Q:

Convergent Sequence Analysis

Suppose that $(a_n)\to a$ with $a \in \mathbb R$
If $A \leq a_n \leq B$ then $A \leq a \leq B$
Proof:
Consider the sequences $(b_n)=A$ and $(c_n) = B$
Thus $(b_n)\to A$ & $(c_n)\to B$
Noting $(b_n) \leq(a_n)\leq(c_n) $ $\forall n \in \mathbb N$ 
Thus $A \leq a \leq B$
$\blacksquare$
is this a valid proof? thanks.

A:

This is not a valid proof. You proved your statement only for two specific sequences, not in the general case.
Suppose that $a<A$. Take $\varepsilon=A-a$. There is a natural $N$ such that$$n\geqslant N\implies\lvert a_n-a\rvert<\varepsilon=A-a.$$But then$$a_N-a\leqslant\lvert a_N-a\rvert<A-A$$and therefore $a_N<A$, which is a contradiction.
By a similar argument, you can't have $a>B$.

