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

Question

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.

I did understand, however denoting the direct image with square brackets is unusual. You normally use round brackets. Note that you can differentiate between the direct image $f(X)$ and the point $f(x)$ by the size of the x. — Jonas Linssen, Feb 15 '20 at 05:37
I see ur confusion, haha. I took the square parentheses practice from my textbook actually lol — Andes Lam, Feb 15 '20 at 05:42
Right. I just haven’t seen this notation before. Normally you denote by square brackets things like polynomials or free objects, so I got confused... — Jonas Linssen, Feb 15 '20 at 05:45
However, the inclusion will be an equality if $f$ is injective. — Geoffrey Trang, Feb 15 '20 at 06:00
In set-theoretical circles it is not at all unusual to use square brackets for images of sets -- on the contrary, I'd argue in this context $f(X)$ might be confusing as $X$ is not in the domain of $f$. OP's usage, in this context, is totally standard. — MacRance, Feb 15 '20 at 06:02
Hello guys, the textbook I am referencing is attached below https://lib.hku.hk/Press/9622090133.pdf P.64/76 — Andes Lam, Feb 15 '20 at 06:04

score 0 · Accepted Answer · answered Feb 15 '20 at 05:42

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 $f(\{1\}\setminus \{0\}) = f(\{1\}) = \{*\} \neq \emptyset$.

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

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