 This video will be our last video for lecture one about our introduction to sets. Now, in terms of Judson's textbook, we're gonna be in section 1.2 for a while, because we're gonna wanna talk about functions and equivalence relationships in much more detail as these are important objects for the study of abstract algebra. But in this concluding video for lecture one, I wanted to list and prove some important properties about sets. These, in the previous video, we learned about some set operations, unions, intersections, compliments, set differences. And so I wanna prove some properties associated to those operations with names so that you can refer to them in the future, right? So the first property is referred to as the idempotency property. Idempotency, that is to say, we say an element as idempotence. When you operate it on an element itself, you get back that original element. So for example, zero plus zero is zero. So zero is an idempotent element with respect to addition. If you take one times one, this is equal to one, or also zero times zero is equal to zero. In terms of multiplication, one and zero are both idempotent elements. When you operate an element with itself, you get back itself. Well, when it comes to unions and intersections, this is also true for sets. A union A is equal to A and A intersect A is equal to A. And this is the idempotency property. In fact, interesting enough, I had a student once who got really interested in this idea of idempotency and actually wrote a poem about it. If you're interested, you can see a link to that in the description. This was something that he wanted to do to kind of express himself and make the world a better place through poetry and abstract algebra. It was kind of fun. He presented it at a coffee shop, open mic night, I think. He actually published it in a literary journal. It was pretty cool. Again, take it out if you're into that type of stuff. Property two, we're gonna call the identity property. That is, there's identities with our operations. Like again, with integers, if I take x plus zero, this always equals to x. And if I take one times x, this always equals to x as well. We have these identity elements for the operations of addition and multiplication. It turns out unions and intersections also have these identities. If you take A union, the empty set, you always get back A. It doesn't change the set by adding the empty set to it. And then if you take A intersect, you use our universe here. You'll always get back A. Because if you want everything that's in A and in the universe, well, everything in A is inside the universe. So the intersection is unaffected when you put the union in there. So again, these are properties we see mimicked by the usual number systems that we're used to. This next one's a little bit different. The absorption property. If we take A union, the universe, that always gives back the universe. Or if we take A intersect, the empty set, we always get back the empty set. Or written in a slightly different way. If you take A intersect, the empty set, you always get back A. So the empty set's kind of like an identity for set differences on the right hand side. Not on the left hand side because it's not a commutative operation. Absorption, right? So if you operate by this dominant element union, the universe with respect to unions or the empty set with respect to intersections, you always get back that dominant element. And it turns out we actually do see things like this with number systems as well. For example, if you take zero times X, this always gives you back zero. It doesn't matter what that element X is. Or on the other hand, if you take like X plus infinity, the infinity in terms of addition has this absorption property. It's this dominant element that something plus infinity equals infinity. So it turns out even with number systems we'd see kind of this absorption property. In terms of complements, this one we mentioned in the previous video, but we'll just mention it again. A union, A complement is always equal to the universe and A intersect A complements equal to the empty set. And sort of a similar relationship. If you take A minus A, that's always gonna give you the empty set. So you get this complementary property. Commutivity, we have things like this. All you guys should mention with complements. This is kind of like additive inverses a little bit like X plus a negative X, right? This always equals zero. It's not exactly the same thing because in this with respect to unions, the universe is actually not an identity. It's the dominant element. So this is not the inverse property, but the complement property is similar to that, that given any element A, there exists some related element that gives you the dominant, not the identity. Again, that's a little bit different about set algebra. Set algebra is commutative, right? A union B is equal to B union A and A intersect B is equal to B intersect A. We're used to the commutivity property with addition and multiplication. Associativity, this has to do with redoing parentheses. A union B union C is the same thing as A union B, union C, right? You can redo the parentheses. And the same is also true for intersection. If you redo the parentheses, that is always gonna be true. So the associative property holds for set operations. You get the distributive properties. These are also some things we're kind of used to. So we see that unions intersect or they distribute over intersections. A union B intersect C is the same thing as A union B intersect A union C. So this is kind of like saying if you take like A times B plus C, this is the same thing as A B plus A C. Now, because of all our operations, we often don't write the parenthesis, but it should be something like this. If you take A times B plus C, this will equal A times B plus A times C. So you get that distributive property. So for numbers, like the real numbers, multiplication distributes over addition. Well, for sets, unions distribute over intersections, but it's also true that intersections distribute over unions. So this is kind of weird when you think of, to like the real numbers, addition does not distribute over multiplication, but for sets, intersections distribute over unions and unions distribute over intersections. It's quite phenomenal there. And then the last property, these are known as the De Morgan laws. This tells you that if you take the complement of A intersect B, this is actually the union of A complement and B complement. And if you take the complement of A union B, this gives you the intersection of A complement and B complement right there. So the De Morgan laws tell you how the complement interacts with these operations of intersections and unions. It'll toggle between them. The complement of a union is an intersection and the complement of intersection is a union. And we're gonna see why this happens for some of these properties here. We're not gonna prove all of them because many of them are left as exercises for you, the student in your homework here. But if you wanted to prove any of these identities, right, how do you prove that A union A is equal to A? That one's a little bit more elementary, but if you wanna see something a little more complicated, how do you prove it to Morgan law like this one? The complement of A intersect B is equal to the complement of A union complement B. What you wanna do is you would show that, you know, to show that two sets are actually equal, what you're gonna do is you're going to prove that A is a subset of B and that B is a subset of A. So that's how most of these proofs are gonna work. And many of them are elementary and quite straightforward. I'm gonna do a few of them. Like let's prove the associative property, one of the associative properties here. So not the associative property. I'm, excuse me, let's do one of the distributive laws. So we'll do the first distributive law together. A union B intersect C is A union B intersect A union C. So how would we prove that? So what we're gonna do is we're gonna start off by trying to prove that the first one is a subset of the other one. So we're gonna show that, so what we wanna do is we're gonna show that A union B intersect C is a subset of A union B intersect A union C, like so. So how do we show the two things are subsets? Well, we're gonna say, well, okay, let X belong to the first set, A union B intersect C, like so. And so if C belongs to this, because of the union symbol, one of two things happens, then we get that X belongs to A or X belongs to B intersect C, like so. So let's then consider the two possibilities that are happening here. So in the first case, I'm gonna color code these things here. So if X belongs to A, then we're gonna get that X belongs to A union B, because if I make the set bigger, it's still gonna be inside of there. And we're gonna get that X belongs to A union C, because again, if you make the set bigger, X will still be in there. And so thus we can see that X belongs to A union B and it belongs to A union C. This is the set on the right hand side we're trying to show here. So okay, in the first possibility, if X belongs to A, we get inside of that set, great. Well, what's the other possibility? We'll color that one blue here. If X belongs to B intersect C, then what that tells us is that X belongs to B and it belongs to C here. So we're just unraveling the definition of these two things here. If you belong to the intersection, you belong to B and you belong to C. Well, then if X belongs to B, that means X belongs to A union B, because if we make the set bigger, then it's gonna be inside there. And if X belongs to C, that means X belongs to A union C, because again, if we make the set bigger, it'll still be inside of there. And so therefore we can then say that X belongs to A union B intersect A union C. And so what we've now shown here is therefore we get that A, which direction we're going again, A union B intersect C, it belongs, it's a subset of A union B intersect A union C. So we've now shown the first direction. So that's what's summarized in this paragraph right here. We expounded it a little bit. Let's go now the other direction, right? So we wanna go the other direction. So now what we wanna show, we wanna show that A union B intersect A union C. This is a subset of A union B intersect C, like so. So let's break this thing apart here. So we're gonna let X be an arbitrary element of this set this time. So we have to go the other direction. A union B intersect A union C, like so. And so in this situation, because we have intersection, this means that then A belongs to A union, sorry, not A, our element was X. X belongs to A union B and we get that X belongs to A union C, like so. And so again, there's some situations to consider. It's in both situations. So if X belongs to A, right? Because when you look at the first one, let's rewrite that. So like note that X belongs to A or X belongs to B. So that's what we get by unraveling this one right here. So if X is in A and A is a subset of A union B intersect C here, in that situation, then we're done, right? So if X is in A, then we're done. So we're gonna assume, and that's okay, right? So assume X is in B. And so by similar reasoning, by similar reasoning, because we have the other situation going on here, right? Because we also have that X belongs to A union C. So if X was for an A, we'd be done again. So by similar reasoning, what we're gonna do is we're gonna assume that X belongs to C. Now, if that's the case, if X belongs to C, then that means that X belongs to B intersect C, which of course is a subset of A union B intersect C, which then got us the direction we wanted to go. And so therefore, right? Therefore, we conclude that A union B intersect A union C is a subset of A union B intersect C, like so. And so by showing that the subset goes in both directions, we then conclude, therefore, that the two sets are actually equal to each other. A union B intersect A union C is equal to A union B intersect C. And so this proves the first of the two distributive laws, but this is how the basic proof would work. We would show that the two sets are subsets of one another, and therefore they're equal. All the other properties can be proven in a similar manner, and those will be left up to you to prove on your own there. Best of luck everyone, and I'll see you next time. 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