show function equality is false

[cateater2000]

Hi there I'm having trouble with this question, any tips would be great.

Show the following is false, by giving an example of a function(s) f:X->A for which the equalities fail:

f(Y&Z)=f(Y)&f(Z);


thanks for your help.

[mathwonk]

try some examples and let us see what you come up with.

remember a function can be any way at all of mapping the elements of one set to the elements of another set. so start with some real simple finite sets and define the maps in various ways.

[cateater2000]

I came up with something like this but it doesn't seem right.

f: (x,y)->{x+1,y-1}

then i had something like
X={(1,2)}
y={(0,0)}

f(X) gives us (2,1)
f(Y) gives us (1,-1)
so f(X) & f(Y) is just (1)

But f(X&Y) i'm confused because X&Y = 0
I'm not sure if I'm on the right track or not, thanks again for the help.

[matt grime]

XnY is empty not zero, they are different.
You're constructing the function so you can make it so that this problem goes away.

Suppose that X and Y are sets of two elements that have one element in common, say

{1,2} and {2,3}

what about it f were the map that takes 1 and 3 to 'a', and 2 to 'b' what happens there?