We have two sets ${E}$ and ${T}$, that are non empty subsets of ${\Bbb R_{}}$ and are bounded above.

How can I prove that,

${E}$ ${\cup}$ ${T}$ has a least upper bound (supremum), and that ${\sup(E\cup T)=\max\{\sup E,\sup T\}}$ ?

  • 1
    $\begingroup$ HINT: prove by contradiction: suppose that $E\cup T$ doesnt have a finite supremum. Then choose $x>\max\{\sup E,\sup F\}$ and observe that it is not possible that $x\in E$ or $x\in F$. $\endgroup$ – Masacroso Nov 26 '16 at 14:24

First show that $E\cup T $ is bounded above. Indeed, since $E$ is bounded there exists a real number $M_1$ such that for all $e \in E$, we have $e \leq M_1$ and, since $F$ is bounded, there exists a real number $M_2$, such that, for all $f \in F$, we have $f \leq F$. Set now $M = Max(M_1,M_2)$ which is finite, and notice that, if $x \in E\cup F$, then $x \leq M$. Therefore $E \cup F$ has a least upper bound (it is non-empty and bounded above).

Now for all $e \in E$ and $f \in F$, we have : $e \leq \sup E $ and $ f \leq \sup F$. Set $S = Max(\sup E, \sup F)$. Then, for all $x \in E \cup F$, we have $x \leq S$, insofar as $x$ lies in $F$ or in $E$. Therefore, by definition $\sup(E \cup F) \leq S$. Now, let $\epsilon >0$ and suppose, without loss of generality, that $S= \sup E $. Consider $S - \epsilon$, which is a real number strictly inferior to $S$, then, by definition, there exists $e \in E$ such that $e > S - \epsilon$. Hence, since for all $x \in E \cup F$, we have $x \leq \sup(E \cup F) $, we must have $\sup(E \cup F) > S- \epsilon$. And, since this must be true for any $\epsilon >0$, we must also have $\sup(E \cup F) \geq S$. Therefore, combining with the other inequality, $S = \sup(E\cup F)$



Let $A=E\cup T, M=\max(\sup E,\sup T)$

we have

$\forall x\in A \;\; x\leq M$

take $\epsilon>0$.

WLOG, we can suppose that $M=\sup E$


$$\exists e\in E \;\; :\;M-\epsilon<e\leq M$$

$$\implies \exists e\in A\;\;:\; M-\epsilon<e\leq M$$



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