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

Mathematics
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Let \[f:A\rightarrow B\] be a given function. Prove that f is one-to-one (injective) \[\leftrightarrow f(C\cap D)=f(C)\cap f(D)\] for every pair of sets C and D in A
\[Let f:A\rightarrow B be a given function. Prove that f is one-to-one (injective) \\\Leftrightarrow f(C\cap D)=f(C)\cap f(D) for every pair of sets C and D in A\]
i was just rewriting so i could read it, i am not sure i know how to do it

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Other answers:

well one way is trivial, since \(f(A\cap B)\subset f(A)\cap f(B)\) for any \(f\)
or does that need clarification as well? we can write it out if you like
yes we need to right it..
ok
why letter is not separated
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 \(z\in f(A)\cap f(B)\)
this shows for any \(f\) you have \(f(A\cap B)\subset f(A)\cap f(B)\)
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 \[f(A)\cap f(B)\subset f(A\cap B)\]
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 \[z\in f(A\cap B)\]
typo there, i meant "likewise there exists \(x_2\in B\) with \(f(x_2)=z\) sorry
so that is the proof one way, that "if \(f\) is injective, then \(f(A\cap B)=f(A)\cap f(B)\)
other way is easier, since a singleton is a set
A={x} B={y} you mean like this one..

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