Let f and g be sequences in a valued field F. Assuming that f converges to 0 and g is bounded, show that fg converges to 0.

Question

I know that through the definitions, $\exists m\in \mathbb{N}$ such that $$f_n \in B(o, \epsilon) \forall n \geq m$$ and $\exists c\in \mathbb{R}^+$ such that $$|g_n|\leq c$$ I'm not quite sure how to put these definitions and information together to prove this. I assume it will be an epsilon proof of some sort, which we haven't gone over much but I assume: $$|f_n - 0| < \epsilon$$ is the epsilon rule for f here.

Answer 1

As $(g_n)_{n\in\mathbb{N}}$ is bounded, there exists an $M\in\mathbb{R}^+$ such that for all $n\in\mathbb{N}$, we have $|g_n|\leq M$.

Let $\varepsilon>0$. As $(f_n)_{n\in\mathbb{N}}$ converges to $0$, then there exists an $N\in\mathbb{N}$ such that for all $n\geq N$, we have $$|f_n|<\frac{\varepsilon}{M}$$

Then consider $|f_n||g_n|$ and see what inequalities you can derive.

Answer 2

As you correctly stated, there exist $c$ and $M$ such that $|g_n|\leq c$ for all $n>M$ and for all $\epsilon>0$ there exists $N$ such that $|f_n|<\epsilon$ for all $n>N$. (this is using $g_n$ bounded and $f_n\rightarrow 0$ respectively).

You want to show that for all $\epsilon>0$ there exists $N$ such that $|f_ng_n|<\epsilon$ for $n>N$. We can find some $N'$ such that $|f_n|<\frac{\epsilon}{c}$ for all $n>N'$ using the fact that $f_n\rightarrow 0$. Then $|f_ng_n|<\epsilon$ for all $n>N=\max(M,N')$.