Set Theory Proof

Anushkaspam

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May 13, 2019
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Prove f(a1∩f^-1(b2)=f(a1)∩b2
 

HallsofIvy

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Jan 27, 2012
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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.
 

Dr.Peterson

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Nov 12, 2017
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Prove f(a1)∩f^-1(b2)=f(a1)∩b2
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.​
 
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