Proof Wrong: Where f(A∩B) ≠ f(A) ∩ f(B)

  • Thread starter woundedtiger4
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In summary, the conversation discusses the difference between f(A∩B) and f(A) ∩ f(B) and how to prove that they are not equal. The main idea is to find a counterexample where f(A)\cap f(B) is not a subset of f(A∩B). This can be done by finding a function f and values p and q, where p is in A and q is in B, such that f(p)=f(q).
  • #1
woundedtiger4
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I know that f(A∩B) is not equal to f(A) ∩ f(B) but i don't know that where am i wrong in the following proof...:( can someone please give me an intuitive example?
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  • #2
In order to prove "[itex]X= Y[/itex]" you must prove both "[itex]X\subseteq Y[/itex]" and [itex]Y\subseteq X[/itex].

You want to prove that [itex]f(A\cap B)[/itex] is not equal to [itex]f(A)\cap f(B)[/itex] and your give proof shows only that [itex]f(A\cap B)\subseteq f(A)\cap f(B)[/itex].

So look for a counter example in which [itex]f(A)\cap f(B)[/itex] is NOT a subset of [itex]f(A\cap B)[/itex].

That is, find a function f and values p and q, p in A, q in B such that f(p)= f(q).
 
  • #3
HallsofIvy said:
In order to prove "[itex]X= Y[/itex]" you must prove both "[itex]X\subseteq Y[/itex]" and [itex]Y\subseteq X[/itex].

You want to prove that [itex]f(A\cap B)[/itex] is not equal to [itex]f(A)\cap f(B)[/itex] and your give proof shows only that [itex]f(A\cap B)\subseteq f(A)\cap f(B)[/itex].

So look for a counter example in which [itex]f(A)\cap f(B)[/itex] is NOT a subset of [itex]f(A\cap B)[/itex].

That is, find a function f and values p and q, p in A, q in B such that f(p)= f(q).
Thank you sir.
I understand your point & I can guess the best example is one to many which is not true, & i have tried to show it in proof.
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1. What is the meaning of "Proof Wrong: Where f(A∩B) ≠ f(A) ∩ f(B)"?

The statement "Proof Wrong: Where f(A∩B) ≠ f(A) ∩ f(B)" means that there exists at least one set where the intersection of the images of two sets (A and B) under a function f is not equal to the image of the intersection of those sets. In other words, there exists at least one case where the function does not preserve the intersection of sets.

2. Can you provide an example to illustrate this statement?

Yes, for instance, consider the function f: R → R where f(x) = x². Let A = {-2, 0, 2} and B = {-1, 0, 1}. The intersection of A and B is {0}, and the intersection of the images of A and B under f is {0}. However, the image of A under f is {4, 0, 4}, and the image of B under f is {1, 0, 1}. Therefore, the intersection of the images of A and B is {0}, which is not equal to the image of the intersection of A and B.

3. Why is this statement important?

This statement is important because it highlights the fact that not all functions preserve the intersection of sets. It also serves as a reminder to be cautious when making assumptions about functions and their properties.

4. Does this statement apply to all functions?

No, this statement does not apply to all functions. In fact, there are many functions that do preserve the intersection of sets. For example, the function f: R → R where f(x) = x is an example of a function that preserves the intersection of sets.

5. How can we prove that this statement is true or false?

We can prove this statement by providing a counterexample where the intersection of the images of two sets under a function is not equal to the image of the intersection of those sets. On the other hand, we can prove this statement is false by showing that for all functions, the intersection of the images of two sets is equal to the image of the intersection of those sets.

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