 So welcome back to another screencast about equivalence classes. In this screencast, we're going to prove three important properties about equivalence classes. So let's jump right into it. First of all, let's remind ourselves what an equivalence class is. If we start with an equivalence relation tilde on a non-empty set A and let little a be an element of capital A, then the equivalence class of little a, determined by tilde, is a subset of capital A. And we denote it with square brackets around the little a. This is the equivalence class of A, and it's merely the set of all points in capital A to which little a is equivalent. So we're not going to prove three important things about equivalence classes. And these all involve proofs, and some of these proofs are fairly lengthy. So I invite you to put on your thinking caps and settle in, because a lot of these proofs cover just about every proof technique we've seen so far in the course. First of all, let's start with something easy, okay? The proposition here says that if tilde is an equivalence relation on a non-empty set A, then for every A in capital A, the equivalence class of A is non-empty. In other words, this is saying that there's no such thing as an empty equivalence class. All equivalence classes contain something. This is so easy that I'm going to prove this right below the proposition statement. So here's my proof, proof. So let's let A be in capital A. Then, since tilde is an equivalence relation, A is equivalent to itself. That's because of the reflexive property of the equivalence relation. Therefore, since A is equivalent to itself, A belongs to its own equivalence class. Remember, this equivalence class right here is this, the set of all points in capital A, such that x is related to A. And A itself is one of those points. Okay, so all equivalence classes are non-empty. They at least contain the thing that is between the square brackets. So that's, that's an important, simple, but important property about equivalence classes is that they're never empty. Now the next property is quite lengthy in its proof. It's a really important concept that will show up in later videos as well. So let's state it carefully and then prove it quite carefully. So if tilde is an equivalence relation on a non-empty set A, then if A and B are any points in capital A, A is equivalent to B if and only if their equivalence classes are equal. So in other words, I could have two different points in A perhaps that give me two equal classes as long as they are related to each other. And that relationship goes both ways. A is equivalent to B if and only if their classes are equal to each other. Now before we start into a proof of this, let's break this down and see how the proof is going to be set up. This is quite a complex statement here. First of all, notice it's an if and only if statement. So I need to prove two directions. I'm just going to list these here. The forward direction of this biconditional statement would be to assume that A is equivalent to B. And then I would want to prove that the class of A equals the equivalence class of B. The backwards direction of this theorem would give me the converse. I would want to assume that the class of A equals the class of B. And then prove that A is equivalent to B. Now not only that, look at the thing I'm going to prove in the forward direction. This itself is two different proofs because I have two sets here, two equivalence classes. Remember, those are sets and I'm claiming that they're equal to each other. So I have to prove two things in this one direction. I have to show that the equivalence class of A is a subset of the equivalence class of B. And then I have to prove that the equivalence class of B is a subset of the equivalence class of A. So this has got it all. This is a biconditional statement. So I have to prove two directions. And in the conclusion of one of those directions, I have to prove a double subset inclusion. So this is going to be a fairly complicated proof. So just pause the video if you need to and just mull over this outline of the proof. Okay, you've mulled, you've paused. Now we're going to move on and try to prove this. Okay, so let's add a new page here. So let's first of all prove the forward direction. So I'm going to let A and B belong to A. And I'm going to assume in this case that A is equivalent to B. And I want to prove, again, that the class of A equals the class of B. And I'm going to do this by showing, so I'm trying to be very clear to my reader. So I'm going to show that the class of A is a subset of the class of B. And the sub, the class of B is a subset of the class of A. Remember, by definition, all of these equivalence classes, these things with the square brackets on them are sets. They are sets of objects in A. And if you need to pause the video again and go back and review that definition or have your book open and review it. So now I have two little subproves to do. So two, I'm just going to label these maybe part one and part two. So in part one, I'm going to show that the equivalence class of A is contained in the equivalence class of B. Now, how do I prove that a set is a subset of another? I'm going to choose an element. So let's let X be some point in A that belongs to the equivalence class of A. And I want to show that X belongs to the equivalence class of B. Okay, so finally, in the end here, I am letting X be an element of the equivalence class of A. So since, let's interpret what that means. So since X belongs to the equivalence class of A, what does that mean? That means that X is equivalent to A. Now, again, I'm going to state what I want to show. You notice in a well-written proof, you're saying what you're going to show a lot. I'm going to prove this. Here I said it here. I'm showing this. I'm showing this. I'm showing this. This is a hallmark of a clearly written proof here. What am I going to show now? I want to show that X belongs to the equivalence class of B, which means I want to show that X is equivalent to B. So in the end, what I have here is I'm assuming that X is equivalent to A, and I want to prove that X is equivalent to B. Now, what I also have going for me here, in my initial assumption, is that A is equivalent to B. So let's put these together. So X is equivalent to A by assumption. And A is equivalent to B also by assumption. Notice that I can use the transitive property here. I have X equivalent to A and A equivalent to B, and this tilde thing here is an equivalence relation. So by transitivity, by transitivity, X is equivalent to B. Okay, and that's what I wanted to show right here. All the other stuff that I wanted to show boils down to showing that X is equivalent to B. I've done it, so that little sub, sub, sub part of the proof is done. So what I've proven here now, let me just say what I've proven since I have so much to prove here. So now, therefore, therefore, the equivalence class of A is a subset of the equivalence class of B. Now what I want to do is prove the other direction of inclusion. Ultimately, I'm trying to prove that the two equivalence classes are equal. So now I want to show in part two here that the equivalence class of B is a subset of the equivalence class of A, and let's prove this similarly. Let's let, so I'm trying to prove that one sets a subset of another. Let's let, let's say Y belong to the equivalence class of B. And I want to show that Y belongs to the equivalence class of A. That is, I want to show that Y is equivalent to A. Now since, this is going to look very similar to part one here. Since Y belongs to the equivalence class of B, what that means by definition is that Y is equivalent to B. But we also know that A is equivalent to B. Now, I can't say that Y is equivalent to A just yet. There's one more thing I need to do, and that's use the symmetry property of the equivalence relation. Remember an equivalence relation is reflexive, symmetric, and transitive. So since A is equivalent to B, and again, that's by our very, very initial assumption way, way up here at the top. Since A is equivalent to B, B is equivalent to A, and now I can use my transitivity. So by transitivity, since Y is equivalent to B, B is equivalent to A. So by the transitive property, transitivity, Y is equivalent to A. And so Y belongs to the class of A. And so the class of B is a subset of the class of A. And that, I don't really have enough room to write anything more down here. So I'll just put a check mark or a little, not on a box. I'm not full, done with the entire proof yet. But notice I am done with this forward direction of the bi-conditional statement. I've assumed that A is equivalent to B, and I've proven that these two classes are equal by proving that they are subsets of each other. So that is the first half of the proof. The second half is considerably shorter than the first half. So it's kind of hard to call it a half. But that's at least the forward direction of this proof. You should really pause the video and really study this proof and make sure you master the flow of the argument here. This is very important and puts a lot of pieces together. So let's move on and look at the backwards direction of this proof here. Okay, so I'm not going to prove the converse. That is, I'm going to suppose or assume that the equivalence class of A equals the equivalence class of B. And I want to prove that A is related to B, or is equivalent to B, related and equivalent to B. So how are we going to do this? Well, from the first proposition, the non-emptiness proposition, we know that the point A is an element of its own equivalence class. And that was something we established early on. And since these two sets are equal, I can also say that A belongs to the equivalence class of B as well, because the equivalence class of A and the equivalence class of B are assumed to be the same thing. So what does that mean? Well, if A belongs to the equivalence class of B, that allows me to conclude simply that A is equivalent to B. And that's what I wanted to prove. This is true by definition. This is what it means to be an element of this equivalence class. So that was a really short one right there. And so that proves the entire proposition that the two equivalence classes are equal if and only if they're representatives, the objects that are inside the square brackets are equivalent to each other. Now let's move on to this final proposition I'd like to look at here. And this is an important one too. It says that if tilde is an equivalence relation on a non-empty set A, then for all points A, B, and A, something interesting happens with the equivalence classes. They are, the equivalence class of A and B are either equal or completely totally disjoint. In other words, if you have two equivalence classes, one for A, one for B, you're never going to get any sort of partial overlapping on here. These two classes are going to be either totally, totally equal or totally, totally separate. I call this the all or nothing intersection property here. So I'm going to prove this by cases, it turns out. So let's move on to make a page for this and look at a proof. So a proof. So let's let A, B, B just any two points out of the set A. Now one of the following two things has to be true. Either A and B are equivalent or they aren't. Either A is related to B equivalent to B or A is not equivalent to B. I'll just sort of invent that little notation there. That's certainly true. So these are two cases that I can consider. I'm just pulling A and B out of the set A. I have no idea what kind of properties they may or may not have. So I need to throw some cases in here or possibly to give a bit more structure to this. Let's set up two cases. One for if A is related to B and one for if it isn't. So in case one, and remember what I'm trying to prove here is that either the equivalence classes are equal or they are disjoint, okay? So in case one, if A is related to B, then we just proved in that really long proposition that A is equivalent to B if and only if their classes are equal. So we're done in this case. If A is related to B under an equivalence relation, we just prove that their equivalence classes must be equal. So the proposition we're working on here is true in that case. In case two, what if they're not equivalent to each other? So what if A is not equivalent to B? Then let's prove by contradiction that the equivalence classes must be disjoint. Then I'll say this. So we prove that A intersect B, or the class of A intersect the class of B is disjoint by contradiction. And here's how we're going to do this. Contradiction. So suppose for a contradiction that the for a contradiction, I'll say it this way. Contradiction. So many long words in this. So for a contradiction, let's assume that the intersection is not empty. Let's assume the opposite of what I'm trying to prove. Let's suppose that the intersection of the class of A and the class of B is not empty. Well that means there's something in it. Let's call it a name. So let, let's call it XB in the intersection, okay? And let's just think about what that means. So X is in the intersection of two sets. So that means that X belongs to the class of A and X belongs to the equivalence class of B. Okay, fantastic, that's what intersection means. Now if X belongs to the equivalence class of A, what does that, what does that mean? Well that means that X is equivalent to A. And X belongs to the equivalence class of B, that means that X is equivalent to B. Now let's see where this leads us. I remember our initial assumption in this case was that A and B are not related to each other. However, we now have that X is related to A by symmetry. What does this mean? By symmetry, we know that A is equivalent to X. And now check out what we have. I have A is equivalent to X and X is equivalent to B. By transitivity, what does this tell me? You see the contradiction? This would give A is equivalent to B, but that's a contradiction to the initial assumption that A is not equivalent to B. So here I have two things that are impo, two things that are completely opposite but happening at the same time. That's impossible. So that's a contradiction. Where did this contradiction arise? It arose from assuming that we had an X in the intersection. So after all, this is over with, the intersection of A, the class of A and the class of B must actually be empty after all. So that proves that in any case, either whether the two points are related or whether they're not, and that covers all possible cases here, that the equivalence classes are either equal or disjoint. And that ends the proof of all those three propositions. And those propositions will come up in the next video and in the future. So those are important. And again, I think the important thing here too is to master the flow of the argument and the proof. So make sure you are doing that. Thanks.