Let M be a set and let X, Y, Z, W ⊂ M. We define the symmetric difference: X △ Y := (X − Y ) ∪ (Y − X) (i) (1 pt) Show that X △ Y = (X ∪ Y ) − (X ∩ Y ). (ii) (1 pt) Show that (M − X) △ (M − Y ) = X △ Y . (iii) (1 pt) Show that the symmetric difference is associative, i.e., (X △ Y ) △ Z = X △ (Y △ Z). (iv) (1 pt) Show that X ∩ (Y △ Z) = (X ∩ Y ) △ (X ∩ Z). (v) (1 pt) Show that X △ Y = Z △ W iff X △ Z = Y △ W. (vi) (1 pt) Indicate the region of X △ Y △ Z in a Venn diagram. (vii) (1 pt) Sketch a Venn diagram for 4 distinct sets.
时间: 2023-03-19 08:22:09 浏览: 113
(i) To show that X △ Y = (X ∪ Y ) − (X ∩ Y ), we need to prove two inclusions:
First, suppose that x ∈ X △ Y. Then either x ∈ X but x ∉ Y or x ∈ Y but x ∉ X. In either case, we have x ∈ X ∪ Y but x ∉ X ∩ Y, so x ∈ (X ∪ Y ) − (X ∩ Y ). Therefore, X △ Y ⊆ (X ∪ Y ) − (X ∩ Y ).
Conversely, suppose that x ∈ (X ∪ Y ) − (X ∩ Y ). Then x ∈ X ∪ Y and x ∉ X ∩ Y. If x ∈ X, then x ∉ Y, so x ∈ X △ Y. Otherwise, x ∈ Y but x ∉ X, so x ∈ Y − X, and therefore x ∈ X △ Y. Therefore, (X ∪ Y ) − (X ∩ Y ) ⊆ X △ Y.
Combining both inclusions, we get X △ Y = (X ∪ Y ) − (X ∩ Y ).
(ii) To show that (M − X) △ (M − Y ) = X △ Y, we can use the result from part (i) twice:
(M − X) △ (M − Y ) = ((M − X) ∪ (M − Y )) − ((M − X) ∩ (M − Y )) = ((M ∩ Y ) − X) ∪ ((M ∩ X) − Y )
= (X ∪ Y ) − (X ∩ Y ) = X △ Y.
(iii) To show that the symmetric difference is associative, we need to prove that (X △ Y ) △ Z = X △ (Y △ Z).
Using the definition of symmetric difference, we have
(X △ Y ) △ Z = ((X − Y ) ∪ (Y − X)) △ Z = (((X − Y ) ∪ (Y − X)) − Z) ∪ (Z − ((X − Y ) ∪ (Y − X)))
= ((X − Y ) ∩ (Z − Y )) ∪ ((Y − X) ∩ (Z − X)),
and
X △ (Y △ Z) = X △ ((Y − Z) ∪ (Z − Y)) = ((X − (Y − Z)) ∪ ((Y − Z) − X)) ∪ ((Y ∩ Z) − (X ∪ (Y − Z)))
= ((X − Y ) ∪ ((Y − Z) ∩ (X − Z))) ∪ (((Y ∩ Z) ∩ X) − (Y − Z)).
To show that these sets are equal, it suffices to show that each is a subset of the other.
Suppose that x ∈ (X △ Y ) △ Z. Then either x ∈ (X − Y ) ∩ (Z − Y ) or x ∈ (Y − X) ∩ (Z − X). In the first case, we have x ∈ X − Y and x ∈ Z − Y, so x ∈ X − Y ⊆ X. Since x ∈ Z − Y, we have x ∉ Y, so x ∈ X △ Y. Similarly, in the second case, we have x ∈ X − Y, x ∉ Z, and x ∈ Y △ Z, so again x ∈ X △ (Y △ Z).
Conversely, suppose