text: # Critique my elementary proof for a set bounded above
  
  Let $A$ and $B$ be two non-empty subsets of $\mathbb{R}$ that are both bounded above.
  
  $(i)$Prove that $A ∪ B$ is bounded above
  
  and prove
  
  $(ii)$ that $\sup(A ∪ B) = \max(\sup(A),\sup(B))$.
  
  for $(i)$ 1st attempt
  
  By the axiom of completeness, $\sup A$ and $\sup B$ exist as $A$ and $B$ are both non-empty sets which are bounded above.
  
  Then $\{A \cup B\} \leq \max(\sup A , \sup B)$ and hence $A \cup B$ are bounded above.
  
  2nd attempt
  
  Define $\alpha$ as some upper bound for $A$ and $\beta$ as some upper bound for $B$.
  
  Then $\forall a \in A, \alpha \geq a \space \text{and} \space \forall b \in B, \beta \geq b$
  
  Then $\max(\alpha,\beta) \geq \text{all} \space x \in A \cup B$ since $x \in A$ or $x \in B$ and so $A \cup B$ is bounded above.
  
  Are either or both of these proofs rigorous?
  
  $(ii)$ We need to show that $\sup (A \cup B) \geq \max(\sup A, \sup B)$ and also $\sup (A \cup B) \leq \max(\sup A, \sup B)$
  
  For all $x \in A \cup B$, either $x \in A \implies x \leq \sup A$ or $x \in B \implies x \leq \sup B$
  
  Then $x \leq \max(\sup A, \sup B)$ and $\sup (A \cup B) \leq \max( \sup A, \sup B)$
  
  Furthermore, it's obvious that $\sup(A \cup B) \geq \sup A$ and $\sup (A \cup B) \geq \sup B \implies \sup (A \cup B) \geq \max(\sup A, \sup B)$ and thus $\sup (A \cup B) = \max(\sup A, \sup B)$
  
  I am still quite new to proof writing so would appreciate any and all criticisms - please be as harsh as possible.
  
  Thanks
  
  In $\{A \cup B\} \leq \max(\sup A , \sup B)$, what does your notation $\{\}$ mean? If it means a set, you cannot say it is less than or equal to a number.
  • Thanks for that, I don't know why I put $A \cup B$ as a set at the time, must have been legless. I appreciate your feedback, do you think anything could be improved? Mar 15 '15 at 23:46