Tugas PDM 4

1. Show that A ∩ B = B ∩ A !

x ∈ A ∩ B

x ∈ A ∧ x ∈ B

x ∈ B ∩ A

so, A ∩ B ⊂ B ∩ A (komutatif) (i)

(ii)

x ∈ B ∩ A

x ∈ B ∧ x ∈ A

x ∈ A ∩ B

so, B ∩ A ⊂ A ∩ B (komutatif) (ii)

from (i) and (ii) we conclude that A ∩ B = B ∩ A

2. Show that (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C) !

x ∈ (A ∩ B) ∩ C

(x ∈ A ∧ x ∈ B) ∧ x ∈ C

x ∈ A ∧ (x ∈ B ∧ x ∈ C)

x ∈ A ∩ (B ∩ C)

so, (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C) (Assosiatif)

(ii)

x ∈ A ∩ (B ∩ C)

x &isin ; A ∧ (x ∈ B ∧ x ∈ C)

( x &isin ; A ∧ x ∈ B) ∧ x ∈ C

x ∈ (A ∩ B) ∩ C

So, A ∩ (B ∩ C) ⊂ (A ∩ B) ∩ C (Assosiatif)

From (i) and (ii) we conclude that (A ∩ B) ∩ C = A ∩ (B ∩ C ∩)

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