 Welcome back to our lecture series Math 31-20, transition to Advanced Mathematics for students at Southern Utah University. As usual, I'll be your professor today, Dr. Andrew Misseldine. In lecture 27, we're going to continue our discussion of equivalence relations that we introduced in lecture 26. In this video, we want to talk a little bit more about the equivalence classes that we had introduced previously. Recall that given an equivalence relation, this is a relation on a set that is reflexive, symmetric, and transitive. This equivalence relation is measuring some notion of sameness that objects in the set are related to each other. They're similar, they're equivalent to each other. Associated to an equivalence relation is the collection of equivalence classes. Remember that an equivalence class for say x here, this is the set of all things inside all the set, all the elements say y inside the set, such that those things are related to x. This is what our equivalence class is. There's a few properties we want to prove about equivalence classes inside of this video. Here, one of which we already stated, we didn't prove, let this symbol be an equivalence relation on our set x, take two elements x and y inside the set x, then it turns out that the equivalence classes of x and y are equal to each other if and only if x is related to each other, x and y are related to each other. And so remember, this is an if and only if statement. We have to prove both directions. We have to prove sufficiency, that is the quality of the sets, the equivalence classes gives you the relation, and we have to prove necessity here that the relation gives us the quality of the sets. Okay, so that's what we're gonna look for here. Remember, this is our definition of an equivalence class in this situation. So we're gonna go the first direction, assume that these equality holds and then we wanna show that the relation happens, okay? So suppose that x is equal to y. Now recall that the definition, like it's above on the screen, you see it there, but I'll see it again now. The definition of an equivalence class is that we're taking all elements z inside of x, such that x is related to z. I switched it up to z now. Since x and y have meanings that are fixed, we're gonna let z be our variable there, okay? Now, this is true for the equivalence class containing x. We can do a similar thing here. The equivalence class containing y is again the collection of all elements z. Such that y is related to z, okay? Now, since this relation is reflexive, this means that y is related to y. Therefore, y is inside of the equivalence class containing, or the equivalence class represented by y, okay? Now, by assumption, the equivalence class containing x and y are equal to each other. So if y is in one of them, y is in the other one. Now, if y is inside the equivalence class represented by x, that means that x is related to y. And this gives us the sufficiency of the argument. We have now the first direction going here. We wanna go the other direction. So this time, we're going to assume that x is related to y. Then we have to argue that the equivalence class represented by x is equal to the equivalence class, which is represented by y. This is set equality, so we have to show that it subsets in both directions. So let's first start off with an arbitrary element of the class represented by x. Well, if z is inside of that class, that means x is related to z. Now, by symmetry, we can flip this around and say that z is related to x, okay? Now, by assumption here, x is related to y, so we can apply the transitivity here. We have z is related to x, x is related to y. Transitivity means we can compose those together and we get that z is related to y. Now, again, by symmetry, if z is related to y, that means y is related to z. And that gives us that z is inside the equivalence class represented by y. And that gives us the first direction here, that the class represented by x is a subset of the class represented by y. The other direction, we might wanna argue a similar, but actually it's easier in the other direction. That's an important thing to point out here. In the other direction, let's take some element z inside of the class represented by y. So this means that y is similar, y is a equivalent to z. But by transitivity, because by assumption, we're still assuming that x is equivalent to y, this gets that since we have x is equivalent to y and y is equivalent to z, we get that x is equivalent to z. So in both directions, we use transitivity, but in the first direction, we have to use symmetry a couple of times. This time, we don't have to. That's what makes it a little bit easier. So since x is related to z, that means z belongs to the class containing x and that gives us the other sets, the subset containment. The class containing y is a subset of the class containing x and this then gives us equality here. So if x is equivalent to y, then they have the same equivalence classes. And conversely, if you have the same equivalence classes, then you are actually related to each other. So the two notions are logically equivalent to each other. Equality of classes means that you're related to each other. Okay, now this gets us to the main result for this theorem. I want to remind the viewer about a topic that we've introduced previously in this lecture series, the notion of a partition, okay? So a partition on a set x is a collection of sets. We'll call them x1, union x2, union x3, all the way up to whatever we end up with, some xn. And honestly, the way I write it, it almost seems like this partition has finitely many cells, that's not necessarily the case. We actually have it as the case that x is equal to the union of the xk's, where this k can range over potentially any set. This could be an infinite union for all I care. So we have x is the collection, it's the union of all these things, but some other things we can do is that each of the xk's are themselves non-empty, okay? We also have that the intersection between two of these, like if I take xi intersect xk, or j whatever you want to call it, is likewise non-empty. So the overlap between these sets is always going to be non-empty, all right? Which honestly, you don't necessarily need this one because if their intersections are non-empty, that means the sets themselves are non-empty, and that's actually how we define this thing, this partition. We use this with counting principles, this came up with like the subtraction principle, the addition principle, the inclusion-exclusion principle when we looked at combinatorial problems. Now it turns out the notion of a partition is actually the same notion as an equivalence relation. Every equivalence relation on a set gives us a partition and every partition gives us an equivalence relation on the set, okay? And let's see exactly what that means here. So given an equivalence relation twiddle on a set x, the equivalence classes of x, that is the things like bracket x, bracket y, bracket z, these equivalence classes form a partition on the set. That is if you take all of the equivalence classes together, this is a partition on our set x. Conversely, if you have a partition, so p equals the collection of all of these xk's, for which of course there's more than one, there's more than one set. It's not a singleton there. It's like the k is allowed to range there. If you take the partition on a set, this gives us an equivalence relation for which we say that two elements are equivalent to each other if they live inside the same equivalence class, all right? So let's prove this. Again, this is an if and only of statement. I didn't phrase it like that, but we have one implication here. Equivalence relations give us partitions, but we also have the reverse direction here that partitions give us equivalence relations so two notions are logically equivalent to each other. Now for the proof. Let's first suppose that we have an equivalence relation on the set x. Notice we need to show that the equivalence classes because we're gonna show the equivalence classes form a partition. We have to show that the equivalence classes are empty because the cells inside a partition have to be, of course, non-empty things. So if you take a particular class like the class represented by x, that thing is non-empty because it contains x itself by the reflexive property. X is equivalent to x, so x is inside that equivalence class. So there's something in there. Maybe there's nothing else in there, but there was at least something x is in there. So these cells are non-empty. The classes are non-empty. Let's also notice that the union of all of the equivalence classes is equal to x. How do we see this? Well, each of these equivalence classes is naturally a subset of x. So if you take the union of all of the classes as you arrange over x here, then clearly this is gonna be a subset of x. But going the other way around, if x is a subset of x, then, sorry, if x is an element of x, then we get that x is inside of an equivalence class, which then tells us that if you take the union of all of the possible equivalence classes, x is gonna be in there as well. So that gives us equality because we get subsets in both directions. We get that x is a subset of the union of the equivalence classes, like so. So we do get that the union of the equivalence classes is equal to the whole set x. Again, this comes from that reflexive property that we were talking about just a moment ago. So then the last thing we need to argue is that we have x equal to the union of all these parts. We have to show that the different parts of the partition are mutually exclusive. So if we take the intersection of two distinct classes, we wanna show you that it's empty. Now, how we're gonna do this is we're gonna assume that the overlap between two classes is non-empty and then argue that they actually have to be equal to each other. That's the option here. We only get that the intersection of distinct classes has to be empty. If there is an overlap, it just means they're equal to each other. So if the intersection between two classes is non-empty, let z be an element inside of that intersection, which that tells us that z is inside of the equivalence class associated to x and z is inside the equivalence class associated to y. So we get that z is related to x and z is related to y. I mean, I'm using the symmetry property there as well potentially. Now, it doesn't actually matter which direction you write them. You could write them as z is related to x or x is related to z because of the symmetric property. So in particular, z is related to x and you put that in either direction. z is related to y, you can put it in either direction. But honestly, this is the, not that one. I want to put it into, nope, that was the right one. I want this direction right here because by transitivity, if x is related to z and z is related to y, that gives us that x is related to y, which if x is related to y, that means x is inside of the equivalence class containing y. And so that gives us this first set containment, subset containment there. The class containing x is contained inside the class containing y. By a similar argument, you can show that the class containing y is a subset of the class containing x. And this then shows us the class containing x is equal to the class containing y. And that then shows us that an equivalence relation gives us a partition. That is the equivalence classes associated to the partition give us, sorry, the equivalence classes associated to the relation give us a partition. And notice, we use all the properties. We use symmetry, we use transitivity and we use reflexivity here. So it is important that this is an equivalence relation. Now let's go the other way around. Let's suppose that we have a partition, we'll call it P inside of, it's a partition of x and we'll say that, well, yeah, we have that partition. We're then now gonna define a relation on the set x by the following rule. We say that two elements x and y are related to each other if x and y are contained inside of the same cell of the partition. So if we were to like draw a little illustration here, we have a bunch of elements of our set x, just kind of drawing them here. Draw, draw, draw, draw, draw, draw, something like this. And so those are the elements of our set x. We're gonna then draw some cells. We get something like the following. So this is a partition into three cells on our set here. And so if we have like little x, little y right here, x and y are related to each other, they're equivalent because they're in the same group as opposed to being in different groups. So that's what we define to be this relation, okay? Now we have to prove that this relation is reflexive, symmetric, and transitive, okay? X is inside of the same cell that contains x, so we have a reflexive relation, okay? Similarly, if x is in the same cell as y, then y is in the same cell as x. Therefore we get if x is related to y, then y is related to x. That gives us the symmetry property. Lastly, if we have a third element, say z, if x is in the same cell as y and y is in the same cell as z, then that means x is in the same cell as z as well. So if x is related to y and y is related to z, then we get that x is related to z, and that's exactly the transitive property. So this is in fact an equivalence relation. But honestly, if you want a much simpler argument, you can honestly take what I have written over here the following. So let x bracket be this class that contains x. Notice that if x is related to y, that means y or x is inside the cell that contains y right there. In particular, x is related to y if and only if they're in the same class. And because the second condition, right, the second condition has to do with equality. Equality is like the poster child of an equivalence relation. It's reflexive, it's symmetric, it's transitive. Those three properties can then be placed onto this one. So honestly, because equality is an equivalence relation, this is gonna be equivalence relation as well, but the specific details you can see on this side of the screen. So as an important corollary of this thing, two equivalence classes of an equivalence relation are either disjoint or they're equal to each other because you form this partition. So whenever you work with partitions, you're working with equivalence relations and vice versa. In our previous combinatorial problems, I wanna mention that every time we introduced a partition, which we did several times, we were essentially, without using the words at the time, introducing an equivalence relation, we were classifying all these problems are equivalent, these problems are equivalent, these problems are equivalent. We count this one, we count this one, we count this one, we add them all together. And so we use equivalence relations all the time. We use them even before we even knew what they were. And this connection to partition solidifies that observation.