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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?
Thank you sir.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).
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.
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.
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.
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.
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.