Definition of set using elementhood test

[el_llavero]

I have a set, A = {1,4,9,16,25,36,49,...}, that I want to write a definition for using an elementhood test. I have written one defintion I'm sure is correct but I'm not sure if the other one is appropriate since there are values that satisfy the conditions in the definitions but produce values that are not elements of A.

I have A = { x $$\in$$ $$\aleph$$ $$^{+}$$ | x $$^{2}$$ }

However I've seen another definition for the same set but I'm not sure it's accurate

A = { x ∈ ℕ $$^{+}$$ | y is a positive odd integer, x + y }

I'm not sure the second defition is correct, take the case where x is 1 and y is 5, then x+y=6, which is not part of the set. Could someone give me their perspective. I'm under the impression that these definitions have to work in all cases.

 

[statdad]

"I'm not sure the second defition is correct, take the case where x is 1 and y is 5, then x+y=6, which is not on the list. Could someone give me their perspective. "

Haven't you answered your own question about the correctness of the second "definition"?

 

[el_llavero]

My question is more about the structure of the second expression used to define the set A.

I've just read something regarding free variables and bound variables. In the case of the second definition, x is a bound variable since the notation {x |...} binds the variable and y is a free variable since you can plug in different values for y, and it's not bound by said notation . Since y is a free variable the statement will be true for some values of y and false for others therefor the second expression is accurate. plugging some values for y will produce an element in the set while others won't but that's ok since y is not a bound variable.

Am I making sense? I going by definitions in this book I'm using.

 

[Tedjn]

I don't know all the relevant definitions and theorems, but since the two sets you described do not contain the same elements, how can they be equal?

 

[el_llavero]

how do you know they don't contain the same elements, y isn't bound so it can stand for all the values that make x+y an element of the original set.