WEBVTT 00:00:00.250 --> 00:00:04.254 Last time, we used four cards to explore what observations can tell 00:00:04.265 --> 00:00:04.482 us. 00:00:05.086 --> 00:00:08.298 A had a dot, B had both features, C had a stripe, 00:00:08.298 --> 00:00:09.455 and D had neither. 00:00:10.409 --> 00:00:13.914 Now we need a precise language for groups of these states. 00:00:14.559 --> 00:00:16.252 That language is sets. 00:00:17.436 --> 00:00:21.333 Our universe is the collection of all the states we are considering. 00:00:21.942 --> 00:00:25.177 For now, it contains A, B, C, and D. 00:00:25.966 --> 00:00:27.526 We will call it omega. 00:00:28.498 --> 00:00:30.484 Today we are describing groups. 00:00:31.149 --> 00:00:34.448 Drawing a group does not mean we have learned which card is 00:00:34.458 --> 00:00:34.915 hidden. 00:00:35.895 --> 00:00:38.847 A set is a collection of distinct elements. 00:00:39.506 --> 00:00:42.024 Here, the elements are card states. 00:00:43.193 --> 00:00:45.808 The dot set contains A and B. 00:00:46.487 --> 00:00:51.305 We write their names inside curly braces: the set containing A and 00:00:51.315 --> 00:00:51.512 B. 00:00:52.766 --> 00:00:54.450 A belongs to this set. 00:00:54.963 --> 00:00:56.103 C does not. 00:00:57.077 --> 00:01:00.016 The membership symbol is read “belongs to.” 00:01:00.558 --> 00:01:02.813 It relates an element to a set. 00:01:03.847 --> 00:01:09.005 Order does not matter: listing B before A describes the same set. 00:01:09.545 --> 00:01:12.682 Repeating A would not add another element. 00:01:13.149 --> 00:01:14.983 Membership is what matters. 00:01:16.123 --> 00:01:19.410 Here is a distinction that will matter throughout the paper. 00:01:20.520 --> 00:01:22.358 A names one state. 00:01:22.824 --> 00:01:27.653 Braces around A name a set containing that state. 00:01:28.362 --> 00:01:31.812 The set has one element, so we call it a singleton. 00:01:32.939 --> 00:01:37.019 Think of the difference between an item and a group containing that 00:01:37.040 --> 00:01:37.474 item. 00:01:38.079 --> 00:01:42.647 Even when the group has only one member, the two descriptions play 00:01:42.668 --> 00:01:43.666 different roles. 00:01:44.828 --> 00:01:48.713 Membership asks whether an element belongs to a set. 00:01:49.285 --> 00:01:54.650 Subset asks whether every element of one set also belongs to another. 00:01:55.677 --> 00:02:00.603 The singleton containing A is a subset of the dot set: its 00:02:00.635 --> 00:02:03.493 only element is in that larger set. 00:02:04.338 --> 00:02:08.103 But the dot set is not a subset of the singleton. 00:02:08.775 --> 00:02:10.537 B would be left out. 00:02:11.414 --> 00:02:14.078 Subset can also include equality. 00:02:14.631 --> 00:02:17.083 Every set is a subset of itself. 00:02:17.717 --> 00:02:21.526 Two sets are equal when they have exactly the same elements. 00:02:22.655 --> 00:02:24.847 The dot set contains A and B. 00:02:25.521 --> 00:02:27.973 The stripe set contains B and C. 00:02:29.197 --> 00:02:31.509 Which states belong to both sets? 00:02:41.823 --> 00:02:42.453 Only B. 00:02:43.143 --> 00:02:44.727 It has both features. 00:02:45.846 --> 00:02:49.568 The intersection of two sets contains their shared elements. 00:02:50.217 --> 00:02:53.732 So the intersection here is the singleton containing B. 00:02:54.926 --> 00:02:56.948 We are calculating a group. 00:02:57.186 --> 00:03:00.540 If we later learn that the hidden card has both a dot 00:03:00.550 --> 00:03:05.020 and a stripe, this calculation tells us that B is the only 00:03:05.053 --> 00:03:06.733 remaining possibility. 00:03:07.596 --> 00:03:09.392 Now ask a different question. 00:03:09.780 --> 00:03:13.198 Which states belong to at least one of these sets? 00:03:23.415 --> 00:03:25.107 A, B, and C. 00:03:26.267 --> 00:03:31.984 Their union contains every element belonging to either set, including elements belonging 00:03:31.984 --> 00:03:32.732 to both. 00:03:33.595 --> 00:03:35.695 B appears just once. 00:03:36.017 --> 00:03:39.349 We do not count it twice because it occurs in both starting 00:03:39.359 --> 00:03:39.848 sets. 00:03:40.972 --> 00:03:44.478 D belongs to neither, so it is outside this union. 00:03:45.582 --> 00:03:48.035 Intersection keeps the shared members. 00:03:48.692 --> 00:03:51.601 Union gathers the members from either set. 00:03:52.493 --> 00:03:55.119 What if two sets share no elements? 00:03:55.784 --> 00:04:00.450 The singleton containing A and the singleton containing C have an empty 00:04:00.450 --> 00:04:01.490 intersection. 00:04:02.353 --> 00:04:03.783 We call them disjoint. 00:04:04.309 --> 00:04:07.320 The empty set contains no elements at all. 00:04:08.230 --> 00:04:09.619 Be careful with card D. 00:04:10.443 --> 00:04:13.854 It has neither feature, but it is still a state. 00:04:14.309 --> 00:04:17.239 A set containing D has one element. 00:04:17.239 --> 00:04:19.166 It is not the empty set. 00:04:20.118 --> 00:04:22.798 Return to our universe of four states. 00:04:23.408 --> 00:04:25.763 The dot set contains A and B. 00:04:26.962 --> 00:04:32.138 Its complement contains everything in the universe that is outside the dot 00:04:32.149 --> 00:04:32.547 set. 00:04:33.063 --> 00:04:35.038 Here, that is C and D. 00:04:36.257 --> 00:04:40.858 You can read the displayed expression as “omega with the dot set 00:04:40.912 --> 00:04:41.633 removed.” 00:04:42.194 --> 00:04:47.185 The backslash means set difference: keep elements of the first set that 00:04:47.185 --> 00:04:48.625 are not in the second. 00:04:49.631 --> 00:04:52.884 We can describe the no-dot region mathematically. 00:04:53.434 --> 00:04:59.558 That does not automatically give our observer a certificate establishing no dot. 00:05:00.638 --> 00:05:04.206 For a separate example, suppose we have learned that the card has 00:05:04.206 --> 00:05:05.074 a stripe. 00:05:05.979 --> 00:05:08.594 Our working universe is now B and C. 00:05:09.244 --> 00:05:14.307 Within that universe, the dot set contains only B, and its complement 00:05:14.329 --> 00:05:15.588 contains only C. 00:05:16.772 --> 00:05:19.937 D does not reappear when we take the complement. 00:05:20.323 --> 00:05:25.198 We are considering only states inside the specified universe. 00:05:25.558 --> 00:05:31.569 Whenever someone says “everything outside this set,” ask: outside it, but inside 00:05:31.569 --> 00:05:32.679 which universe? 00:05:33.613 --> 00:05:36.803 Return to the original four-state universe. 00:05:37.393 --> 00:05:40.025 We can also collect sets themselves. 00:05:40.474 --> 00:05:44.260 Put the dot set and the stripe set into a collection. 00:05:44.850 --> 00:05:49.667 This collection has two members: one set containing A and B, and 00:05:49.667 --> 00:05:51.425 another containing B and C. 00:05:52.109 --> 00:05:54.432 Its members are those two sets. 00:05:55.046 --> 00:05:57.999 They are not the individual states A, B and C. 00:05:59.213 --> 00:06:03.586 Taking their union would instead give one set containing three states. 00:06:04.705 --> 00:06:08.608 Keep track of the level: a state, a set of states, or 00:06:08.618 --> 00:06:10.311 a collection of such sets. 00:06:10.916 --> 00:06:15.300 Later, a topology will be a particular collection of subsets of a 00:06:15.332 --> 00:06:16.348 universe. 00:06:17.116 --> 00:06:21.154 Try two different groups: A and C in one set, C and 00:06:21.154 --> 00:06:22.081 D in the other. 00:06:23.046 --> 00:06:24.554 What is their intersection? 00:06:24.576 --> 00:06:25.955 And what is their union? 00:06:40.008 --> 00:06:42.265 The intersection contains only C. 00:06:43.010 --> 00:06:45.255 The union contains A, C, and D. 00:06:46.484 --> 00:06:50.697 We now have a language for grouping states and comparing those groups. 00:06:51.831 --> 00:06:56.172 Next, we will learn how logical statements are evaluated, using the same 00:06:56.172 --> 00:06:58.322 four cards and a truth table.