Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

yes we need to right it..

Answer 7

ok

Answer 8

why letter is not separated

Answer 9

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 \(z\in f(A)\cap f(B)\)

Answer 10

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

Answer 11

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
\[f(A)\cap f(B)\subset f(A\cap B)\]

Answer 12

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
\[z\in f(A\cap B)\]

Answer 13

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

Answer 14

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

Answer 15

other way is easier, since a singleton is a set

Answer 16

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