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Below is the proof I wrote. Can you please let me know if this is correct? Also, any suggestions for improving my proof writing style would be appreciated (I am new to proofs and still learning):

Without loss of generality, assume $\sup (A) \geq \sup (B)$. We know $\forall \ b \in B,\ \sup(B) \geq b$ $\implies \ \sup (A) \geq b \implies \sup (A)$ is an upper bound for $B$ (and for $A$ by definition). Now we know that $c \in A \cup B \iff c \in A \text{ or } c \in B \implies \not\exists \ c \in A \cup B \ \text{ s.t. } \ c > \sup (A)$. Thus, we have established that $\max\{\sup(A), \sup(B) \}$ is an upper bound for $A \cup B$. It also follows that $\sup (B)$ is not an upper bound for $A \cup B$ if $\max \{ \sup (A),\sup (B) \} \neq \sup (B)$ because this implies that $\exists \ a \in A \in A \cup B \text{ s.t } a>\sup (B)$.

We now show that $\max\{\sup(A), \sup(B) \}$ is the least upper bound for $A \cup B$. Suppose $ \sup(A \cup B) < \max\{\sup(A), \sup(B) \} = \sup(A)$. This is a contradiction since $A \subseteq A \cup B \implies \sup (A) \leq \sup (A \cup B)$.

Therefore, $\sup (A \cup B) = \max \{ \sup(A), \sup (B) \}$.

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marked as duplicate by Arnaud D., Xander Henderson, SchrodingersCat, Shailesh, Simply Beautiful Art Oct 7 '17 at 0:36

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  • $\begingroup$ Looks mostly good. The "$ \exists \ \sup(A \cup B) < \max\{\sup(A), \sup(B) \} = \sup(A)$" in the third paragraph is weird. What does $ \exists \ \sup(A \cup B) $ mean? $\endgroup$ – erfink Apr 1 '17 at 20:50
  • $\begingroup$ Also, your "without loss of generality" gives me a slight pause. What if $\sup (A) = \sup (B)$? $\endgroup$ – erfink Apr 1 '17 at 20:51
  • $\begingroup$ Sup is obviously monotone increasing with respect to the subset relation. Thus the sup is at least the indicated value. But it is easily seen to be an upper bound. $\endgroup$ – Jacob Wakem Apr 1 '17 at 21:05
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    $\begingroup$ @erfink I edited the last paragraph so that it doesn't have the "$\exists \ \sup(A \cup B)$" anymore. While saying there exists a sup doesn't make the sentence wrong, I realize it is redundant. $\endgroup$ – lasoon Apr 1 '17 at 21:55
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    $\begingroup$ @erfink I also changed the inequality in the WLOG statement from $>$ to $\geq$ so that it now reads "WLOG, assume $\sup (A) \geq \sup (B)$". This seems correct to me since regardless of whether you assume $\sup (A) \geq \sup (B)$ or $\sup (B) \geq \sup (A)$, the proof should follow and give the same result. $\endgroup$ – lasoon Apr 1 '17 at 22:04
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sup{supA,supB}>=supA>=a for all a in A. This holds similarly for B and b. Therefore sup{supA,supB} is an upper bound. But anything less would be less than (WLOG) supA and thus some element of A would be greater than it. Thus sup{supA,supB} is the least upper bound.

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  • $\begingroup$ Thank you @Alephnull. This is a very nice and concise proof. My attempt, by comparison, appears to include a lot of unnecessary details. $\endgroup$ – lasoon Apr 2 '17 at 0:03

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