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cinar Group Title

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

  • 2 years ago
  • 2 years ago

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  1. satellite73 Group Title
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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

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

    • 2 years ago
  3. satellite73 Group Title
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    i was just rewriting so i could read it, i am not sure i know how to do it

    • 2 years ago
  4. satellite73 Group Title
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    well one way is trivial, since \(f(A\cap B)\subset f(A)\cap f(B)\) for any \(f\)

    • 2 years ago
  5. satellite73 Group Title
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    or does that need clarification as well? we can write it out if you like

    • 2 years ago
  6. cinar Group Title
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    yes we need to right it..

    • 2 years ago
  7. satellite73 Group Title
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    ok

    • 2 years ago
  8. cinar Group Title
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    why letter is not separated

    • 2 years ago
  9. satellite73 Group Title
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    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 \(z\in f(A)\cap f(B)\)

    • 2 years ago
  10. satellite73 Group Title
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    this shows for any \(f\) you have \(f(A\cap B)\subset f(A)\cap f(B)\)

    • 2 years ago
  11. satellite73 Group Title
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    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 \[f(A)\cap f(B)\subset f(A\cap B)\]

    • 2 years ago
  12. satellite73 Group Title
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    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 \[z\in f(A\cap B)\]

    • 2 years ago
  13. satellite73 Group Title
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    typo there, i meant "likewise there exists \(x_2\in B\) with \(f(x_2)=z\) sorry

    • 2 years ago
  14. satellite73 Group Title
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    so that is the proof one way, that "if \(f\) is injective, then \(f(A\cap B)=f(A)\cap f(B)\)

    • 2 years ago
  15. satellite73 Group Title
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    other way is easier, since a singleton is a set

    • 2 years ago
  16. cinar Group Title
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    A={x} B={y} you mean like this one..

    • 2 years ago
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