Prove that the following statements are equivalent for sets...

Prove that the following statements are equivalent for sets A, B ? U. For some parts an element chasing argument is easier than using the algebra of sets. (i) A ? B = ?. (ii) A ? B^c. (iii) B ? A^c. (iv) A^c ? B^c = U. Prove the following statements for A, B, C contained in a universe U. (a) If C ? A ? B and C ? A = ?, then C ? B. (b) If A ? B = A ? C and A^c ? B = A^c ? C, then B = C. (c) If C ? A, then A ? (B ? C) = (A ? B) ? C. (d) If A ? B = ? and A ? B = C, then A = C ? B. (e) A ? (B ? C) = (A ? B) ? C if and only if A ? C.

Transcript

00:01 So for any sets a and b, we'd like to provide an element argument that a intersect b union a intersect b complement is equal to a.

00:09 So let's do that real quick.

00:11 We'll let a in a.

00:15 First, we're going to show that a is a subset of this guy.

00:20 A intersect b union a intersect b complement.

00:28 Now we can do casework here.

00:30 If a is in b as well, then a is going to be in a intersect b.

00:38 By definition, the union here, that means that a is in a union b.

00:47 Union usually just gets an ordinary u.

00:49 A union b, or a intersect b, excuse me.

00:54 Union, whatever you want.

00:56 Which in particular means that it's in, a is in a union b, a intersect b union a intersect b complement.

01:11 Great, so to recap there, a is an arbitrary element of a.

01:14 If it's in b, then it's in this whole thing because it's in the first part of the union.

01:21 And conversely, if a is not in b, then it's in b complement.

01:27 So it's in a intersect b complement because it is in both a and b complement.

01:32 Likewise, if it was in b, then it's in both a and b, so it's in the intersection.

01:37 Which means that it is in a intersect b union a intersect b complement.

01:44 Because in particular, it's in the second part of the union.

01:48 In either case, we conclude that a is in, in either case, we conclude that this arbitrary element of a is in this union expression.

01:58 Which means that a is a subset of a cap b cap a cap b complement.

02:08 Conversely, let's suppose that a is in the complicated expression, a intersect b union a intersect b complement.

02:20 Then that means that either it is in the first intersection, a intersect b, or it's in a intersect b complement.

02:35 That's what this union means.

02:36 If it's in the union of two sets, then it has to be in one of the two sets.

02:41 If it is in the first intersection, then it is in both, a is in a and it's in b.

02:52 In particular, it's in a.

02:54 So we conclude that it's in a.

02:57 If a is instead in a intersect b complement, then a is in a, and let's write this out formally, a is in b complement.

03:07 It is in a and it's in b.

03:11 Here a is in a and it's in b complement.

03:13 In particular, it's in a, which means that this thing is in fact a subset of a.

03:19 These two sets are subsets of each other, therefore they are equal to each other.

03:25 We'll try this again here.

03:29 We want to show that a intersect b complement is the union of the complements.

03:36 I'm going to put this up here so i have some empty space.

03:40 Let's let a be in the complement of the intersection.

03:48 A intersect b complement.

03:51 That implies that a is not in the intersection of a and b, which in particular means that a is not in a or a is not in b.

04:05 If it's not in a, then it's not in the intersection.

04:08 If it's not in b, then it's not in the intersection.

04:10 It might be in neither, but in particular we only need it to not be in one of them, by definition of the intersection.

04:16 In particular, we conclude that a is either in a complement or it's in b complement, which means that a is in a complement union b complement.

04:29 Thus, this guy is a subset of this guy.

04:32 We can actually follow these same steps in reverse, for the other direction.

04:37 If it is in the union of the complements, then it's in one of the complements.

04:41 If it's in one of the complements, then it's not contained in one of the sets originally...

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