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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...

md2perpe
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PiE
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  • To do proof by contradiction, assume $t<s$ and choose $0<\varepsilon<s-t$. Then the last inequality you obtained is a contradiction. – Kangyeon Moon Jul 29 '23 at 13:46
  • But $s<t \implies s\le t$ – Tim Jul 29 '23 at 13:47
  • No epsilon may be 0. for example starting at n=100 all sn=s and all tn=t – trula Jul 29 '23 at 13:47
  • That's what I was thinking as well, but is the above proof correct? - it just seems a lot easier than a proof by contradiction IMO - but something does not seem right IMO – PiE Jul 29 '23 at 13:48
  • Can we change this statement "Since $\varepsilon>0$ is arbitrary" to this "Since $\varepsilon$ is arbitrary"??? – PiE Jul 29 '23 at 13:51
  • If $s>t$, you can let $\epsilon =( s-t )/2$, it will lead to a contradiction – Tim Jul 29 '23 at 13:51
  • What do you mean with "$s\leq t$ never really happens"? If $s_n=0$ and $t_n=1$ for all $n$ then $s=0$ and $t=1$ so $s\leq t$ clearly happens. Even if you also take $t_n=0$ for all $n$ then $s\leq t$ clearly happens. – md2perpe Jul 29 '23 at 14:15
  • I am no expert here - I'm just trying to understand - $\varepsilon$ is never really $0$ as long as $\varepsilon > 0$. So wouldn't you have to add something like "in the limit as $\varepsilon \rightarrow 0$", then $s \leq t$ --- at the end of the proof? Would that be Ok to do? – PiE Jul 29 '23 at 14:22

1 Answers1

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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:

Let $s, t \in \mathbf{R}$. Suppose that $s \leq t+\varepsilon$ for all $\varepsilon>0$. Then $s \leq t$.

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!!

Tim
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