Problem 8 (10 points). Let f : X ? Y be a function. If V-X, we define the image
ID: 3907005 • Letter: P
Question
Problem 8 (10 points). Let f : X ? Y be a function. If V-X, we define the image of V under f as f(v) = {f(z) | zen. Prove that f is injective if and only if f(A- B)-f(A)f(B) or all A, B CX Hints: When showing f is injective implies f(A B) f(A) f(B), show that y Ef(A B) implies y E f(A) and conclude that f(A - B) C f(A) f(B). Then use the fact that f is injective to show that f(A) f(B) Sf(A - B). To show that f(A - B) f(A) - f (B) implies f is injective, let z1,r2 E X be arbitrary and assume that A-i) and B-2 (for simplicity), which implies that f(A)-If(xi)) and f(B)-{f(??)). Conclude something about f(A-B) and show that this implies that zi-Z2Explanation / Answer
We need to prove if f : X ? Y is an injective function, then f(A?B) = f(A) - f(B) for all subsets A, B ? X, and we need to prove that if f(A - B) ? f(A) - f(B) for all subsets A, B ? X, then f is injective.
We first prove the first implication. Assume f : X ? Y is injective. Let y ? f(A-B). Thus y ? f(A), so y = f(a) for some a ? A, and y ? f(B), so y = f(b) for some b ? B.
But f is injective, so f(a) = f(b) implies a = b, hence a = b ? A-B, so y = f(a) = f(b) ? f(A - B).
Thus f(A) - f(B) ? f(A - B).
Now we’ll prove the second implication. Assume f(A) - f(B) = f(A - B) for all subsets A, B ? X.
Let v1, v2 ? X. Let A = {v1} and B = {v2}.
If f(A) is equal to f(B), then f(A) - f(B) = ?, so f(A - B) = f(A) - f(B) = f(?) = ?; i.e., f(A-B) = ?, so x1 is equal x2, so f is injective.
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