Anushkaspam
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Prove f(a1∩f^-1(b2)=f(a1)∩b2
Set Theory Proof
- Thread starter Anushkaspam
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HallsofIvy
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You missed a parenthesis- I think you mean f(a1)∩f^-1(b2)=f(a1)∩b2.
But now I have a problem with the domain and range of f. In order that the left side be non-empty there must be x that is in both f(a1) and in f^-1(b2). a1 must be a subset of the domain of f so f(x) is in the range of f. But b2 must be a subset of the range of f so that f^-1(x) is in the domain of f. That is, f must map some set onto itself.
But now I have a problem with the domain and range of f. In order that the left side be non-empty there must be x that is in both f(a1) and in f^-1(b2). a1 must be a subset of the domain of f so f(x) is in the range of f. But b2 must be a subset of the range of f so that f^-1(x) is in the domain of f. That is, f must map some set onto itself.
Dr.Peterson
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This is not a complete statement of the problem. Please state everything; I expect to see something like this:
Prove that, for any function f:A -> A, and any subsets a1 and b1 of A, f(a1)∩f-1(b2)=f(a1)∩b2.