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Real Analysis Proof

  • Thread starter Tangent...
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  • #1

Homework Statement


I'm trying to show equivalence of two statements:

Let f:S-->T be a function, show that f is 1-1 (injective) is equivalent to f(A n B) = f(A) n f(B) for all A,B subsets of S.


The Attempt at a Solution


I know equivalence means iff, so I started by assuming f is 1-1 and showing f(A n B) = f(A) n f(B) by showing containment both ways (I think I did that part right, since f(A n B) subset of f(A) n f(B) is easy, and f(A) n f(B) subset of f(A n B) uses the fact that f is 1-1).

Now I assume f(A n B) = f(A) n f(B) and try to show f is 1-1. I let x,y be elements of S such that f(x) = f(y). And I don't know where to go from here. I guess I don't know how to combine f(x) = f(y) and f(A n B) = f(A) n f(B) but I'm pretty sure I have to, somehow. Any tips would be greatly appreciated. Thanks!
 

Answers and Replies

  • #2
vela
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Perhaps let A={x} and B={y}?
 
  • #3
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Or maybe A={x} and B=S-{x}.
 
  • #4
Thanks for the replies, I think I figured it out a way to do it. Since it is p iff q, and I can prove if p then q, for the if q then p I just proved the contrapositive, if not p then not q. This should be valid, right? Logic is confusing :rolleyes:
 

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