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I'm trying to give a shorter alternative proof to this result, i.e.,

Theorem Let $A,B$ be non-empty and $f:A \times B \to \mathbb R \cup \{\pm\infty\}$. Then $$ \sup_{a \in A} \sup_{b \in B} f(a,b) = \sup_{(a,b) \in A \times B} f(a,b)= \sup_{b \in B} \sup_{a \in A} f(a,b) . $$

Could you have a check on my below attempt?

Proof By symmetry, it suffices to prove that $$ \sup_{a \in A} \sup_{b \in B} f(a,b) =: \alpha = \beta:=\sup_{(a,b) \in A \times B} f(a,b). $$

  1. Let $(a_n) \subset A$ such that $$ \sup_{b \in B} f(a_n,b) \uparrow \alpha \quad \text{as} \quad n \to \infty. $$

For each $a_n$, let $(b_n^m)_m \subset B$ such that $$ f(a_n,b_n^m) \uparrow \sup_{b \in B} f(a_n,b) \quad \text{as} \quad m \to \infty. $$

This implies $$ \lim_n \lim_m f(a_n, b_n^m) = \alpha. $$

We have $f(a_n, b_n^m) \le \beta$ for all $n, m \in \mathbb N$, so $\alpha \le \beta$.

  1. Let $(a'_n, b'_n) \subset A \times B$ such that $$ f(a_n', b'_n) \uparrow \beta \quad \text{as} \quad n \to \infty. $$

We have $$ f(a_n', b'_n) \le \sup_{b \in B} f(a'_n,b) \le \alpha \quad \forall n \in \mathbb N, $$ so $\beta \le \alpha$. This completes the proof.

Analyst
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1 Answers1

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Let $M=\sup_{(a,b)\in A\times B}f(a,b)$. For each $a\in A$, let $M_{a}=\sup_{b\in B}f(a,b)$.

To prove that $M\leq\sup_{a\in A}M_{a}$: Let $a\in A$ and $b\in B$ be arbitrary. Clearly $f(a,b)\leq M_{a}\leq\sup_{a\in A}M_{a}$. This shows that $\sup_{a\in A}M_{a}$ is an upper bound of $\{f(a,b)\mid a\in A,b\in B\}$. Therefore $\sup_{(a,b)\in A\times B}f(a,b)\leq\sup_{a\in A}M_{a}$.

To prove that $\sup_{a\in A}M_{a}\leq M$. Let $a\in A$ be arbitrary. Since $\{f(a,b)\mid b\in B\}\subseteq\{f(a,b)\mid a\in A,b\in B\}$ and $M$ is a upper bound of $\{f(a,b)\mid a\in A,b\in B\}$, $M$ is also an upper bound of $\{f(a,b)\mid b\in B\}$. Therefore, $\sup\{f(a,b)\mid b\in B\}\leq M$. That is, $M_{a}\leq M$. Since $a$ is arbitrary, it follows that $\sup_{a\in A}M_{a}\leq M$.

Danny Pak-Keung Chan
  • 12,250