Exercises 0.2 #4

I) t ∈ α(A ∪ B) ⇒ there is an s ∈ A ∪ B such that α(s) = t ⇒ there is an s ∈ A or s ∈ B such that α(s) = t ⇒ t ∈ α(A) or t ∈ α(B). Thus α(A ∪ B) ⊂ α(A) ∪ α(B). t ∈ α(A) ∪ α(B) ⇒ there is an s ∈ A or s ∈ B such that α(s) = t ⇒ there is an s ∈ A ∪ B such that α(s) = t ⇒ t ∈ α(A ∪ B). Thus α(A) ∪ α(B) ⊂ α(A ∪ B).

II) t ∈ α(A ∩ B) ⇒ there is an s ∈ A ∩ B such that α(s) = t ⇒ there is an s ∈ A and s ∈ B such that α(s) = t ⇒ t ∈ α(A) and t ∈ α(B). Thus α(A ∩ B) ⊂ α(A) ∩ α(B).

III) α = { (s1, t1), (s2, t0), (s3, t0) }. A ⊂ S = { s1, s2 }. B ⊂ S = { s1, s3 }. α(A ∩ B) = { t1 }. α(A) ∩ α(B) = { t1, t0 }.

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