Suppose $(a_n) \rightarrow a.$ If $a_n \geq 0$ for all $n$ then $a \geq 0$. Prove this result.

Question

I have decided to assume that $a<0$ and let $\epsilon=-a>0$. Then I want to come to a contradiction.

$|a_n-a|<-a$.

Since we have $a_n>a$ from our assumption,

$a_n-a<-a$

so $a_n<0$ which is a contradiction.

Answer 1

If $a<0$ then since $a_n\to a$, there is an integer $N$ such that whenever $n>N$, we must have $a_n\in (a,0)$, i.e. $a_n<0.$ And this is a contradiction.

In fact, the same reasoning shows that if $a_n>0$ for all $n$, then $any$ limit point $a$ of $(a_n)$ must satisfy $a\ge 0.$

Answer 2

This proof has a flaw since there is no restriction that $a_n>a$ (consider $a_n=(-{1\over 2})^n+1$ and $a=1$) but it can be simply fixed. Just conclude from $|a_n-a|<-a$ that $$a<a_n-a<-a$$ or equivalently $$2a<a_n<0$$ which is now a contradiction. Except that, your proof is fine.