 Welcome to the screencast about identities about sets. What we're about to do in this video is establish some ground rules and some basic properties about sets and the way that we operate sets together that will help us make future computations and proofs about sets a lot easier. When we talk about identities, one of the things that might bring to mind are things like trigonometric identities, like the old trig identity. Sine squared x plus cosine squared x equals one, for example. That's an identity that you learned in trigonometry. And what that allowed you to do is make certain things simpler. And there are a whole host of trigonometric identities that are all aimed at making life simpler for us. Sets have identities too. And what I'm about to show you here is a couple of slides worth of identities. These are all theorems that we can prove and some of them we will prove. But for the most part, what we will do is take them as being true and work with them. Here's one set of set identities here. Some of these have to do with the empty set and some of them have to do with the universal set. This one, for example, says that A intersects to the empty set as always the empty set. And A union the empty set as always itself. This one says that A intersects to A is equal to itself. This one says A union A is equal to itself. This one says that I can intersect sets in any order I wish or union sets in any order I wish. The associative laws here say that I can group intersections and unions arbitrarily, as long as they're all intersections or all unions, the distributive properties here, distributive laws, say, give me some rules for how to distribute an intersection across a union and we get something like this. Now this last one, the distributive law might remind you of a logical equivalence law that was also called the distributive law where we're distributing a conjunction over a disjunction. And they're very similar and logically related to each other. Here's another set of identities that we're also going to use that are a little bit more off the beaten path. This one here for example says that the complement of A complement is A. The A minus B is A intersect B complement. This one is so useful that we're going to prove it in just a minute using some things we learned from the last section. And some other things too. I just want to also point your attention to De Morgan's laws. That's another logical equivalence law that we saw that had to do with negating a disjunction or negating a conjunction. This, in this context, it has to do with taking the complement of an intersection or the complement of a union. The complement of an intersection is the union of the complements and vice versa. So the way that we establish these theorems, they're not axioms. We don't just accept them without proof. We prove each one of them using the choose an element method. So what I'd like to do now is focus in on this basic property here that A minus B is equal to A intersect B complement and prove it. So I'll set it up as a proposition. Let's let A and B be sets in some universal set U. And then A minus B is equal to A intersect B complement. And I just want to point out that you can prove, you can set up a similar proof for every single one of the set identities that you just saw on those two slides. Although there's a lot of them, they're not just handed down as axioms. They are provable and some of the exercises you should go through and prove them. So let's set up a proof of this. And how are we going to proceed? Well, this is a proposition about two sets being equal to each other. So as we saw in the last few videos, that's going to involve two steps. First of all, I'm going to need to prove that A minus B is a subset of A intersect B complement. And then I need to turn around and prove the other direction. I'll have to prove that A intersect B complement is a subset of A minus B. Remember that when we're proving that two sets are equal to each other, that means I have to do two things. I have to do this double subset inclusion type argument here. So let's get to it. So to prove that A intersect A minus B is equal to A intersect B complement, let's first prove, get my pencil back, let's first prove that show that A minus B is a subset of A intersect B complement. Okay? Now to show that one subset is a subset of another, we're going to use the choose an element idea. So the first thing we're going to do is let X be an element of A minus B. And what we want to show, and I'll put this in red because I don't know it yet, I want to show this, I want to show that X belongs to A intersect B complement. I don't know that it is yet, I only know that it belongs to A minus B, but that's where we're headed with all this. So let's work it out. So if X belongs to A minus B, what does that mean? Well, X belonging to A minus B means that X belongs to A and X does not belong to B. That's just by the definition of set difference. The set difference of A and B is the set of all things that do belong to A but do not belong to B. I've subtracted out all the B stuff. And so what does that tell me? Well, if X belongs to A and X does not belong to B, in particular if X does not belong to B, what that means is that X belongs to its complement. Okay, there's a key point there. If X does not belong to B, then what it does belong to is the complement of B. Now let's just see what we have. I have X belonging to A and belonging to B complement. So what that tells me is that therefore X belongs to the intersection of A and B complement. And that is what we were trying to show back up in the first line. I wanted to show that X belongs to that intersection. So therefore I've proven that A minus B is a subset of A intersect B complement. Now I've got to flip this around and go the other way. So let's set up a new pager. I'm going to go the other way. What I want to do is show that A intersect B complement is a subset of A minus B. This will look really similar to the forward direction of this proof. So to do this, let's choose an element. We're going to let, let's say Y be an element of A intersect B complement. And again, let me just write down what I want to show here. Do it in red. I want to show that Y belongs to A minus B. Very good. And switch back to the right color. Now if Y belongs to A intersect B complement, what that tells me is that Y belongs to A and Y belongs to B complement. Just definition of intersection. Now if Y belongs to A, then I'm just going to write this down. Y belongs to A. And in particular if Y belongs to B complement, that means that Y is not an element of B. And so what this means is that therefore, I'll just say then again Y belongs to A minus B because that's the definition of set difference. That's what I wanted to show here. And so I'm done. And now I've shown that A intersect B complement is a subset of A minus B. And so therefore the entire theorem is proven. I've shown that one set is included in the other and vice versa. So therefore A intersect B complement is actually equal to A minus B. So this is a fact that we can take to the bank. Anytime I need to, it's a very handy way of rewriting differences. Okay, anytime I see A minus B or X minus Y or anything of that nature in set theory, I can rewrite that as an intersection involving a complement. And that becomes very, very useful. We'll see this show up in later screencasts. Before we leave, I just want to prove one extra identity that I kind of feel ought to have been on this list, but isn't. And it's really, really simple, but I want to establish it now because we're going to need it later. It says that if A is any set that lives inside some universal set U, then A intersect its complement is empty. So this is saying that A and its complement are disjoint, in other words. So let's think about how we would prove that. Well, oftentimes we've, you know, we've mentioned before that many times we've proved that two sets are disjoint by contradiction. So let's try to prove by contradiction to show that A and its complement are disjoint. So we're going to show A intersect A complement is empty. So let's suppose for a contradiction that there is something in that set. So for a contradiction, that's let X be in A and A complement at the same time. So suppose that the intersection is not empty, but rather there's something in it, and let's call it X. Then what that means is that X is in A and X is in A complement because that's the definition of intersection. But what does it mean for X to be in the complement? I already know that X belongs to A. If X is in A complement, that means X does not belong to A. And here's where our contradiction comes from, pretty obviously. I have X in A and not in A at the same time, and that's a contradiction. So therefore, that X can't exist. Therefore, the intersection of A and A complement must actually be empty in the first place. So there's a couple of quick proofs of set identities. And again, you can go back to every single one of those set identities and prove them all. And it's a good chance to try your hand at some of those things. Thanks for watching.