Prove f(A)-f(B) C f(A-B) (the C is underlined - codomain)?

Question

also, show by counterexamples that the inclusions cannot be changed to inequalities.
This is part of relations and functions, which im no good at! help would be very much appreciated!!!!!

edit: this is all im given, the only symbol i get is C, where it is underlined for the codomain

Answer 1

Homework, but so necessary that I'll do it as an example.

You are asked to show that f(A)-f(B) is a subset of f(A-B). That's that that C shaped thing means: 'is a subset of'.

So, take something in f(A)-f(B). Call that thing y. Then y is in f(A) and not in f(B) [that's what set minus means]. Since y is in f(A), there is an x in A with y=f(x) [that's how the set f(A) is defined). But then, if x were in B, y=f(x) would be in f(B), which we know it isn't. Hence x is in A and not in B, so x is in A-B. But then y=f(x) is in f(A-B).

It turns out that the other set containment is not always true, so you should find a counter-example.

Answer 2

The "C" is supposed to be the symbol for set inclusion. My guess is whoever wrote the question did not know how to typeset the correct symbol. The "C" is underlined to mean that the set f(A) - f(B) is contained in and possibly equal to the set f(A-B). The second part of the question asks you to show that, in fact, f(A)-f(B) is not equal to f(A-B) in general.

Answer 3

Symbols are required.