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Let $X,Y$ be some sets and $f,g:X\times Y\to [0,1]$ be two bounded functions defined over the product set. Suppose that $Y_n\uparrow Y$ is an increasing sequence of sets whose union is $Y$, and that for any $n\in \mathbb N$ and $x\in X$ there exists $z(n,x) $ such that $$ f(x,y) = g(z(n,x),y) \quad\text{ for all }y\in Y_n. $$ Does it mean that for any $x$ there exists $z'(x)$ which satisfies $$ f(x,y) = g(z'(x),y) \quad\text{ for all }y\in Y. $$ I doubt that this is true, and I was looking for the counterexample. Certainly, the set $X$ has to be infinite, since the statement is true for finite $X$.

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Did you mean to exclude $z(n,x)=x$ and $z'(x)=x$? If not, those choices trivially satisfy both conditions. Also, why the supremum formulation; isn't this equivalent to saying that $f(z(n,x),y)=f(x,y)$ for all $y\in Y_n$? – joriki Aug 27 '12 at 15:35
@joriki Some $f$s should be $g$s... – Did Aug 27 '12 at 15:46
@joriki: as did has mentioned, the second function is meant to be $g$. Fixed it, as well as restated the problem without using $\sup$. – Ilya Aug 27 '12 at 15:59
up vote 4 down vote accepted

Let's construct a counter-example.

Take $X=Y=\Bbb N$ and $Y_n=\{1,2,\ldots,n\}$. Define $f$ and $g$ by the equations $f(x,y)=\frac1y$ and $g(x,y)=\frac1{\min\{x,y\}}$. We may now define $z$ by the equation $z(n,x)=n$.

This satisfies the first condition: $$f(x,y)=\frac1y=\frac1{\min\{n,y\}}=g(z(n,x),y)\text{ for all }y\in Y_n$$

But there is no function $z'$ which satisfies the second condition: suppose there is such a function and let $x$ be fixed. Then $$\frac1y=f(x,y)=g(z'(x),y)=\frac1{\min\{z'(x),y\}}\text{ for all }y\in Y$$ But this means that $y=\min\{z'(x),y\}\le z'(x)$ for all $y\in Y$. But then $Y = \Bbb N$ is bounded, which is a contradiction.

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This is nice, but needs a minor repair to satisfy $f,g$ being bounded on $X \times Y$. perhaps take $\arctan \circ f$ and $\arctan \circ g$? – copper.hat Aug 27 '12 at 20:33
@copper.hat: I think it should be ok now. Thanks for the comment. – Dejan Govc Aug 27 '12 at 20:47
Even better! Nice. – copper.hat Aug 27 '12 at 20:47
Thanks a lot for the nice answer! – Ilya Aug 28 '12 at 5:38

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