more convincing! 
Suppose by contradiction that $\bigcap A_i$ is not contained in $(-\infty, y]$, then there exists $z \in (y,\infty) \cap \bigcap A_i$. By definition this implies that $z > y$ and $z \in A_i$ for every $i$. To get a contradiction notice that if $p = z - y$ then $z \notin A_i$ for any $i$ such that $i^{-1} < p$.
A: I don't agree that solution neither. Here is another solution for $\bigcap_{i=1}^{\infty}A_{i} \subset (-\infty,y].$
Suppose not, there is a real number $x\in\bigcap_{i=1}^{\infty}A_{i}$ greater than $y$.As $x>y$, there is positive number $\epsilon$ such that $x=y+\epsilon$. But there is large natural number $N$ such that $\frac{1}{N} <\epsilon$. It means $x \notin A_N$. So there is no such $x$. Done.
A: Let $a\in\bigcap_{i=1}^{\infty}A_i$ ans assume negativly that $a>y$. therefore $\epsilon=a-y>0$. Now, since $\lim_{n\rightarrow\infty}{1/n}=0$, There's some $N$ such that $1/N<\epsilon$, which means:
$y+1/N<y+\epsilon=a$
Therefore, $a\notin(-\infty,y+1/N]=A_N$, and this is contradiction.
that means that $a\le y$, which means $a\in(-\infty,y]$
From all of that, $\bigcap_{i=1}^{\infty}A_i \subseteq (-\infty, y]$
