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anonymous
 4 years ago
\[ Let f:A \rightarrow B $ be a given function. Prove that f is onetoone (injective) $ \Leftrightarrow f(C\cap D)=f(C)\cap f(D) $ for every pair of sets C and D in A $\]
anonymous
 4 years ago
\[ Let f:A \rightarrow B $ be a given function. Prove that f is onetoone (injective) $ \Leftrightarrow f(C\cap D)=f(C)\cap f(D) $ for every pair of sets C and D in A $\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let \[f:A\rightarrow B\] be a given function. Prove that f is onetoone (injective) \[\leftrightarrow f(C\cap D)=f(C)\cap f(D)\] for every pair of sets C and D in A

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[Let f:A\rightarrow B be a given function. Prove that f is onetoone (injective) \\\Leftrightarrow f(C\cap D)=f(C)\cap f(D) for every pair of sets C and D in A\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i was just rewriting so i could read it, i am not sure i know how to do it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well one way is trivial, since \(f(A\cap B)\subset f(A)\cap f(B)\) for any \(f\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or does that need clarification as well? we can write it out if you like

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes we need to right it..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why letter is not separated

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0suppose \(z\in f(A\cap B)\) then \(z=f(x)\) for some \(x\in A\cap B\) making \(x\in A\) and \(x\in B\) so \(z\in f(A)\) and \(z\in f(B)\) therefore \(z\in f(A)\cap f(B)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this shows for any \(f\) you have \(f(A\cap B)\subset f(A)\cap f(B)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now we need to prove that if \(f\) in injective, we have \(f(A\cap B)=f(A)\cap f(B)\) since we already have containment one way, this amounts to showing \[f(A)\cap f(B)\subset f(A\cap B)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0pick a \(z\in f(A)\cap f(B)\) so there exists a \(x_1\) in \(A\) with \(f(x_1)=z\) and likewise there is a \(x_2\) in \( B\) with \(z=x_2\) now comes the "injective" part since \(f\) is injective, and \(f(x_1)=f(x_2)=z\) we know \(x_1=x_2\) and so \[z\in f(A\cap B)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0typo there, i meant "likewise there exists \(x_2\in B\) with \(f(x_2)=z\) sorry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so that is the proof one way, that "if \(f\) is injective, then \(f(A\cap B)=f(A)\cap f(B)\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0other way is easier, since a singleton is a set

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0A={x} B={y} you mean like this one..
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