anonymous
  • anonymous
\[ Let f:A \rightarrow B $ be a given function. Prove that f is one-to-one (injective) $ \Leftrightarrow f(C\cap D)=f(C)\cap f(D) $ for every pair of sets C and D in A $\]
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
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
anonymous
  • anonymous
\[Let f:A\rightarrow B be a given function. Prove that f is one-to-one (injective) \\\Leftrightarrow f(C\cap D)=f(C)\cap f(D) for every pair of sets C and D in A\]
anonymous
  • anonymous
i was just rewriting so i could read it, i am not sure i know how to do it

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
well one way is trivial, since \(f(A\cap B)\subset f(A)\cap f(B)\) for any \(f\)
anonymous
  • anonymous
or does that need clarification as well? we can write it out if you like
anonymous
  • anonymous
yes we need to right it..
anonymous
  • anonymous
ok
anonymous
  • anonymous
why letter is not separated
anonymous
  • anonymous
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 \(z\in f(A)\cap f(B)\)
anonymous
  • anonymous
this shows for any \(f\) you have \(f(A\cap B)\subset f(A)\cap f(B)\)
anonymous
  • anonymous
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 \[f(A)\cap f(B)\subset f(A\cap B)\]
anonymous
  • anonymous
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 \[z\in f(A\cap B)\]
anonymous
  • anonymous
typo there, i meant "likewise there exists \(x_2\in B\) with \(f(x_2)=z\) sorry
anonymous
  • anonymous
so that is the proof one way, that "if \(f\) is injective, then \(f(A\cap B)=f(A)\cap f(B)\)
anonymous
  • anonymous
other way is easier, since a singleton is a set
anonymous
  • anonymous
A={x} B={y} you mean like this one..

Looking for something else?

Not the answer you are looking for? Search for more explanations.