Prove $ f(A1) \setminus f(A2) \subseteq f(A1 \setminus A2) $

Question

here is my proof but it only works for functions that have an inverse:

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Please put a little effort in and write up a post with MathJax. :) Its hard to read your work. — , Oct 02 '14 at 14:07
Heres a quick tutorial: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — , Oct 02 '14 at 14:16
See also: http://math.stackexchange.com/questions/511662/proving-fc-setminus-fd-subseteq-fc-setminus-d-and-disproving-equality — Martin Sleziak, Oct 02 '14 at 14:24
Possible duplicate of Proving $f(C) \setminus f(D) \subseteq f(C \setminus D)$ and disproving equality — mle, May 01 '17 at 20:01

score 3 · Accepted Answer · answered Oct 02 '14 at 14:16

I can prove it by showing that all elements from $f(A1) - f(A2)$ belong to $f(A1 - A2)$. (Let me use the symbol $-$ for the exclusion of sets.)

Let's consider a $y \in f(A1) - f(A2)$. This implies two consequences:

$y \in f(A1) \Rightarrow \exists x \in A1 \text{ s.t. } f(x) = y$.
$y \not\in f(A2) \Rightarrow \forall x' \in A2,\: f(x') \not= y$.

Since $f(x) = y \not= f(x')$, we derive that $x \not= x'$ for any $x' \in A_2$. Therefore, $x \not\in A_2$.

In conclusion, $\exists x \in A1 - A2 \text{ s.t. } f(x) = y$, so $y \in f(A1 - A2)$.

score 1 · Answer 2 · answered Oct 02 '14 at 14:18

1

Or $f(A_1)\setminus f(A_2)\subset f(A_1\setminus A_2)\Leftrightarrow f(A_1)\subset f(A_1\setminus A_2)\cup f(A_2)=f((A_1\setminus A_2)\cup A_2)=f((A_1\cup A_2)\cap (A_2^{c} \cup A_2))=f(A_1\cup A_2)$ and $A_1\subset A_1\cup A_2$

answered Oct 02 '14 at 14:18

Haha

5,648

score 1 · Answer 3 · answered Oct 02 '14 at 14:41

Claim $(\forall x)(x \in f(A1) \setminus f(A2) => x \in f(A1\setminus A2)$.

Let $x \in f(A1) \setminus f(A2)$.

$\Rightarrow x \in f(A1)\land x \notin f(A2) $

$\Rightarrow (\exists y \in A1) f(y) = x \land (\not \exists z \in A2) f(z) = x $

$\Rightarrow (\exists y \in A1) f(y) = x \land ( \forall z \in A2) f(z) \neq x $

$ \Rightarrow (\forall z \in A2) (\exists y \in A1) f(y) = x \land{} f(z) \neq x $

$f$ a function $ \Rightarrow (\forall z \in A2) (\exists y \in A1) f(y) = x \land f(z) \neq x \land z \neq y $

$ \Rightarrow (\forall z \in A2) (\exists y \in A1) f(y) = x \land f(z) \neq x \land (\forall z \in A2) z \neq y $

$ \Rightarrow (\exists y \in A1) f(y) = x \land y \notin A2 $

$ \Rightarrow (\exists y) y \in A1 \land y \notin A2 \land f(y) = x $

$ \Rightarrow (\exists y \in A1 \setminus A2) f(y) = x $

$ \Rightarrow x \in f(A1 \setminus A2) $

Prove $ f(A1) \setminus f(A2) \subseteq f(A1 \setminus A2) $

3 Answers3