1

I'm trying to prove that difference of range is always bigger than range of difference, under some assumptions. Actually it's quite intuitive, but I want to prove it without venn diagram, using the definition of subset. How can I prove it?

Assumptions:

  1. $f:A\to B$;

  2. $C\subseteq A$.

Claim: $$f(A) \setminus f(C)\subseteq f(A\setminus C)\,.$$

Asaf Karagila
  • 393,674
Ashe
  • 11
  • 1

1 Answers1

1

First, write down explicitly $$f(A)\setminus f(C) = \{y\in B\text{ such that } \exists x\in A, f(x)=y \text{ and } \forall z\in C, f(x)\neq y\}\,.$$ Now let $y \in f(A)\setminus f(C)$. For any $x\in A$ such that $f(x)=y$, we must have that $x\notin C$ (as for all $x\in C$ we have that $f(x)\neq y$). Hence, we can say that $y$ lies in the set $$\{y\in B\text{ such that } \exists x\in A, f(x)=y \text{ and } x\notin C\}\,.$$ But this set is exactly $f(A\setminus C)$. We thus have proved that $$f(A)\setminus f(C)\subseteq f(A\setminus C)\,.$$

ECL
  • 2,960