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Problem statement:

Assume there are two constants $\alpha, \beta \in \mathbb{R}$ and two functions $$f: A\times B \rightarrow \mathbb{R} \text{ and } g: A\times B \rightarrow \mathbb{R}$$ non-negative, continuous, and bounded, $A$ and $B$ non-empty, such that $\forall b \in B, \exists a\in A$ with $\alpha \leq f(a,b)$ and $\forall a \in A, \exists b \in B$ with $\beta \leq g(a,b)$, then there exists a pair $(a,b) \in A \times B$ such that $$\alpha \leq f(a,b) \text{ and } \beta \leq g(a,b).$$

My attempt:

The statement seems like it should be trivial, but I find myself making what seems like a circular argument.

Proof:

Let $b\in B$, then by assumption there is an $a_b \in A$ such that $\alpha \leq f(a_b, b)$. Now, with $a_b$, there exists a $b_{a_b} \in B$ such that $\beta \leq g(a_b, b_{a_b})$. If $b=b_{a_b}$, then $g(a_b, b) = g(a_b, b_{a_b})$ and $$\alpha \leq f(a_b, b) \text{ and } \beta \leq g(a_b, b)$$ and so $(a_b, b)$ is a pair that satisfies the statement.

Comment:

My issue is, can I just write "if $b = b_{a_b}$"? Nothing prevents this from happening and since I just need existence this argument seems valid...

Stephen Diadamo
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  • If $A$ and $B$ are empty, then the preconditions hold vacuously but the consequence fails as there simply isn't any pair in $A\times B$. – Derek Elkins left SE Jul 18 '18 at 19:16
  • For your comment, you can certainly say "if $b=b_{a_b}$ then" but you have to also handle the $b\neq b_{a_b}$ case, or you have to give an argument for what that case can't happen. You can't just say "if it is the case that..." because the obvious retort is then, "well, what if it isn't the case?" – Derek Elkins left SE Jul 18 '18 at 19:25
  • Let $\alpha=\beta=1$ and $A=B={0,1}$. Define $f(a,b)=a+b\pmod 2$ and $g(a,b)=1- f(a,b)\pmod 2$. $f$ and $g$ satisfy the preconditions ($f(a,1-a)=1$ and $g(a,a)=1$), but $f+g$ is the constantly $1$ function and thus less than $\alpha+\beta=2$. Perhaps you also have an unstated continuity assumption. – Derek Elkins left SE Jul 18 '18 at 20:11
  • I modified the statement slightly, sorry for the unnecessary confusion. – Stephen Diadamo Jul 18 '18 at 20:35
  • Well, my counter-example was non-negative and bounded. So you will need to make use of the continuity assumption. The fact that your proof attempt would, if it worked, work just as well for non-continuous functions shows that something is missing/wrong and suggests a direction to look to find out what may need to be added. – Derek Elkins left SE Jul 18 '18 at 20:38
  • I see what you mean... Thanks. I will have to think about it more. – Stephen Diadamo Jul 18 '18 at 21:11

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