# 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
The text passage contains a mathematical problem involving the proof of properties of the union of two bounded sets, specifically proving that the union is bounded above and that the supremum of the union is the maximum of the suprema of the individual sets. The passage also includes two attempts at proving the first part and a proof for the second part, followed by a request for feedback on the proofs.
no_question

the passage does not contain a direct mathematical question that requires a numerical or calculative answer. instead, it presents proofs and asks for critiques, which is more of a request for feedback rather than a question in the traditional sense.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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
user
The text passage contains a mathematical problem involving the proof of properties of the union of two bounded sets, specifically proving that the union is bounded above and that the supremum of the union is the maximum of the suprema of the individual sets. The passage also includes two attempts at proving the first part and a proof for the second part, followed by a request for feedback on the proofs.

TYPE: NO_QUESTION

The passage does not contain a direct mathematical question that requires a numerical or calculative answer. Instead, it presents proofs and asks for critiques, which is more of a request for feedback rather than a question in the traditional sense.