# Just a matter of rigor

• Aug 28th 2010, 03:59 AM
Hinatico
Just a matter of rigor
My teacher really takes account about the rigor of a proof. The following is the proposition and a proof given by myself. Can someone please check it whether it is rigorous enough?

Proposition: Let $(s_n)$ be a sequence that converges to $s\in \mathbb{R}$. Show that if $s_n\ge a$ for all $n>N$, then $s\ge a$.

Proof: Let $\epsilon>0$, then there exists $N_0$ such that $n>N_0$ implies $|s_n-s|<\epsilon$. Now take $N_1=\text{max}\{N_0, N\}$ and $n>N_1$ implies
$|s_n-s|<\epsilon$
$|a-s|<\epsilon$
Since $x\le |x|$, we must have
$a-s<\epsilon$
Since the difference can be taken arbitrarily small, or even negative. This proves that $s\ge a$.

I really doubt the rigor of this proof. And after I type in the proof, I doubt the validity.(Doh)
• Aug 28th 2010, 06:05 AM
Plato
Here is an easy proof.
If it were true that $s then $\varepsilon = a-s$.
From that we get $\left| {s_N-s} \right| < \varepsilon \, \Rightarrow \,s_N.
What is wrong to that?