Prove that $f(A \backslash S) = B \backslash f(S)$

Question

If $f:A \rightarrow B$ is a bijection and $S \subseteq A$, prove that $f(A \backslash S) = B \backslash f(S)$

Does this answer your question? Overview of basic results about images and preimages, and see also this question — Arnaud D., Nov 02 '20 at 15:21

score 1 · Accepted Answer · answered Nov 02 '20 at 15:07

1

Let $x \in B$. One has $$x \in f(A \setminus S) \Longleftrightarrow f^{-1}(x) \in A \setminus S \Longleftrightarrow f^{-1}(x) \notin S \Longleftrightarrow x \notin f(S) \Longleftrightarrow x \in B \setminus f(S)$$

answered Nov 02 '20 at 15:07

TheSilverDoe

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I think instead of $\Leftrightarrow$ (iff) you meant $\Rightarrow$, right? Because that's not the only case where x is not in f(A\S). – code06 Nov 02 '20 at 15:20
No, I mean $\Longleftrightarrow$, i.e. "iff". What bothers you ? – TheSilverDoe Nov 02 '20 at 15:23
Oh, sorry, I see why now. Thanks. – code06 Nov 02 '20 at 15:24

Prove that $f(A \backslash S) = B \backslash f(S)$

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