Real Analysis: Proving Equivalence of f 1-1 & f(A n B) = f(A) n f(B)

In summary, to show equivalence of two statements, "f is 1-1" and "f(A n B) = f(A) n f(B) for all subsets A,B of S," the attempts involved assuming one statement and proving the other, and vice versa. The key was to use the fact that f is 1-1 and to consider specific subsets A and B in the proof.
  • #1
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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!
 
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  • #2
Perhaps let A={x} and B={y}?
 
  • #3
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:
 

1. What is Real Analysis?

Real Analysis is a branch of mathematics that deals with the study of real numbers, functions, and their properties. It involves using rigorous mathematical techniques to analyze and understand the behavior of continuous functions and their derivatives.

2. What does it mean for a function to be one-to-one (1-1)?

A function is considered one-to-one if each element in the domain is mapped to a unique element in the range. This means that no two elements in the domain are mapped to the same element in the range.

3. How do you prove the equivalence of f 1-1 and f(A n B) = f(A) n f(B)?

To prove the equivalence of f 1-1 and f(A n B) = f(A) n f(B), you need to show that both statements are logically equivalent. This can be done by proving that if f is 1-1, then f(A n B) = f(A) n f(B) and vice versa.

4. What is the significance of proving the equivalence of f 1-1 and f(A n B) = f(A) n f(B)?

This equivalence is important because it allows us to use either condition to prove the other. In other words, if we know that a function is 1-1, we can use it to prove that f(A n B) = f(A) n f(B) and vice versa. This can be useful in solving problems involving functions and sets.

5. Can this equivalence be applied to any type of function?

Yes, this equivalence can be applied to any type of function, as long as the function is well-defined over the given sets. However, it is most commonly used for continuous functions as they have many important properties that can be proven using this equivalence.

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