I wonder if using any one of its linear order extension is completely enough for later discussion. ]]>

When I said “It is not correct” I mean that the second definition that you wrote is not correct. If you compare your definition of $(f^{-1}(E))_x$ with mine, you will realize that they are different.

However, I really appreciate your help and I will be alert to your answers.

]]>When I said “It is not correct” I mean that the second definition that you wrote is not correct. If you compare your definition of $(f^{-1}(E))_x$ with mine, you will realize that they are different.

However, I really appreciate your help and I will be alert to your answers.

]]>When I said “It is not correct” I mean that the second definition that you wrote is not correct. If you compare your definition of $(f^{-1}(E))_x$ with mine, you will realize that they are different.

However, I really appreciate your help and I will be alert to your answers.

]]>I don’t really see how my answer is “not correct”. Maybe you’re upset that I didn’t do *all* of the work for you? Seems a funny way to show the gratitude you start off by expressing.

However, I think it is not correct. Look:

E_x=\{y\in Y\vert(x,y)\in E\}

\left(f^{-1}(E)\right)_x=\{y\in Y\vert(x,y)\in f^{-1}(E)\}=\{y\in Y\vert f(x,y)\in E\}

I don’t see any relation between those two sets.

Maybe if we impose some condition on the function f, we can compare the sets.

I hope you read this and leave some comments. Best regards. Camilo Chaparro.

]]>[ (1 + 1/r)^n – ( 1 – 1/r)^n]^1/n = 2^1/n [ (*n/1)^1/r + (*n/3)^1/r^3 ….]^1/n.

Let r = 1 so that p = q. This deliberately converts the 3 term assumption into 2 terms.

There can only be equality with 2 terms not the 3 terms assumed in Fermat’s Last Theorem. End ]]>