# Image of difference of sets

#### 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?

#### Plato

MHF Helper
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 $\subseteq$

• 1 person

#### johng

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.

• 1 person