Hi, this is my first post here so please tell me if there is something wrong about it...

I just started to read "Notes on set theory" by Yiannis Moschovakis which is available online through my university library and now I am very unsure even about the first few introductory exercise. I would really appreciate some verification of my way to think here, especially since I am not very good in math and haven't read any thorough courses in logic et.c. before. Because this is just an online book which I picked up and started reading, there are no lecturers or such to ask either, hence I turn to mathhelpforum =)

Also, sorry about this wall of text! But I guess I should post my entire attempt...

...in the end of the post I try to be a bit more specific!

I realize that I may be way off with this attempt and there may very well be MAJOR misconceptions and errors. I appreciate all input, big thx in advance!

An example of a problem is :"Show that for everyinjectionf : X $\displaystyle \rightarrow$ Y, and all $\displaystyle A,B \subseteq$X

f[A $\displaystyle \cap$ B] = f[A] $\displaystyle \cap$ f[B].

Also show that this is identity does not always hold if f is not an injection."

First attempt at solution:

____________________start of first attempt________________________________

if x $\displaystyle \epsilon$ f[A $\displaystyle \cap$ B], then there is some y$\displaystyle \epsilon$A$\displaystyle \cap$ B such that f(y)=x.

y$\displaystyle \epsilon$(A$\displaystyle \cap$ B) $\displaystyle \Leftrightarrow$ y$\displaystyle \epsilon$A $\displaystyle \wedge$ y$\displaystyle \epsilon$B

Hence, x $\displaystyle \epsilon$ f[A $\displaystyle \cap$ B]$\displaystyle \Leftrightarrow$ y$\displaystyle \epsilon$A $\displaystyle \wedge$ y$\displaystyle \epsilon$B

If x$\displaystyle \epsilon$ f[A] $\displaystyle \cap$ f[B], then there is some y´$\displaystyle \epsilon$ A $\displaystyle \wedge$ y´$\displaystyle \epsilon$ B such that f(y´) = x.

That is, x$\displaystyle \epsilon$ f[A] $\displaystyle \cap$ f[B] $\displaystyle \Leftrightarrow$ y´$\displaystyle \epsilon$ A $\displaystyle \wedge$ y´$\displaystyle \epsilon$ B

Since f is an injection (stated in the problem formulation); x=f(y)=f(y´) $\displaystyle \Leftrightarrow$ y´= y

Using this to evolve the result from above:

x$\displaystyle \epsilon$ f[A] $\displaystyle \cap$ f[B] $\displaystyle \Leftrightarrow$ y´$\displaystyle \epsilon$ A $\displaystyle \wedge$ y´$\displaystyle \epsilon$ B $\displaystyle \Leftrightarrow$ y$\displaystyle \epsilon$A $\displaystyle \wedge$ y$\displaystyle \epsilon$B$\displaystyle \Leftrightarrow$x $\displaystyle \epsilon$ f[A $\displaystyle \cap$B]

f[A $\displaystyle \cap$ B] = f[A] $\displaystyle \cap$ f[B]

If f isn't an injection, then x=f(y)=f(y´) does not imply y=y´.

________________________end of first attempt_____________________________

Then I started to think about an example wof when the relation does not hold and I realized that my solution might not be very good.

Hence, I tried to adjust my first attempt:

____________________start of augmentation___________________________

If x$\displaystyle \epsilon$ f[A] $\displaystyle \cap$ f[B], then there is some y´$\displaystyle \epsilon$ A such that f(y´) = x and y´´ $\displaystyle \epsilon$B such that f(y´´)=x.

Since f is an injection, x=f(y´)=f(y´´) implies y´=y´´.

y´$\displaystyle \epsilon$A $\displaystyle \wedge$ y´ $\displaystyle \epsilon$ B

___________________________end of augmentation_____________________

Would this augmentation make the solution correct?

Then if y´ is not equal to y´´ , I think it would be quite easy to make a specific example of when the identity does not hold for a non-injective fcn. (for instance say f(x)=x^2 and y´=-y´´, right?)

What I´m concerned about is probably mainly this:

x f[A] f[B] y´ A y´ B yA yB

by the argument that x=f(y)=f(y´) implies y=y´for an injection.

Thereby, both

x f[A B] yA yB

and

x f[A] f[B] yA yB.

is this even correct?? necessary?

Big thx in advance for any input! // HeadmasterEel