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cinar

  • 2 years ago

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

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  1. satellite73
    • 2 years ago
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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. cinar
    • 2 years ago
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    \[Let f:A\rightarrow B be a given function. Prove that f is one-to-one (injective) \\\Leftrightarrow f(C\cap D)=f(C)\cap f(D) for every pair of sets C and D in A\]

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

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

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

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

  7. satellite73
    • 2 years ago
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    ok

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

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

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

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

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

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

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

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

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

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