Addendum: When I said f(A∩B) ⊆ f(A)∩f(B), I forgot to justify this statement. Obviously, if x ∈ A ∩ B, x ∈ A and x ∈ B is true. Since f is a function, you don't need to worry about x having more than one corresponding value. We know already that f(x) ∈ f(A ∩ B). And if this is true, it must follow that f(x) ∈ f(A) and f(x) ∈ f(B) is true. Therefore, f(A∩B) ⊆ f(A)∩f(B). Note that the converse statement, f(A)∩f(B) ⊆ f(A∩B), is not always true. It only holds true if the function is injective.
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Doing god's work mah dude. ❤
Addendum: When I said f(A∩B) ⊆ f(A)∩f(B), I forgot to justify this statement. Obviously, if x ∈ A ∩ B, x ∈ A and x ∈ B is true. Since f is a function, you don't need to worry about x having more than one corresponding value. We know already that f(x) ∈ f(A ∩ B). And if this is true, it must follow that f(x) ∈ f(A) and f(x) ∈ f(B) is true. Therefore, f(A∩B) ⊆ f(A)∩f(B).
Note that the converse statement, f(A)∩f(B) ⊆ f(A∩B), is not always true. It only holds true if the function is injective.
good luck kyle :>