Clarification: Prove there exists a number $N$ such that $n > N$ implies $s_n >a$

Question

Below is the proof that I have been working on and the solution provided by the professor.

Let $(s_n)$ be a convergent sequence, and suppose $\lim s_n > a$. Prove there exists a number $N$ such that $n > N$ implies $s_n >a$.

We will prove that $ \exists \ N$ such that $n > N \Rightarrow s_n >a$.

Assume $s_n \rightarrow s+a$, since $s_n \rightarrow s$ and $s>a$.

Choose $\epsilon = \frac{s-a}{2}$.

Then $\exists \ N$ such that $n>N \Rightarrow |s_n - s|< \epsilon$. Thus, by choice of $\epsilon$, this gives us $s_n > a$.

I would like help understanding this proof. Specifically, I am not sure how we could choose $\epsilon = \frac{s-a}{2}$. I am also unclear about the last line, and why it serves as a completion to the proof (why is "by choice of $\epsilon$" sufficient?).

Answer 1

You can choose $\varepsilon$ as you please, then you determine an $N$ such that &c.

Completion of the proof:

$$\lvert s_n-s\rvert<\varepsilon\iff s-\varepsilon<s_n<s+\varepsilon.$$ In particuler, $\; s_n>s-\varepsilon=s-\dfrac{s-a}2=\dfrac{s+a}2>\dfrac{2a}2=a.$

Answer 2

Do you know that for a chosen $\epsilon>0$ there is $N$ s.t. for all $n\ge N$ the terms of the sequence $s_n$ belong to the interval $(s-\epsilon,s+\epsilon)$? Now to answer the question it suffices to choose $\epsilon$ such that $s-\epsilon>a$. Draw a picture!