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f: X → Y, A ⊆ X and B ⊆ X

X = {1, 2}, Y = {0}

A = {1}

B = {2}

Function f: X → Y takes values: f(a) = f(b) = 0.

f(A\B) = f(A) ={0}, but f(A)\f(B) = ∅.

  • I'm trying to make the containment fail by showing the f is not injective. By the definition of injectivity, this means that f(a1) = f(a2), then a1 ≠ a2 or also by contrapositive f(a1) ≠ f(a2), then a1 = a2.

Let me know what you think of the numerical calculation. Any alternative approaches are welcomed too!

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    This example shows that $f(A\setminus B) \subset f(A) \setminus f(B)$ fails. The claim $f(A) \setminus f(B) \subset f(A\setminus B) $ is always true. Check this link for details: https://www.quora.com/How-do-I-prove-this-logic-and-set-theoretic-question/answer/Amit-Goyal-135?srid=apAO – 2017-02-15
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    @Amit how is f(A) \ f(B) = ∅? how is f(A) \ f(B) = empty set? I understand the f(A\B) = f(A) since x is an element of A and x is not an element of B so we consider f(A\B) as only the x elements found in A, hence f(A\B) = f(A) and I think f(A) = f(1) = 0? – 2017-02-15

1 Answers 1

1

f: X → Y, A ⊆ X and B ⊆ X

X = {1, 2}, Y = {0}

A = {1}

B = {2}

Function f: X → Y takes values: f(a) = f(b) = 0.

f(A\B) = f(A) ={0}, but f(A)\f(B) = ∅.

Answer by Amit Goyal (https://math.stackexchange.com/users/378131/amit) in Quora (https://www.quora.com/How-do-I-prove-this-logic-and-set-theoretic-question/answer/Amit-Goyal-135). I don't take credit for this answer. All credits to Amit Goyal. The answer has been found.