I'm just practicing doing proofs that involve supremums and infimums to better understand them for my elementary analysis class. Could someone please correct any errors in my proof. I found this proof here. The proof in question is example 2. I'm just wondering if I could prove the second part of the proof using contradiction.
$A=(m,n)$ implies $\{x\in R: m\lt x\lt n\}$ In addition, we can see that there is no minimum or maximum since this is an open interval.
Now, in order to prove that $\inf(A)=m$, I need to 1) Show m is a lower bound: this is done by using the definition of what the set A is. By definition: $m\lt x:\forall x\in A$
Thus, m is a lower bound of A.
2) By contradiction: assume $m_0$ is the largest lower bound of A which implies that $m_0\gt m$ and $m_0\lt x:\forall x\in$ A. Then, $m\lt m_0 \lt x \lt n$ $\Rightarrow$ $m_0 \in A$. Since $m_0\in A \Rightarrow \frac{m+m_0}{2}\in A \Rightarrow \exists$ x such that $x\lt m_0$. Which is a direct contradiction to $m_0\lt x$. Thus, m is the largest lower bound.
Since these two conditions were satisfied, $\inf(A)=m$
As always, thank you for your time.
It's easier to just do definitions.
$A=(m,n)=\{x\in \mathbb R | m <x <n\} $.
1) for all $x \in A $, $x <m $ by definition. So $m$ is lower bound.
2) If $r>m $ then $m < \min (n,r) $ and by the archimedian principle there is an $y$ so that $m<y <\min(n,r)$. (If you want you can set $y=\min (\frac {m+r}2,\frac {m+n}2) $... but you don't have to.
Then $m<y <\min (n,r)\le r $ so $r$ is not an lower bound. So $m $ is greatest lower bound.
If you want to do a prove by contradiction, you can do 2 differently.
2') Suppose $m $ is not greatest lower bound. Then there is an lower bound $m_0; m <m_0$ (it doesn't have to be the greatest lower bound).
Let $y $ be so $m < y <m_0$. The for any $x\in A $ we have: $m_0 \le x; m <x < n ; m < y <m_0$. So $m <y <n $. So $y \in A $. This contradicts $m_0$ is lower bound of $A $.
So $m $ is greastest lower bound.
A third way and fourth use the theorem: if $x \le y \le x+\epsilon $ for all $\epsilon > 0$ then $y=x $. (Have you proven that? If you haven't you should. It's a good exercise.)
2'') $m $ is a lower bound. So if $m_0$ is greatest lower bound then $m \le m_0$. Let $(n-m) > \epsilon >0$. Then $m <m+\epsilon < n $ so $m+\epsilon \in A $. So as $m_0 $ is lower bound $m_0 \le m+\epsilon$. So $m\le m_0\le m+\epsilon $.
So $m=m_0$
2''') let $m_0$ be greatest lower bound. $m_0 <n $ as $m_0 <x <n $ for all $x\in A $. Let $\epsilon >0$. $m_0 - \epsilon < \inf A=m_0$ so $m_0-\epsilon \not \in A $ and $m_0-\epsilon <m_0 <n$. So $m_0-\epsilon \le m $.
And as $m $ is lower bound $m\le m_0$. So $m_0-\epsilon\le m\le m_0$.
So $m=m_0$.