If a sequence $s_n$ of real numbers converges to $s>a$ then $s_n>a$

Question

Let $s_n$ be a sequence of real numbers that converges to $s>a$. Then there exists an $N>0$ such that $n>N\implies s_n>a$.

PROOF: By definition of convergence in sequences we have that for all $\varepsilon>0$, there exists $N>0$ such that $n>N$ implies $|s_n-s|<\varepsilon$.

Choose $\varepsilon=s-a$. Then $|s_n-s|<s-a$. By properties of absolute values and inequalities we have $-s+a<s_n-s<s-a$ which implies $s_n>a$.

Is my proof rigorous, correct? I'd appreciate any comments for improvement.

Answer 1

I would just add, after “Choose $\varepsilon=s-a$”, that there is a $N\in\mathbb N$ such that $n>N\implies\lvert s_n-s\rvert<\varepsilon$.