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.



(i) First attempt:

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.

Second attempt is correct.

(ii) It is correct.

  • $\begingroup$ 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? $\endgroup$
    – elbarto
    Mar 15 '15 at 23:46
  • $\begingroup$ Nope. They look very good to me. $\endgroup$
    – KittyL
    Mar 16 '15 at 9:13

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