Convergence of General Sequence

Question

I am presented with the following problem;

"Suppose that a sequence $b_{n}$ converges to $B\ne0$. Show using only the definition of convergence: There exists a natural number $n_{0}$ such that for all $n\geq n_{0}$ $$|b_{n}|\geq \frac{1}{2}|B|.$$ Hint: Use that $|x-y|\geq |x|-|y|$."

So far I have used the definition to show that $$|b_{n}-B|<\epsilon\\|b_{n}|-|B|<\epsilon$$ From this point onwards I'm unsure as to how I should manipulate this inequality to get be the result above.

Answer 1

Take $\varepsilon=\frac12|B|$. This makes sense, since $\frac12|B|>0$. So, there is some $N\in\Bbb N$ such that$$n\geqslant N\implies|b_n-B|\leqslant\frac12|B|.$$But, if $|b_n-B|\leqslant\frac12|B|$, then\begin{align}|b_n|&=|b_n-B+B|\\&\geqslant\bigl||b_n-B|-|B|\bigr|\\&=|B|-|b_n-B|\text{ (since $|b_n-B|<|B|$)}\\&\geqslant|B|-\frac12|B|\\&=\frac12|B|.\end{align}

Answer 2

Take $\epsilon =\frac {|B|} 2$. Then, for $n$ sufficiently large, $|B| \leq |B-b_n|+|b_n|<\epsilon+|b_n|$ so $|b_n| \geq |B|-\epsilon =\frac {|B|} 2$.

Answer 3

$|B|\not =0;$

Assume for all $n_0 \in \mathbb{N}$ there exists a $n \ge n_0$ s.t.

$|b_n| <(1/2)|B|;$ i.e. there is a subsequence

$|b_{n_k}| <(1/2)|B|$.

Since $b_n$ converges, every subsequence $b_{n_k}$ converges to the same limit $B.$

Then

$ |B|= \lim_{k \rightarrow \infty}|b_{n_k}| \le(1/2)|B|$,

a contradiction .

Used: $b_n \rightarrow B$ implies $|b_n| \rightarrow |B|$ (Reverse triangle inequality).