Trouble proving $\bigcap_{n=1}^\infty (-\frac1n,1+\frac1n) = [0,1]$ I learned in class that the finite intersection of open sets is open. With that "big picture" in mind I want to prove the following:
Define $ A:=\bigcap\limits_{n=1}^{\infty} I_n $, where $I_n:=\left(-\frac{1}{n}, 1+\frac{1}{n}\right). \text{Then } A = [0,1]$
It seems very obvious that A equals $ [0,1] $, but I am having trouble proving that the two sets are equal to each other. I think I have to show that $ A \subset [0.1]$ and $ [0,1] \subset A $. It is very irritating because I don't know where to begin.
Which direction should I take to prove the two sets are equal? Any help or hint in any form would be very much appreciated. I believe this question would be helpful for novices like me, because the answers I found from the "similar questions" tab do not discuss how to prove that the sets are identical.
 A: Start by "proving" (actually it is evident) that $[0,1]\subseteq I_n$ for every $n$ so that consequently: $$[0,1]\subseteq A\tag1$$
Then for $x\notin[0,1]$ prove that a positive integer $n$ exists with $x\notin I_n$, hence $x\notin A$
That proves: $[0,1]^{\complement}\subseteq A^{\complement}$ or equivalently $$A\subseteq [0,1]\tag2$$
Taking $(1)$ and $(2)$ together we end up with:$$[0,1]=A$$qed
A: Since $(\forall n\in\mathbb{N}):A\subset\left(-\frac1n,1+\frac1n\right)$,$$A\subset\bigcap_{n\in\mathbb{N}}\left(-\frac1n,1+\frac1n\right).\tag1$$In order to prove that $(1)$ is actually an equality, take $x\in A^\complement$ and then prove that $x\notin\left(-\frac1n,1+\frac1n\right)$, for some $n\in\mathbb N$.
A: If $x \in [0,1]$ then for any $n$: $-\frac1n < 0 \le x \le 1 < 1+\frac1n$, so $x \in I_n$ for all $n$, hence $x \in \bigcap_n I_n = A$. So $[0,1] \subseteq A$. 
To see that $A \subseteq [0,1]$, let $x$ be any point of $A$. Then suppose (for a contradiction) that $x \notin [0,1]$ so either $x <0$ or $x > 1$. If $x < 0$, then as $-\frac1n \to 0$, there is some $n_0$ such that $x < -\frac1{n_0}$ so $x \notin I_{n_0}$ which contradicts $x \in A$. If $x >1$ we find an $n_1$ such that $1+\frac1{n_1} < x$, as $1+\frac1n \to 1$, and then $x \notin I_{n_1}$, another contradiction with $x \in A$. This shows that either way $x \notin [0,1]$ leads to a contradiction, so $x \in [0,1]$ as required. So we have equality $[0,1] = A$.
The intersection of open sets can be any set at all (need not be closed): note that all sets of the form $\mathbb{R}\setminus \{x\}$ are open and $$A = \bigcap \{\mathbb{R} \setminus \{x\}: x \notin A\}$$
Subsets that are countable intersections of open sets are called $G_\delta$ sets and among those we have the irrationals, and yes, also all closed subsets (of metric spaces at least) can be written that way. But these sets can also be open, if we take $O_n = O$ for some fixed open set $O$, then also $\bigcap_n O_n = O$ etc.
A: If $x>0, x<1$ then it is in all $I_n$. Hence, $(0,1)\subseteq A$.
Assume, $a < 0$. Then there is a $n >0 $ such that, $a <-\frac{1}{n}$. That $a \not  \in I_n$. Hence it will not be in $A$.
Further, similarly, $a>1$ is not in $A$.
Now consider $0$. It is in every $I_n$. So it is in A. Similarly for $1$.
Thus, $A=\{0\}\cup(0,1)\cup\{1\}=[0,1]$.
