Prove: $\sup (A \cup B) = \max\{\sup(A),\sup(B) \}$ where $A,B \subseteq \mathbb{R}$

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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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lasoon
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lasoon

Updated on August 01, 2022

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  • lasoon
    lasoon over 1 year

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

    • erfink
      erfink over 6 years
      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?
    • erfink
      erfink over 6 years
      Also, your "without loss of generality" gives me a slight pause. What if $\sup (A) = \sup (B)$?
    • Jacob Wakem
      Jacob Wakem over 6 years
      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.
    • lasoon
      lasoon over 6 years
      @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.
    • lasoon
      lasoon over 6 years
      @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.
  • lasoon
    lasoon over 6 years
    Thank you @Alephnull. This is a very nice and concise proof. My attempt, by comparison, appears to include a lot of unnecessary details.