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I'm having some difficulty with this proof. I want to show $ \sup (E_1 \cup E_2) = \max \{ \sup E_1, \sup E_2 \}$, where $E_1, E_2$ are two bounded, non-empty subsets of $\mathbb{R}$. Naturally, the supremum must exist by the least upper bound property.

I know by definition that $\sup (E_1 \cup E_2) \geq x \ \forall x \in E$ and that $\sup (E_1 \cup E_2) \leq \beta$ where $\beta$ is the set of all upper bounds for E. I know the first condition satisfies one of the conditions for the maximum, but my problem is that I don't know how to show that $ \sup (E_1 \cup E_2)$ is in an element of E.

As for $\max \{\sup E_1, \sup E_2\}$, I have proved that it is equal to a supremum by the following:

Let $\beta _0$ denote $\max \{\sup E_1, \sup E_2\}$. Assume $\beta$ is the set of upper bounds for E, and $ x \in E$. Then we have that:

If $x > \beta_0$, then by definition of upper bound $x \notin E$

If $\beta_0$ > $\beta$, then $\beta$ cannot be an upper bound since $\beta_0 \in E$

Therefore, $\beta_0 = \sup (\max \{\sup E_1, \sup E_2\})$

Again, this proof is incomplete because I have not shown that $\beta_0$ lives in the supremum of the union.

Nikitau
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1 Answers1

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Let $\beta_0=\max\{\sup E_1,\sup E_2\}.$ We have to show that $\beta_0=\sup(E_1\cup E_2),$ i.e., that $\beta_0$ is the least upper bound for $E_1\cup E_2.$

I. Suppose $x\in E_1\cup E_2.$ Then $x\in E_i$ for some $i\in\{1,2\},$ so $$x\le\sup E_i\le\max\{\sup E_1,\sup E_2\}=\beta_0,$$ which shows that $x$ is an upper bound for $E_1\cup E_2.$

II. Let $\beta$ be any upper bound for $E_1\cup E_2.$ Then $\beta$ is an upper bound for $E_1,$ so $\beta\ge\sup E_1.$ Likewise $\beta$ is an upper bound for $E_2,$ so $\beta\ge\sup E_2.$ From $\beta\ge\sup E_1$ and $\beta\ge\sup E_2$ it follows that $$\beta\ge\max\{\sup E_1,\sup E_2\}=\beta_0,$$ which shows that $\beta_0$ is the least upper bound for $E_1\cup E_2.$

bof
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