Q:

Prove $f[X]\backslash f[X']\subset f[X\backslash X']$

edit: $f[X]=\{f(x):x\in X\}$
$f[X]\backslash f[X']\subset f[X\backslash X']$
Simple: Proof by definition. What I would like to provide is a counter-example and intuition to the reverse of this question. Please correct me if I am wrong. If it is correct, I hope this could serve as a reference to the textbook which I can't find its solution in the internet.
$f[X\backslash X']\subsetneq f[X]\backslash F[X']$
Counter-example: constant mapping

$X\cap X'=\emptyset,A=X\cup X',B:\{y\}$
$f:A\rightarrow B\;s.t.f(x)=y\;\forall x\in A$
$f[X\backslash X']=f[X]=\{y\}=f[X']$
$f[X]\backslash f[X']=\{y\}\backslash \{y\}=\emptyset$
$\{y\}\subsetneq\;\emptyset$

Intuition:

$A\subset\emptyset\equiv\;False$
$f[X]=f[X']\rightarrow f[X]\backslash f[X']=\emptyset$
$X\cap X'=\emptyset\rightarrow X\backslash X'=X\;and\;X\neq\emptyset\;and\;X\neq\emptyset$
$f[X]\subsetneq\emptyset$

So it all depends on which is which--either you find constant mapping or manipulation on statement calculus mechanical.

A:

Take $f:\{0,1\}\rightarrow \{*\}$. Note that there is no choice to make, you have to send $0$ and $1$ to $*$. Now $f(\{0\}) = \{*\} = f(\{1\})$, hence $f(\{1\})\setminus f(\{0\}) = \emptyset$, however $f(\{1\}\setminus \{0\}) = f(\{1\}) = \{*\} \neq \emptyset$.

