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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 $a < b$?"

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    $\begingroup$ Did you assume b<+oo ? Otherwise, that's correct $\endgroup$
    – Anthony
    Apr 25, 2020 at 18:53
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    $\begingroup$ $b<\infty$ by construction as $E \in \mathbb R$ and $\infty \not\in \mathbb R$ $\endgroup$
    – user779041
    Apr 26, 2020 at 1:55
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    $\begingroup$ 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. $\endgroup$
    – miracle173
    Apr 26, 2020 at 5:47
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    $\begingroup$ "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. $\endgroup$
    – miracle173
    Apr 26, 2020 at 5:57
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    $\begingroup$ 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. $\endgroup$
    – miracle173
    Apr 26, 2020 at 6:00

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Correct, yes.

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.

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