I saw this proof here.
But, I am not sure if it's totally correct. Here is the full question:
Suppose that $(s_n)$ converges to $s$, $(t_n)$ converges to $t$, and $s_n \leq t_n \: \forall \: n$. Prove that $s \leq t$.
Here is a proof without contradiction:
For every $\varepsilon>0$, there exists $N$ large enough so that $s-\varepsilon/2\leqslant s_{n}\leqslant s+\varepsilon/2$ and $t-\varepsilon/2\leqslant t_{n}\leqslant t+\varepsilon/2$ for all $n>N$.
Therefore, from the hypothesis that $s_{n}\leqslant t_{n}$, $$ s-\varepsilon/2\leqslant s_{n}\leqslant t_{n}\leqslant t+\varepsilon/2, $$ from which it can be concluded that $$ s\leqslant t+\varepsilon. $$ Since $\varepsilon>0$ is arbitrary, the proof is done.
The problem I see is that since $\epsilon > 0$, then $s \leq t$ never really happens because $\epsilon$ is always greater than 0 - no matter how small it becomes so $s \leq t$ is never really proved.
Is this a correct analysis?
I am not an expert in analysis, but this seems a bit wonky to me...
Thanks for your input...
In that proof, it concludes for all $\varepsilon>0$, $$s\leq t+\varepsilon$$
I think you agree with his/her proof, but you are confused about why this can conclude $s \leq t$.
That is the following statement:
Here I provides a simple proof by contradiction.
Suppose the contrary that , $s > t$.
Now, let $\varepsilon = \frac{s-t}{2}$, since $s>t$, we have $\varepsilon>0$.
By assumption, we have $s \leq t+\varepsilon$, $$s \leq t + \frac{s-t}{2} = \frac{s+t}{2}<s$$
which is a contradiction!!