$A$ and $B$ are closed subset of $\mathbb R$. Show that $A\cap B$ is also closed in $\mathbb R$. 
Definition - A subset $S$ of $\mathbb R$ is said to be closed provided that if ${\{a_n}\}$ is a sequence in $S$ that converges to a number $a$, then the limit $a$ also belongs to $S$.

Actually, the exercise was two-part; first part was proof of the closedness of $A\cup B$ which is easy, but I can't prove for $A\cap B$. 


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*Suppose ${\{a_n}\}$ is a sequence in $A$. Its limit $a\in A$, so is $a\in A\cup B$; and if ${\{b_n}\}$ is a sequence in $B$ then its limit $b\in B$, so is $b\in A\cup B$. Q.E.D.     


Would someone please guide me how to prove it only based on the mentioned definition. 
Than kyou.  
 A: HINT
For the intersection, you have to start with a (converging) sequence $\{a_n\} \subseteq A\cap B$, and you have to show that $\lim a_n \in A\cap B$.
Now, remember that by the definition of intersection we have both $\{a_n\} \subseteq A$ and $\{a_n\} \subseteq B$. 
Can you take it from here?
For the union: let $\{a_n\} \subseteq A\cup B$. This sequence has infinitely many elements, some of the elements are in $A$ and some are in $B$. One of the sets must contain infinitely many elements from the sequence (why?), and let's assume (WLOG) that it's $A$. 
So there are infinitely many indices $j$ for which $a_j \in A$. Thus we can construct a subsequence $a_{n_k}$ of $a_n$  that lies entirely in $A$. Now, since $a_n$ converges, so does its subsequence  $a_{n_k}$; and since $A$ is closed...
A: Hint for $A\cap B$: let $(a_n)_{n\in\mathbb N}$ be a sequence in $A\cap B$ with $a_n\rightarrow a~(n\to\infty)$. We need to show: $a\in A\cap B$.
We can definitely say that $a_n\in A$, so...can you take it from here?
