#### Alfredo Dawlabany

##### New member

- Joined
- Aug 22, 2017

- Messages
- 45

My teacher solved it in this way:

\(\displaystyle \forall n,m,\frac{1}{n}+\frac{1}{m}\leqslant 1+1\leqslant 2\)

Then 2 is an upper bound of E and since 2 belongs to E for n=1 and m=1 then 2=sup E (sup E is done)

\(\displaystyle \forall n,m\in \mathbb{N^*},\frac{1}{n}+\frac{1}{m}> 0\Rightarrow 0\) is a lower bound of E

but \(\displaystyle 0\notin E\)

(for now everything is clear)

But after that we have to see if 0 is the greatest lower bound so that 0=inf E .

Here my teacher said that we have to prove that \(\displaystyle \forall \varepsilon >0\) , \(\displaystyle \exists p\in E\) such that \(\displaystyle 0<p<0+\varepsilon\)

and that \(\displaystyle \exists n_0,m_0\in \mathbb{N^*}\) such that \(\displaystyle 0<\frac{1}{n_0}+\frac{1}{m_0}<\varepsilon\)

Then the proof went like this

\(\displaystyle \forall \varepsilon >0,\exists n_0\) such that \(\displaystyle \frac{1}{n_0}<\frac{\varepsilon }{2}\)

\(\displaystyle \forall \varepsilon >0,\exists m_0\) such that \(\displaystyle \frac{1}{m_0}<\frac{\varepsilon }{2}\)

\(\displaystyle \Rightarrow \frac{1}{n_0}+\frac{1}{m_0}< \varepsilon \Rightarrow 0+\varepsilon\) is not a lower bound of E (contradiction)

then 0=inf E

Now, the problem is that I didn't know why we have to prove this and also the I didn't understand the proof