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  1. MHB Apprentice

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    #1
    1.Prove f(A⋂B) ⊂ f(A) ⋂ f(B)
    2.Prove f(A) ⋂ f(B) ⊂ f(A⋂B) ⟺ f is an injection

    I've solved the first question , as for the second I started with f(A) ⋂ f(B) ⊂ f(A⋂B) ⇒ f is an injection this way :
    Let's suppose f(a) = f(b) = p
    If we consider A = {a} and B = {b} then f(A) = f(B) = p
    then f(A) ⋂ f(B) = p
    then f(A⋂B) = p (because f(A) ⋂ f(B) ⊂ f(A⋂B) from the supposition and f(A⋂B) ⊂ f(A) ⋂ f(B) from the first question)
    then A⋂B ≠ ∅
    then a=b
    then f is an injection .

    But I don't know how to solve "f is an injection ⇒ f(A) ⋂ f(B) ⊂ f(A⋂B)"

  2. Indicium Physicus
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    #2
    $(\implies)$ Assume $f(A)\cap f(B) \subset f(A\cap B)$ for all sets $A$ and $B$. Suppose $A=\{x\}$ and $B=\{y\}$ are singleton sets. Further assume that $f(x)=f(y)$. Then $f(A)=f(B)$, so $f(A)\cap f(B)$ is nonempty. By assumption, $f(A\cap B)$ is nonempty, since $f(A)\cap f(B)\subset f(A\cap B)$. This implies $A\cap B$ is nonempty. Since $A$ and $B$ are singletons, this implies $A=B$, or $\{x\}=\{y\}$, or $x=y$. Hence, $f$ is an injection.

    $(\impliedby)$ This is a "what do you know" kind of proof. Can you write it out?

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