countable intersection of open set argument I know that countable intersection of open sets could be closed or open. But I was wondering what is wrong with my argument here:
Given an element say x in a countable intersection of open sets say denoted as 
$\bigcap_{i\in\mathbb{N}}A_{n}$ where each of $A_n$ is an open set.
Then since x is an element that belongs to each of $A_i$, so for each $A_i$, we have an $\epsilon_i>0$ such that ($x-\epsilon_i, x+\epsilon_i$) is entirely contained in $A_i$. Then out of all the $\epsilon_i$, we can pick the minimum of all those $\epsilon_i$ say denoted as $\epsilon^*$ such that $\epsilon^* \leq \epsilon_i$. Then we have $(x-\epsilon^*, x+\epsilon^*)$ within each of the $A_i$. Since we can do it for arbitrary element of x in $\bigcap_{i\in\mathbb{N}}A_{n}$ , then we can conclude that $\bigcap_{i\in\mathbb{N}}A_{n}$  is open.
I know it is incorrect, but which part is incorrect?
Is it because of the "countable intersection" part? I know that instead of coutable, if I have "finite intersection", then the finite intersection of those open sets will be open. But not sure how to get the concept correct for the "countable intersection" to understand that "countable intersection" of open interval is not necessarily open without using Counterexample. I saw many counter example that shows intersection of open set could be closed. But I just want to know which part of my thinking above is incorrect.  Thank you.

The problem is that while a finite set of positive real numbers has a minimum (which is then positive), an infinite set of positive real numbers may not. Instead it is only guaranteed to have an infimum, which could be zero. 
In your example, consider the case where $A_i=(-\frac{1}{i},\frac{1}{i})$ and $x=0$. Then we could take $\epsilon_i=\frac{1}{i}$, but then no minimum $\epsilon^*$ exists. And indeed $\cap_{i}A_i=\{0\}$ is not open.
