Union and intersection, and what it means for two sets to miss each other entirely

0:00Add five and thirteen and you get eighteen. Multiply the same two and you get sixty five. Two numbers go in, one number comes out, and that is what an operation is. So here is the question. What does an operation on sets look like? Two sets go in, and one set comes out. There are two such operations, and between them they are most of what you can do to a pair of sets.

0:28The surprising part is not what they are. It is that neither of them has an idea of its own. Each one is a word from ordinary English, turned into a set. And once you see which word, every property either of them has stops being something to memorise. Here is the first one. Take two sets, and ask about each object in turn. Are you in the first one, or are you in the second one?

0:59Everything that answers yes goes into a new set, and that set is called the union. Now notice what the question does not ask. It does not ask how many of the two sets you are in. An object in both of them answers yes, and it is the same yes as an object in only one of them. So it goes in once. This is where people trip, and it is worth being exact about why.

1:26Membership is a verdict, not a tally. There is nowhere in the question, is this object in the set, for a count to live. Write a shared object twice and you have not got two members, you have written the same member down twice. Take the set holding two, four, six and eight. And the set holding six, eight, ten and twelve. Walk the objects and ask the or question of each one.

1:56Two, yes. Four, yes. Six answers yes from both sides at once, and eight does the same. Ten, yes. Twelve, yes. So the union holds two, four, six, eight, ten and twelve. That is six members. Four and four would have been eight. The two that are missing are exactly the two that were shared. And that is not a coincidence about this pair. Count the union, count the shared part, add those two numbers, and you always get the sizes of the original two sets added together.

2:32Every object in either set is counted once in the union, and counted a second time in the shared part only if it was in both. Now a case worth predicting before you compute it. The five vowels, a, e, i, o and u. And the set holding just a, i and u. Every member of the second one is already a member of the first. Ask the or question of each vowel in turn.

3:02a is in the first set, so yes. e is in the first set, so yes, and so on down the list. Nothing new ever arrives, because there was nothing in the second set that was not already in the first. The union is the five vowels, the larger set, unchanged. That is the general rule. When one set sits inside another, their union is the larger one, and you do not have to compute it - you can read it off the containment.

3:35Here is a union that means something. A hockey team of three, Ana, Ben and Chidi. A football team of three, Dana, Ana and Elias. Ana plays both. The union holds Ana, Ben, Chidi, Dana and Elias. Five students, not six. And now say aloud what that set actually is. It is the students who play at least one of the two sports. That is what or means here - at least one, and not exactly one.

4:07The shared part is coming back later, and it will mean something too. Union arrives with a list of properties, and a list like that invites memorising. Do not memorise it. Every one of them is a fact about the word or. Order does not matter, because p or q has the same answer as q or p, and there are only four cases to check. Grouping does not matter, for the same kind of reason.

4:38A set unioned with itself is itself, because asking p or p is just asking p. The empty set changes nothing, because p or false is p - a second chance that always fails is not a chance. And the surrounding set swallows everything, because p or true is true, whatever p was. Five properties, one word. If you ever forget one of them, ask the question about two answers instead of about two sets, and it falls out.

5:12Which brings up the word people reach for, and it is the wrong word. A union is not an addition. Add four to four and you get eight. Union a four member set with itself and you get the same four members back. Arithmetic has no law saying that a plus a is a. Union has exactly that law, and it has it because or, repeated, is still or. Adding counts things.

5:40Union asks a question about them. Those are different operations, they behave differently, and the six against eight we just did is where the difference shows. Now the second operation, and it is the same idea with one word changed. Ask each object, are you in the first set and in the second? Everything that answers yes goes into a new set, called the intersection. Drawn, it is the part where the two curves overlap, and nothing else.

6:14Take the same pair as before. Two is in the first and not in the second, so no. Six is in both, so yes, and eight is in both, so yes. Ten and twelve are in the second only, so they fail the second half of the question. The intersection holds six and eight. Two members, out of a four and a four. Here is the mirror of the containment case.

6:41Take the whole numbers from one to ten. And the set holding two, three, five and seven. Those four are exactly the primes up to ten. Every one of them is already among the first ten numbers. So the and question, asked of each of them, comes back yes on both halves. And asked of any other number in the ten, it fails on the second half. The intersection is the smaller set itself.

7:09So when one set sits inside another, the union is the larger one and the intersection is the smaller one. Two predictions, both read off the containment, and neither of them computed. Now the case that a whole word is reserved for. Take the set holding two, four, six and eight, and the set holding one, three, five and seven. Ask the and question of anything at all. Nothing passes.

7:40Their intersection is empty, and two sets like that are called disjoint. Drawn, it is two curves that do not touch. But notice what the drawing tells you, and what it does not. It shows you that nothing overlaps in this picture. The reason is better than the picture. Every member of the first set is even, and every member of the second is odd. And no whole number is both, because halving it leaves a remainder of nought or of one, never both at once.

8:13So it is not that nothing happens to be shared here. Nothing could be. Three things get believed about disjoint sets, and all three are wrong. The first is that disjoint means unequal. It does not. The two sets from the union example are unequal, and they share two members. Unequal is easy, and sharing nothing at all is a far stronger thing to say. The second is that an empty intersection means one of the sets was empty.

8:48The evens and the odds we just used have four members each. Neither of them is empty, and their intersection is. The third is that the intersection is always the smaller of the two sets. It can be. It can also be empty, and it can be one of them exactly, as it was with the primes sitting inside the ten. The intersection is the set of shared members, and how big that is depends entirely on the two sets.

9:20Intersection has its own list of properties, and they come from the word and, exactly as union's came from the word or. Order does not matter. Grouping does not matter. A set met with itself is itself. And then something worth stopping on. For union, the harmless set was the empty one. For intersection, the harmless set is the surrounding one. They swap, and the reason is sitting in the two words.

9:51p or false leaves p alone, because a chance that never fires cannot help. p and true leaves p alone, because a demand that is always met cannot hurt. False is harmless to or, and true is harmless to and. Turn those two sentences into sets and you have both of the identity laws, and you will not mix them up again. The other two swap as well - the surrounding set swallows a union, and the empty set swallows an intersection.

10:25One law uses both operations, and it is worth checking rather than believing. Take three sets. Meet the first with the union of the other two. Now do it the other way round. Meet the first with the second, meet the first with the third, and union those two results. The claim is that you land on the same set. You can watch it happen in five panels. Tint the second or the third.

10:54Then tint the part of the first that meets it, and hold on to that shape. Start again. Tint the first with the second, then the first with the third, then both of those together. The last panel carries the same shading as the one we held on to. And the reason is the connective again - p and, q or r, has the same answer as p and q, or p and r, on all eight rows.

11:24Once you have both operations you can write expressions, and reading one is a skill of its own. Take four sets. Three, five, seven, nine and eleven. Seven, nine, eleven and thirteen. Eleven, thirteen and fifteen. And fifteen and seventeen. Meet the first with the union of the second and the third, and you get seven, nine and eleven. Meet the first with the union of the second and the fourth, and you get seven, nine and eleven again.

11:58The fourth set contributed nothing, because it shares nothing with the first. Now change one thing. Union the first with the fourth, and meet that result with the union of the second and the third. Now you get seven, nine, eleven, and fifteen. Fifteen could never reach the answer while the fourth set was on the inside of the expression, and it can now that it is on the outside.

12:27Work from the inside out, ask the question the connective names, and expressions like that stop being a puzzle. The last case is the one where the sets never end. The natural numbers. The even ones, the odd ones, and the primes. Meet the naturals with any of the other three and you get that other one back, because all three sit inside it. Meet the evens with the odds and you get nothing, for the reason we already have.

12:59Meet the evens with the primes and you get exactly one number, and the number is two. Every other even number has two as a divisor, and a number with a divisor like that is not prime. So a whole endless family contributes a single member. Meet the odds with the primes and you get every prime except two. One operation, on families nobody can finish listing, settled by asking the and question about the conditions instead of about the members.

13:34So here is what these two operations are. Union is the word or, made into a set. Intersection is the word and, made into a set. Everything else followed from that. A shared object gets written once, because membership is a verdict with no count inside it. Order and grouping do not matter, because they do not matter to the connectives. Each operation is unchanged by repetition, for the same reason.

14:06The empty set is harmless to one and the surrounding set is harmless to the other, because false is harmless to or and true is harmless to and. And disjoint is not a fact about size, it is a verdict - the and question can never pass. Neither operation had an idea of its own to teach you. They borrowed two, from a language you already speak.