Need help with "Power set" and "Bijection"

[FunnyMaths]

Good afternoon, I dont even know where to start for this question.

Let A, B be sets and let f : A -> B be a function.
Define a function h : P(B) -> P(A) by declaring that, for Y E P(B), h(Y) = {x E A : f(x) is not a member of Y}.
P(A) = the power set of A.

Show that if f is bijection, then h is bijection.
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What I know:
h is bijection iff h is 1-to-1 and onto;
1 - 1: f(a)=f(a') => a=a'
onto: for all b E B, there exists a E A such that f(a) = b
P(A), domain, codomain.
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Any hints/ ideas would be appreciated.

Thank you very much.

 

[daon2]

One problem in doing mathematics is "seeing" what the problem is saying. Sometimes this is difficult because authors seem to enjoy being terse and mysterious, but let's see what we can do here.

(1) \(\displaystyle f :A \to B\) is any function.

(2) \(\displaystyle h (B) \to P(A)\) is such that \(\displaystyle H(Y\subseteq B) = \{x\in A;\,\, f(x)\notin Y\}\)

(1) is clear enough, but what is (2) saying? Given a subset Y of B, h returns to us the set of elements that f does not take to Y. Another way to write this set is as the preimage of \(\displaystyle Im(f) - Y\).

Now, assuming f is a bijection this simplifies to the preimage of \(\displaystyle B-Y = Y^c\), as since \(\displaystyle f\) is onto, \(\displaystyle Im(f)=B\)

So \(\displaystyle h(Y) = f^{-1}(Y^c) = f^{-1}(Y)^c\). This is easier to work with.

Suppose \(\displaystyle h(X)=h(Y)\). Then \(\displaystyle f^{-1}(X)^c=f^{-1}(Y)^c\). But basic set theory tells us that if sets have the same complement (relative to the same containing set), they are the same sets! So \(\displaystyle f^{-1}(X)=f^{-1}(Y)\). But f is a bijection, and so is f^(-1), so...

Think you can do onto?

 

[FunnyMaths]

Now I have a better understanding of the question

Thank you so much for the starting guide, I will try my best to work through it,,

FunnyMaths