Is f-1(A ∪ B) the same as f-1(A) ∪ f-1(B)?

  • Thread starter brookey86
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In summary, The conversation is discussing how to prove that two sets are equal, specifically f^{-1}(A\cup B)\subset f^{-1}(A)\cup f^{-1}(B). The method suggested is to show that each set is a subset of the other, by using mathematical symbols and proving that for any x in one set, it is also in the other set. It is noted that this does not require f to have an inverse.
  • #1
brookey86
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Homework Statement

Homework Equations


The Attempt at a Solution

 
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  • #2
Yes, you are. But I'd feel more comfortable if you tried to prove it rather than taking as an assumption.
 
  • #3
Have you tried to prove it's true?
 
  • #4
SammyS said:
Have you tried to prove it's true?

I can prove it using words, not quite there using mathematical symbols, but that part is out of the scope of my class. Thanks guys!
 
  • #5
You prove two sets are equal by proving that each is a subset of the other. You prove "A" is a subset of "B" by saying "let [itex]x\in A[/itex]", then show "[itex]x\in B[/itex]".

Here, to show that [itex]f^{-1}(A\cup B)\subset f^{-1}(A)\cup f^{-1}(B)[/itex], start by saying "let [itex]x\in f^{-1}(A\cup B)[/itex]". Then [itex]y= f(x)\in A\cup B[/itex]. And that, in turn, means that either [itex]y\in A[/itex] or [itex]y \in B[/itex]. Consider each of those.

Note, by the way, that we are considering the inverse image of sets. None of this implies or requires that f actually have an "inverse".
 

FAQ: Is f-1(A ∪ B) the same as f-1(A) ∪ f-1(B)?

What is the meaning of f-1(A ∪ B)?

f-1(A ∪ B) means the preimage of the union of sets A and B. In other words, it is the set of all elements in the domain of the function f that map to any element in the set A or B.

What is the difference between f-1(A ∪ B) and f-1(A) ∪ f-1(B)?

The difference is in the order of operations. f-1(A ∪ B) first takes the preimage of the union of sets A and B, and then combines the resulting sets. f-1(A) ∪ f-1(B) first takes the preimage of set A and the preimage of set B separately, and then combines the two resulting sets.

Are f-1(A ∪ B) and f-1(A) ∪ f-1(B) always equal?

No, they are not always equal. Depending on the function f and the sets A and B, the two may result in different sets. It is important to consider the order of operations and the specific elements in each set when determining equality.

Can f-1(A ∪ B) be simplified to f-1(A) ∪ B?

No, f-1(A ∪ B) cannot be simplified to f-1(A) ∪ B. The preimage operation only applies to sets, not individual elements. Therefore, f-1(A) ∪ B is not a valid operation.

How can I determine if f-1(A ∪ B) is equal to f-1(A) ∪ f-1(B)?

To determine if the two sets are equal, you can use set equality properties such as the subset or equality tests. You can also use specific examples and the definition of preimage to compare the resulting sets. If the two sets have the same elements, they are equal.

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