# Thread: Image of difference of sets

1. ## Image of difference of sets

Hey,

I want to show that $\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)$

I have done the following:

Assume that $\displaystyle y \in f(A\backslash B)$.
This means that $\displaystyle \exists x\in A\backslash B: f(x)=y$ (by definition).
This means that $\displaystyle x\in A \land x\notin B$.
This means that $\displaystyle f(x) \in f(A) \land f(x) \notin B$.
This means that $\displaystyle f(x) \in f(A) \backslash f(B)$

Therefore $\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)$.

Is the above correct?

2. ## Re: Image of difference of sets

Originally Posted by detalosi
Hey,

I want to show that $\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)$

I have done the following:

Assume that $\displaystyle y \in f(A\backslash B)$.
This means that $\displaystyle \exists x\in A\backslash B: f(x)=y$ (by definition).
This means that $\displaystyle x\in A \land x\notin B$.
This means that $\displaystyle f(x) \in f(A) \land f(x) \notin B$.
This means that $\displaystyle f(x) \in f(A) \backslash f(B)$

Therefore $\displaystyle f(A\backslash B) \subset f(A) \backslash f(B)$.

Is the above correct?
Yes it is completely correct. Good work.

One comment: I would have preferred using \subseteq $\subseteq$

3. ## Re: Image of difference of sets

The statement is not true for arbitrary functions. For example, let A be the set of positive reals, B the set of negative reals and f the squaring function. Your conclusion says $A\subseteq\emptyset$ ??

One hypothesis to make the statement true is that f is one to one. in the 4th line of your proof, you need some reason why $f(x)\not\in f(B)$ is true; I hope you understand why f one to one guarantees this to be true.