Proof-verification: $(x_n)$ bounded and $y_n \to 0$ implies
$(x_ny_n)\to 0$
My attempted proof:
Let $(x_n)$ be bounded, hence there exists $A\in \mathbb{R}$ such that $|x_n|\leq A$. Furthermore let $y_n$ be a null-sequence; thus for an arbitrary $\varepsilon >0$ there exisits $N\in \mathbb{N}$ such that $|y_n|<\varepsilon$ for all $n\geq N$.
$$|x_n\cdot y_n|\leq |A\cdot y_n|=|A|\cdot |y_n|<|A|\cdot \varepsilon$$ Since $\varepsilon$ does not depend on $|A|$, one can conclude that the claim is indeed true. $\Box$
is that okay?
Best Answer
If we correct the obvious typo then this is fine; exactly this sort of argument is given all the time.
But it's somewhat "informal", in that the conclusion does not look like what's required by the definition. A more formally correct version would read "Since $y_n\to0$, given $\epsilon>0$ there exists $N$ such that $|y_n|<\epsilon/A$ for all $n>N$..."
So whether it's actually "right" depends on the context. In a research paper, or some other context where the point is to convince the reader it's true, what you wrote is fine - people do write this sort of thing all the time. In a calculus class it's different - there the point is to convince the reader that you know exactly how to prove it, and in that context you're much better off getting a literal $|x_ny_n|\dots<\epsilon$ at the end.
Note: Of course @pre-kidney is also correct in saying that the last sentence is missing the point; that issue doesn't come up if we change it as suggested.