 Now that we've introduced the idea of a subset for our logic portion of lecture three, we're going to talk about how does one prove that a set is a subset of another? And this is actually a very important proof template that we're going to do. So remember, by definition, A is a subset of B if every element of A is an element of B. So this definition leads itself to a method for showing whether A is a subset of another set B. And it's exactly this bolded statement right here to show that A is a subset of B. We assume that A, little A is an element of A and then argue that this same element A is also an element of B. In other words, we take an arbitrary element of the first set and argue that that arbitrary element belongs to B as well. Okay, now if we take an arbitrary element of A, the only thing we know about that element is that it belongs to A. And therefore, since it belongs to A, it satisfies the defining property of A, the set, and that's the only thing we know about it. But then when it comes to how do you show that A belongs to B, this is something we talked about in lecture two, to show that an element belongs to a set. You have to show that that object satisfies the defining property of that set. And so that's how you show this, because if little A is arbitrary in this whole, that means every element of A belongs to B. And so therefore, we have a subset. And so let's provide some examples of such a thing. So for this first example, let's take set A to be all of the integers which are divisible by 18. So we're looking for multiples of 18 inside of the set of integers. And set B is, likewise, going to be the set of integers which are divisible by 6. That is, we're looking for all of the multiples of 6. And so we're going to show that the set A is a subset of B. I want you to be aware that what we're showing is equivalent to showing that all multiples of 18 are multiples of 6. Which it sounds fairly straightforward when you phrase it that way, but we're going to do it in the language of set theory for the sake of practice here. All right. Now the first thing you do is you take an arbitrary element of A. Now some people, when they write proofs like this, they like to redefine the set like, oh, let A be this and let B be this. You don't need to do that because it's already on the page. A is right here, B is right here. You don't need to tell me what they are twice. Once is enough. So the proof starts with this. Suppose little A is an arbitrary element of A. Suppose little A belongs to capital A. Now because little A belongs to A, the only thing I know about it is it belongs to the set A. So I then apply the definition of capital A. What does it mean to be inside of capital A? Since little A belongs to big A, there exists some integer K such that A equals 18K. Remember, to be inside of big A, this means that 18 divides A. And what does it mean for 18 divide A? It means exactly this right here. So to be inside of capital A means that there's some integer K such that A equals 18K. This is what it means to be inside of capital A. Now note that the number 18 can factor as 3 times 6. This is just a fact inside of the arithmetic of integers. This needs no further justification because we, as we're reading this proof, are all people who can accept basic arithmetic and algebraic properties of integers. So I don't need to justify that 18 can factor as 3 times 6. My audience already knows that. Now since K is an arbitrary integer, we don't know what it is, right? Because we don't know what A is, so we can't predict what K is. But since A is divisible by 18, there has to be some number K and it's going to be an integer. So since K is an integer, we can also infer that 3K is also an integer. Again, this is one of those algebraic properties of integers. You can assume without any further proof that the sum, the difference, and the product of two integers is always an integer. You do have to watch out for multiplication because like 1 divided by 2 is not an integer. It is a rational number, but it's not an integer. But you can assume by just properties of integers that the sum, difference, and product of integers is an integer itself. If you have any confusion with that, well, continue on and take abstract algebra. That's where you talk about those into more detail. But we can take for granted that 3K is an integer because K is an integer. It's a property of integers. Hence, if we take A equals 18K, which we know that to be true because little A belongs to A, we can then factor 18 like we did above and 18K becomes 6 times 3K, where K is an integer. Using algebraic properties of integers and particularly the associative property, this is the same thing as 6 times 3K. And like we observed earlier, 3K is an integer. So 6 times 3K is 6 times an integer. Now, if you're 6 times an integer, you're divisible by 6. So this number right here is divisible by 6. Now, if you're divisible by 6, that means you belong to B. Hence the statement we have right here. So A is a multiple of 6, so it belongs to B. Now, as A was an arbitrary element, we have to then conclude that every element of capital A belongs to B. Therefore, A is a subset of B. And that's how you show that a set is a subset of another. Choose an arbitrary element of the first set and then argue that that element also belongs to the second set. Let's do another example of this. This is something we referred to in the previous video. We talked about interval notation there. For any real numbers A and B, prove that the open interval A to B is a subset of the closed interval A to B. It seems geometrically obvious, but nothing should be taken as obvious here. We should be able to prove every assertion we make in mathematics, even the quote-unquote obvious ones. Now to prove that the open interval is a subset of the closed interval, we take an arbitrary element of the open interval and then argue it belongs to the closed interval. So that's exactly what we're going to do. Take an arbitrary element of the open interval, call it X. Because X belongs to the open interval, we apply the definition of this interval notation. X belongs to the open interval A to B because X is greater than A and because X is less than B. X is greater than A and X is less than B. Since X is greater than A, it's also true that X is greater than or equal to A. Greater than or equal to has two options, greater than or equal. I can't assume that X doesn't equal A because this symbol doesn't allow for equality. But if X is greater than A, then it is greater than or equal to, even if I know the second option isn't available to me. So I can generalize to this statement. Likewise, since X is less than B, then it's true that X is less than or equal to B. And so since X is greater than or equal to A and X is less than or equal to B, that means that X belongs to the closed interval A to B. And therefore this shows that the open interval A to B is a subset of the closed interval A to B. It might seem like an elementary thing geometrically, but it does deserve a proof. But the proof is actually fairly elementary because of the observations we just provided in this proof. It's a good example of the template. Now, what we're going to do is we're going to extend the proof we just did, the proof template, I should say. We've just argued how we can prove a subset is a, how a set is a subset of another. Very related to this is the idea of proving that two sets are equal to each other. We defined this in the previous video as well. By definition, two sets are equal to each other if they have exactly the same subsets, they have exactly the same elements, excuse me. Now, phrased a little bit differently, two sets are equal to each other exactly when they are subsets of each other. That is A equals B if and only if A is a subset of B and B is a subset of A. Now, to justify that, think of the following. If A and B are sets that have exactly the same elements, then this means that every element of A belongs to B. And that's what it means for A to be a subset. And likewise, if B has all the same elements of, as A, then every element of B belongs to A and so B is a subset of A. To B equal, they have exactly the same elements. Well, if every element of A belongs to B, then boom, we get this. If every element of B belongs, I think I said that backwards, I'm sorry. If every element of A belongs to B, we get this one. If every element of B belongs to A, we get this one. So equality of sets actually means that they're subsets of each other. And this is one of the very reasons why we actually include the improper subset as a subset. It seems weird, but in this situation, A and B are the same set. So we're showing they're subsets of each other. So at the beginning, we might not know that it's the same set. We're showing that they're subsets of each other. So it might not be known that this is a improper subset until the very end of the argument. So we allow it to show that sets are equal to each other. So the template is then illustrated here in bold. To show that two sets are equal to each other, we show that they are subsets of each other. Usually then, how do you show their subsets of each other? What we did on the previous slide. So in practice, when you want to show that two sets are equal to each other, what you do is you take an arbitrary element of the first set, argue that it's in the second set, and you take an arbitrary element of the second set and show that it's contained inside the first set. And so let me provide you some examples of this right here. All right, this time we're going to take the sets A and B, where A is defined to be all multiples of 35, integer multiples, and we're going to define B to be integers which are multiples of 5 and multiples of 7. So what you can think of is this right here, the set B, we're taking the set of common multiples of 5 and 7. Compare that to the set of multiples of 35. We argue that these sets are actually equal to each other. We're going to prove they're equal. So to prove that they're equal, you have to first show that A is a subset of B, and then we're going to show that B is a subset of A. Now to show that A is a subset of B, you take an arbitrary element of A. So take an integer which is a multiple of 35. Since A is divisible by 35, there exists some integer k such that A equals 35k. Now note that 35 can be factored as 5 times 7. So using properties of integers and their multiplication, if A equals 35k, then you can factor that as 5 times 7k. Now since 7k is an integer, you have 5 times an integer, this gives you that 5 divides A. So every multiple of 35 is a multiple of 5. Much like we did earlier when we talked about multiples of 6 versus multiples of 18. It's the same basic argument here. But likewise, we're doing basically the same argument again with a slight tweak. If you're a multiple of 35, because of this factorization, you're also a multiple of 7 because 5 times an integer is an integer. So 7 times 5k is 7 times an integer. And so this also shows that 7 divides A. So notice we've shown here that if you're a multiple of 35, you're a multiple of 5 and a multiple of 7, and so we see that this element A belongs to the set B. Now as A was chosen arbitrarily, this shows what we are coming to prove here that A is a subset of B. Now let's go in the other direction. Suppose we have an arbitrary element of B. So what does it mean to belong to B? If it belongs to B, then it means your element is divisible by 5 and divisible by 7. Now if you're divisible by 5, you're a multiple of 5. That means there's some integer n such that B equals 5 times n. Now likewise, if 7 divides B, that means there's some integer m such that B equals 7m. Notice here that I use two different symbols, m and n. I don't assume that these are the same integers. They could be, but I don't have any evidence to support that. I probably argued they're not the same. But regardless, I mean because they could both be 0. Like our number B could be 0. That's a multiple of 5 and 7. So we can't suppose they're different, but we should use a different symbol because it is a possibility. So B equals 5n and B also equals 7m because they're multiples of 5 and 7. So since B of course is the same number, this gives us that 5n equals 7m. Now it might be tempting to try to solve this equation, but if you start dividing by 5 and 7 or 7, we don't necessarily know that n or m are divisible by 7 and these are integers not necessarily fractions. So we have to be very careful how we proceed with this. What I can say is the following here. As 5 divides 5n, sorry that's a typo right there, 5 divides B, right? That was something we assumed earlier, and B is equal to 7m here. So we have that since 5 divides this side of the equation, 5 has to also divide this side of the equation as well. Now 5 is a prime number. There's a significance about that right here. 5 is a prime number. So if 5 divides 7m, it must be true that 5 either divides 7 or 5 divides m. This fact is known as Euclid's lemma. We'll talk more about it in the future, but we're going to use it right here. Euclid's lemma states that if a prime number divides a product, then it actually divides one of the factors. So since 5 is a prime number, Euclid's lemma tells us that 5 either divides 7 or 5 divides m. But 7 itself is a prime number, so we know that 5 doesn't divide 7. So the only other possibility then would be that 5 divides m. So since 5 divides m, we don't know what m is other than it's an integer, but since 5 divides m, there has to be some other integer k so that m equals 5k. Thus, b which equals 7m and m which equals 5k, we then end up with 7 times 5 times k, but 7 times 5 is 35, b is equal to 35. This shows that b is a multiple of 35, and hence b belongs to capital A. And this shows us the other containment that capital B is a subset of A. Now, since we show that A is a subset of b and b is a subset of A, we then conclude that A is equal to b. So what we've now shown is that the multiples of 35 are exactly the common multiples of 5 and 7. And this we could then leverage if we want to do some number theory to argue that 35 is the least common multiple of 5 and 7. But we'll worry about that another time. Let's focus on proving that two sets are equal to each other. Let's do one more example. Let's say that the set of elements of the form 2 to the x where x is an arbitrary real number, we want to prove that this is the same as the set 4 to the y where y is an arbitrary real number as well. I intentionally used a different letter x and y here so it doesn't put any assumption that x and y are the same. But that isn't necessary. You could have also described this set as 4 to the x, x. Those two sets, the x's, there's nothing that says the x's have to be equal to each other even though I reused the symbol. We've overloaded the symbol but it's okay. But for the sake of us beginners, I did put a y there so there's no confusion whatsoever. Now these sets don't have any abbreviated names and I don't want to write them over and over again. So because of that, I'm going to start my proof this time by declaring the sets. Let A be the set of the exponential expressions 2 to the x and let B be the set of the form 4 to the y for any real number there. So I'm only mentioning the sets because they didn't have names and I want to give them names so I can refer to them in the future. Alright, so what we're asked to prove is that the set A equals the set B. That's what we're tasked to do. Now to prove that A and B are equal, we have to show that A is a subset of B and we have to show that B is a subset of A. And so to show that A is a subset of B, we take an arbitrary element of A and argue that it belongs to B. That's our plan right now. So suppose A is an element of capital A. Now if A belongs to A, that means there exists some real number x such that little A equals 2 to the x because that's what things in A look like. Now using exponential rules from algebra here, A if you take 2 to the x here, 2 to the x you can rewrite as 2 to the 2 times x over 2. That's perfectly valid and you can then rewrite that as 2 squared to the x over 2, completely justifiable by exponential laws. 2 squared is the same thing as 4 and so you get 4 to the x over 2. Now since x was a real number, x divided by 2 is likewise a real number. Hey, not every quotient of integers is an integer, but every quotient of real numbers is a real number as long as the denominator is not 0, but it's 2 in this case, we're fine. So this power of 2 is some power of 4, a different power, but still a power of 4. Therefore, since x over 2 is a real number, this means that 2 to the x belongs to B. And as this was an arbitrary element of A, we see that A is a subset of B. That's the first direction. We want to next go in the second direction. We want to show that B is a subset of A. To do that, we take an arbitrary element of B, which if you take an element of B, then there exists some real number y such that B equals 4 to the y. Now, since you have 4 to the y, we can do the same trick with exponential rules here. 4 to the y is the same thing as 4 to the 2 times y over 2, for which then I can factor the exponent as 4 to the 1 half times 2 to the y. Exponential laws justify equality here. And then the square root of 4 is actually 2, and so this becomes 2 to the 2y. Since y was a real number, 2 times y is likewise a real number. So this is a power of 2 to a real number. This gives us that B belongs to A right here. And therefore the set B is a subset of A. And then we can conclude, since we prove both directions, the sets A equal B. And so this is the general template. How do you show that two sets are equal? You show that they are subsets of each other. This is one of the most important proof templates we will learn this semester, which is in this course, which is why we learned it so early on in our lecture series. Thank you.