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
explanation: The document is a request for critique of a proof in real analysis. It contains two parts of a proof: (i) proving that the union of two sets bounded above is also bounded above, and (ii) proving that the supremum of the union of two sets is the maximum of their individual suprema. The author provides two attempts for part (i) and one attempt for part (ii), and asks for feedback on the rigor of these proofs.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a request for critique of a proof in real analysis. It contains two parts of a proof: (i) proving that the union of two sets bounded above is also bounded above, and (ii) proving that the supremum of the union of two sets is the maximum of their individual suprema. The author provides two attempts for part (i) and one attempt for part (ii), and asks for feedback on the rigor of these proofs.
  
  Label: question-and-answer