## cinar 3 years ago $Let f:A \rightarrow B  be a given function. Prove that f is one-to-one (injective)  \Leftrightarrow f(C\cap D)=f(C)\cap f(D)  for every pair of sets C and D in A$

1. satellite73

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

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

3. satellite73

i was just rewriting so i could read it, i am not sure i know how to do it

4. satellite73

well one way is trivial, since $$f(A\cap B)\subset f(A)\cap f(B)$$ for any $$f$$

5. satellite73

or does that need clarification as well? we can write it out if you like

6. cinar

yes we need to right it..

7. satellite73

ok

8. cinar

why letter is not separated

9. satellite73

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

this shows for any $$f$$ you have $$f(A\cap B)\subset f(A)\cap f(B)$$

11. satellite73

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

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

typo there, i meant "likewise there exists $$x_2\in B$$ with $$f(x_2)=z$$ sorry

14. satellite73

so that is the proof one way, that "if $$f$$ is injective, then $$f(A\cap B)=f(A)\cap f(B)$$

15. satellite73

other way is easier, since a singleton is a set

16. cinar

A={x} B={y} you mean like this one..

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