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

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satellite73Best 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
 one year ago

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

satellite73Best ResponseYou've already chosen the best response.0
i was just rewriting so i could read it, i am not sure i know how to do it
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
well one way is trivial, since \(f(A\cap B)\subset f(A)\cap f(B)\) for any \(f\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
or does that need clarification as well? we can write it out if you like
 one year ago

cinarBest ResponseYou've already chosen the best response.0
yes we need to right it..
 one year ago

cinarBest ResponseYou've already chosen the best response.0
why letter is not separated
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
suppose \(z\in f(A\cap B)\) 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 \(z\in f(A)\cap f(B)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
this shows for any \(f\) you have \(f(A\cap B)\subset f(A)\cap f(B)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
now 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 \[f(A)\cap f(B)\subset f(A\cap B)\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
pick 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 \[z\in f(A\cap B)\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
typo there, i meant "likewise there exists \(x_2\in B\) with \(f(x_2)=z\) sorry
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
so that is the proof one way, that "if \(f\) is injective, then \(f(A\cap B)=f(A)\cap f(B)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
other way is easier, since a singleton is a set
 one year ago

cinarBest ResponseYou've already chosen the best response.0
A={x} B={y} you mean like this one..
 one year ago
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