Let $f:A\rightarrow B$. Prove that if $ X\subseteq A$, and $f$ is one to one, then $f(A)-f(X) \subseteq f(A-X)$. Let $f:A\rightarrow B$. Prove that if $ X\subseteq A$, and $f$ is one to one, then $f(A)-f(X) \subseteq f(A-X)$.
Could anyone please guide me through this problem? I got stuck and don't know if what I'm doing is right. Thanks.
 A: My attempt: assume that $y\in f(A)-f(X)$. Then $y\in f(A)$, $y\not\in f(X)$. Since $f$ is surjective over $f(A)$ and injective by definition, there exists only one $x\in A$ such that $f(x)=y$. Now, if $x\in X$, $f(x)=y\in f(X)$ which is false. Therefore $x\not\in X$, that is, $x\in A-X$ and thus $f(x)\in f(A-X)$. This shows that $f(A)-f(X)\subset f(A-X)$.
A: Let $f:A\rightarrow B$. Prove that if $ X\subseteq A$, and $f$ is one to one, then $f(A)-f(X) \subseteq f(A-X)$.
My attempt:
Definition : $f(X)$= { $y\in B | y=f(x), \exists x\in X $ }
Let $y\in f(A)-f(X)$ | NTS : $ y\in f(A-X)$
$y\in f(A)$ and $y\notin f(X)$
Here is where I'm not sure if I took the right approach;
$\exists x\in A $ such that $y=f(x)$
but since $y\notin f(X)$ , then $ x \notin X$
Thus $x\in A-X$
Therefore $ y\in f(A-X)$
A: This is actually an extended comment on your attempt to answer your question. First, I agree completely with Gustavo’s advice to use more words and fewer symbols. A proof must first of all be correct, of course, but after that your main goal should be clarity; typically this means using enough words to make the flow of the argument clear to the reader. Thus, the first two lines of your argument would be clearer if written out like this:

Let $y\in f[A]\setminus f[X]$; we want to show that $y\in f[A\setminus X]$, and we know that $y\in f[A]$ and $y\notin f[X]$.

Your next two lines are contradictory: on the one hand you assert that there is an $x\in X$ with a certain property, and on the other you say that this $x$ cannot be in $X$. I think that you have the right basic idea in mind, but you’ve not expressed it very clearly. I think that this is what you’re trying to convey:

Since $y\in f[A]$, there is an $x\in X$ such that $y=f(x)$. But $y\notin f[X]$, so $x$ cannot belong to $X$, and therefore $x\in A\setminus X$. Thus, $y=f(x)\in f[A\setminus X]$, as desired.

Added: I see that while I was writing this, you corrected the problem; what you have now is much better, though it could still benefit from more verbal ‘connective tissue’. Your proof is now correct.
