Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $X$.

share|improve this question
1  
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
1  
@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

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.