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I'm getting that we can neither prove nor disprove it, but it seems there's a definitive answer for this. How to prove or disprove this?

user0102
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codeg
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  • Welcome to MSE. What are your thoughts? – ArsenBerk Dec 21 '20 at 17:03
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    You need to clarify your question. What are $A$ and $B$, and what kind of function is $f$? And does the minus sign mean subtraction (like in a vector space), or is it set difference? – Nick Dec 21 '20 at 17:03

1 Answers1

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This is true since preimage commutes with set difference.

Theorem: Let $A$ and $B$ be sets, and consider a function $f: A \rightarrow B .$ Let $B_{1} \subseteq B$ and $B_{2} \subseteq B$ be any subsets. Then we have $$ f^{-1}\left(B_{1} \backslash B_{2}\right)=f^{-1}\left(B_{1}\right) \backslash f^{-1}\left(B_{2}\right) $$ Proof: For any given $x \in A$, we have $$ \begin{aligned} x \in f^{-1}\left(B_{1} \backslash B_{2}\right) & \Longleftrightarrow f(x) \in B_{1} \backslash B_{2} \\ & \Longleftrightarrow f(x) \in B_{1} \wedge f(x) \notin B_{2} \\ & \Longleftrightarrow x \in f^{-1}\left(B_{1}\right) \wedge x \notin f^{-1}\left(B_{2}\right) \\ & \Longleftrightarrow x \in f^{-1}\left(B_{1}\right) \backslash f^{-1}\left(B_{2}\right) \end{aligned} $$ and therefore, since they comprise the same elements of $A$, $$ f^{-1}\left(B_{1} \backslash B_{2}\right)=f^{-1}\left(B_{1}\right) \backslash f^{-1}\left(B_{2}\right) $$

Son Gohan
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