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 $X$.
Counterexample for the difference between two bounded function
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real-analysis
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0@joriki: as did has mentioned, the second function is meant to be $g$. Fixed it, as well as restated the problem without using $\sup$. – 2012-08-27
1 Answers
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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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0Thanks a lot for the nice answer! – 2012-08-28