$\bigcap_\limits{i\in I}{A_i}\in \{A_i\}_{i\in I}$ is open proof 
Let $(X,d)$ be a metric space.
Given a family of open sets $\{A_i\}_{i\in I}$
a)Prove that $\bigcup_\limits{i\in I}{A_i}$ is an open set.
b)Prove that $\bigcap_\limits{i\in I}{A_i}$ is an open set.

a)Given the fact the family sets are all open.If $x\in\bigcup_\limits{i\in I}{A_i} $, then for a $j\in I$, $x\in A_j$. There exists $\epsilon>0$ so that $B(x,\epsilon)\subset A_j$ Therefore $B(x,\epsilon)\subset A_j\subset\bigcup_\limits{i\in I}{A_i} $, which proves the assertion. 
Question:
When it comes to b) it is trivial to me that $\bigcap_\limits{i\in I}{A_i}\in \{A_i\}_{i\in I}$ so that it is open. How can I prove b)?
Thanks in advance!
 A: b) is false in general. Consider for example $I = \mathbb N$ and $A_i = (- \frac{1}{i}, \frac{1}{i})$ as a subset of $\mathbb R$ with its standard topology.
$A_i$ is open for all $i \in I$ but
$$
\bigcap_{i \in I} A_i = \{0\}
$$
is not open.
A: Counter example for $(b)$: $\bigcap_{n=1}^{\infty} (-1/n,1/n) = \{0\}$
A: For (a) simply note, that any metric space $(X, d)$, induces a topology $\mathcal{T}$, the open sets of $(X, d)$ are the elements of this topology.
Since $A_i \in \mathcal{T}$ for each $i \in I$, it follows that the union $$\bigcup_{i \in I} A_i \in \mathcal{T}$$ by the definition of a topology. Hence $\bigcup_{i \in I} A_i$ is open in $(X, d)$

Also (b) only holds if $I$ is finite (again this is by the definition of a topology). (b) is false in general, as other answers have pointed out.
A: Let $A\subseteq X$. You may recall that $A$ is open iff $X\setminus A$ is closed. Therefore, you may conclude (by De Morgan's laws) that an arbitrary intersection of closed sets is closed.
Only a finite intersection of open sets is necessarely open and consequently only a finite union of closed sets is necessarely closed.
