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

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

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