More specifically if for some function f: X → Y for which A ⊆ X and B ⊆ X. How does it affect f(A)\f(B) ⊆ f(A\B)? If it is not injective then does the containment fail?
What happens when f is injective for the containment of some functions?
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0Have you tried some examples? Look at some maps from $\{1, 2\}$ to $\{1, 2\}$ . . . – 2017-02-13
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1@user400188 - "all functions are injective"? Not true. $f(x) = x^2$ is a function, but it is not injective. – 2017-02-13
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0if $y \in f(A)\setminus f(B)$, what does that mean? It means that there is some $a$ in $A$ such that $f(a) = y$, but there is no $b$ in $B$ such that $y = f(b)$. If $y \in f(A\setminus B)$, what does that mean? It means that there is some $a \in A$ such that $a \notin B$ and $y = f(a)$. Now, does every $y$ that meets the first criterion also meet the second? More particularly, Is it possible that the $a$ found in the first criterion will not meet the conditions of the second? – 2017-02-13
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0Currently working on where the containment fails with a numerical example. I'd appreciate your comment in this post --> http://math.stackexchange.com/questions/2145303/how-can-i-find-that-the-containment-for-fa-fb-⊆-fa-b-fails-given-that-f-x – 2017-02-15