Let A and B be non-empty sets, E and F be subsets of A, and G and H be subsets of H. Considera function f : A → B.(i) Present a counterexample to disprove the following statement:f (E ∩ F ) = f (E) ∩ f (F )

Sure, let's consider the following counterexample:

Let A = B = {1, 2, 3}, E = {1, 2}, F = {2, 3}, and let f be a function defined as f(x) = 1 for all x in A.

Then, E ∩ F = {2}, so f(E ∩ F) = {1}.

However, f(E) = {1} and f(F) = {1}, so f(E) ∩ f(F) = {1}.

Therefore, in this case, f(E ∩ F) = f(E) ∩ f(F) holds true.

But if we change the function f to be f(x) = x, then f(E ∩ F) = {2}, while f(E) = {1, 2} and f(F) = {2, 3}, so f(E) ∩ f(F) = {2}.

So, in this case, f(E ∩ F) = f(E) ∩ f(F) also holds true.

However, if we change the sets to E = {1, 2}, F = {3}, then E ∩ F = {}, so f(E ∩ F) = {}. But f(E) = {1, 2} and f(F) = {3}, so f(E) ∩ f(F) = {}.

So, in this case, f(E ∩ F) = f(E) ∩ f(F) holds true.

But if we change the function f to be f(x) = 1 for x = 1, 2 and f(x) = 2 for x = 3, then f(E ∩ F) = {}, while f(E) = {1} and f(F) = {2}, so f(E) ∩ f(F) = {}.

So, in this case, f(E ∩ F) ≠ f(E) ∩ f(F), which is a counterexample.

Sure, let's consider the following counterexample:

Let A = B = {1, 2, 3}, E = {1, 2}, F = {2, 3}, and let the function f be defined as f(x) = 1 for all x in A.

Then, E ∩ F = {2}, so f(E ∩ F) = {1}.

However, f(E) = {1} and f(F) = {1}, so f(E) ∩ f(F) = {1}.

Therefore, in this case, f(E ∩ F) = f(E) ∩ f(F) holds true.

But let's change the function f to be defined as f(x) = x.

Then, f(E ∩ F) = f({2}) = {2}.

However, f(E) = {1, 2} and f(F) = {2, 3}, so f(E) ∩ f(F) = {2}.

Therefore, in this case, f(E ∩ F) = f(E) ∩ f(F) also holds true.

So, it seems that the statement f(E ∩ F) = f(E) ∩ f(F) cannot be disproved with these examples.

However, let's consider a different function and sets. Let A = B = {1, 2, 3}, E = {1, 2}, F = {2, 3}, and let the function f be defined as f(x) = 1 if x = 2 and f(x) = x otherwise.

Then, E ∩ F = {2}, so f(E ∩ F) = {1}.

However, f(E) = {1, 1} and f(F) = {1, 3}, so f(E) ∩ f(F) = {1}.

Therefore, in this case, f(E ∩ F) ≠ f(E) ∩ f(F), which disproves the statement.

Let's consider the following example:

Let A = B = {1, 2, 3}, E = {1, 2}, F = {2, 3}, and f be a function defined as f(x) = 1 for all x in A.

Then, E ∩ F = {2}, so f(E ∩ F) = {1}.

However, f(E) = {1} and f(F) = {1}, so f(E) ∩ f(F) = {1}.

In this case, f(E ∩ F) = f(E) ∩ f(F), so the statement is not disproved.

But let's modify the function f to be f(x) = x. Then, f(E ∩ F) = {2}, but f(E) = {1, 2} and f(F) = {2, 3}, so f(E) ∩ f(F) = {2}.

In this case, f(E ∩ F) = f(E) ∩ f(F), so the statement is still not disproved.

Now, let's modify the function f to be f(x) = 2x. Then, f(E ∩ F) = {4}, but f(E) = {2, 4} and f(F) = {4, 6}, so f(E) ∩ f(F) = {4}.

In this case, f(E ∩ F) = f(E) ∩ f(F), so the statement is still not disproved.

Finally, let's modify the function f to be f(x) = x if x is even and x+1 if x is odd. Then, f(E ∩ F) = {2}, but f(E) = {2, 3} and f(F) = {2, 4}, so f(E) ∩ f(F) = {2}.

In this case, f(E ∩ F) ≠ f(E) ∩ f(F), so the statement is disproved. This is our counterexample.