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## Main Question or Discussion Point

Alright, I would like to pose a question on the validity of this argument, I seem to be caught on whether or not to believe it.

The proposition is, if, for any sets A and B, [itex]f(A \cap B) = f(A) \cap f(B)[/itex], then [itex]f[/itex] is injective.

Proposed proof:

Let's look at the contrapositive, that is if [itex]f[/itex] is not injective, then [itex]f(A \cap B) \neq f(A) \cap f(B)[/itex]. Since we already know that [itex]f(A \cap B) \subseteq f(A) \cap f(B)[/itex] by some earlier exercise. Let us assume for the sake of contradiction that [itex]f(A) \cap f(B) \subseteq f(A \cap B)[/itex]. Now consider, A = {1, 2, 3}, B = {1, 2} and f = {(1,1), (2,2), (3,3)}, so that [itex]f(A) \cap f(B) = \{1,2,3\} \not\subseteq \{1,2\} = f(A \cap B)[/itex]. Hence [itex]f(A \cap B)[/itex] cannot possibly be equal to [itex]f(A) \cap f(B)[/itex]

The proposition is, if, for any sets A and B, [itex]f(A \cap B) = f(A) \cap f(B)[/itex], then [itex]f[/itex] is injective.

Proposed proof:

Let's look at the contrapositive, that is if [itex]f[/itex] is not injective, then [itex]f(A \cap B) \neq f(A) \cap f(B)[/itex]. Since we already know that [itex]f(A \cap B) \subseteq f(A) \cap f(B)[/itex] by some earlier exercise. Let us assume for the sake of contradiction that [itex]f(A) \cap f(B) \subseteq f(A \cap B)[/itex]. Now consider, A = {1, 2, 3}, B = {1, 2} and f = {(1,1), (2,2), (3,3)}, so that [itex]f(A) \cap f(B) = \{1,2,3\} \not\subseteq \{1,2\} = f(A \cap B)[/itex]. Hence [itex]f(A \cap B)[/itex] cannot possibly be equal to [itex]f(A) \cap f(B)[/itex]