# Proof that $\sup (a,b) = b$.

I was hoping someone could check my proof.

Thm. Define the open interval $$E := (a,b)$$ in $$\Bbb R$$. Prove that $$\sup E = b$$.

Proof. Since $$\emptyset \ne E\subset\Bbb R$$ is bounded above (e.g., by $$b$$), $$\exists\sup E$$ since $$\Bbb R$$ possesses the $$LUB$$ property. Since $$b$$ is an upper bound, we have $$x \leq b\;\forall x\in E$$. By the definition of supremum, $$\sup E \leq b$$. For a contradiction, suppose $$\sup E \neq b$$. Then consider the element $$\beta = \frac{\sup E + b}{2}.$$ So $$a \leqslant \sup E < \beta < b$$, so $$\beta \in E$$. But this implies that $$\sup E$$ is not an upper bound of $$E$$, a contradiction. Hence, $$\sup E = b$$.

Update. I have come across one additional question about thinking about the result more. If $$a = b$$, $$(a,b) = \emptyset$$, and $$\sup E$$ does not exist. Can I begin the proof with "without loss of generality, suppose $$a < b$$?"

• Did you assume b<+oo ? Otherwise, that's correct Apr 25, 2020 at 18:53
• $b<\infty$ by construction as $E \in \mathbb R$ and $\infty \not\in \mathbb R$
– user779041
Apr 26, 2020 at 1:55
• I would assume that the definition of open interval assumes that $a,b \in \mathbb R$ and $a<b$. But if you have to include the special case "a=b" in your interval definition then the statement sup(a,b)=b is not correct for all open intervals. Apr 26, 2020 at 5:47
• "without loss of generality." means that all other cases can be transformed to the case you investigate , e.g by renaming, But the case a=b cannot be transformed to the case a<b. It is simple a different case with a different result. Apr 26, 2020 at 5:57
• But now I m not sure if my interpretation of "without loss of generality." is correct. Maybe one can say "if a=b the supremum does not exist so without loss of generality assume a<b" But I wouldn't use this in this way. Apr 26, 2020 at 6:00

To nitpick, I would delete the second sentence; you never use it, and in the first sentence you already asserted (in the parenthetical) that $$b$$ is an upper bound, so it's implied that you/your reader already know the definition.