 I'm Zor. Welcome to Unisor Education. I would like to present a problem number three for the set zero. Set problem. Well, the previous two problems, number one and number two, were not really problems. They were kind of illustrations. Right now, this seems to be like a little bit more interesting problem, and here it is. Let's consider you have a set which contains n elements. They're all different. As an example, you can have, for instance, a set of first n natural numbers from one to n or anything else. We can actually kind of denote these elements as x1, x2, et cetera, xn. Okay. These are my elements. Now, this set obviously contains subsets. If you went through exercise number one, problem number one for set theory, I was actually enumerating all the subsets over one particular set which contains four attributes, and they found that there are 16 different subsets. I just listed one after another all the different subsets. Basically, what I'm asking right now is, how many different subsets this particular set has? In case of n is equal to four, as I was talking about in exercise number one, we had 16 subsets. This is a more general problem. We have n completely different elements, and the question is how many subsets does this have? This is a good moment to press the pause button and really think about this because this is a real problem. Try to come up with your own solution. You can actually, as a hint, you can actually do the following. Try to take as an example a couple of different sets like I did, for instance, for n equals four in exercise one. You can use something similar to that and try to enumerate all the different subsets of a set with three elements or two elements, and then try to guess the formula. After you guessed the formula, it might be a little easier to prove it using some logic. In this particular case, I will even present the proof using the method of mathematical induction, which is another topic in one of the lectures which are there. Again, press the pause button, and when you are ready, play it again, and I will continue with the solution to this problem. As I said, let's first examine a few examples. Now, what if n is equal to zero? That means we have an empty set, no elements at all. How many subsets does this particular set have? Well, think about this. If this is an empty set, and obviously it has one subset which is empty, it also has a subset which is a full set, but it's exactly the same thing, the empty set. So no matter how we count, we can have only one subset of the set which has n equals to zero elements. We can have only one subset. So for an empty set, we have one and only subset which is empty. All right, great. Let's say n is equal to one. So we have only element x1. Let's think about how many subsets we have. Well, again, we can start with two trivial subsets. One is empty and another is the whole set. Empty and the whole set are always there. Are always two subsets. Well, in case of n equals zero, they are the same. But in case of n equals one, it's already two different sets. One is empty and another is the whole set, which is x1, right? So we have two and nothing else. Because the subset which contains only one element is exactly the same x1 which we have already counted. So no matter how we count, we have an empty set as a subset and we have an x1 as containing one and only element as another subset and there are no others. Okay, next, n equals to two. So we have x1 and x2. Well, that's still easy. I mean it's easier than the exercise number one which we did where we had four elements, right? So we have two elements. Well, what are the subsets? Empty, number one, x1, x2, and then both together, x1 and x2, which is a complete set. So how many? Zero elements, only x1, only x2, and x1 plus x2, four elements we have. Okay, n is equal to three. We have x1, x2, x3. Okay, this I probably can do some writing. x1, x2, and x3. So what do we have? Obviously we can count every subset based on the number of elements it has. How many number of elements subsets of this particular set can have? Well, they can have zero elements, which is empty set. They can have one element and we have three different subsets with one element. It's x1, x2, and x3. How many subsets with two elements are? Again, in this case, it's easier to count instead of counting pair, we can count what's remaining. So we can have a pair which does not contain x1, a pair which does not contain x2, and a pair which does not contain x3. This pair is actually x2, x3. This pair is x1, x3, which does not contain x2, and this pair is x1, x2, which does not contain x3. Next, subsets which contain three elements. Well, there's only one subset, obviously, which is all of them, right? Okay, how many altogether? One, two, three, four, five, six, seven, eight. Okay, we got that number. And if you go back to problem number one, if you remember, we had four elements and we counted 16. Well, obviously you understand that this is a sequence which can be presented as this general formula. Two to the zero's degrees, one to the first degrees, two, et cetera. All right, so we guessed the formula. Obviously now the problem is to prove it. Okay, let's try to do it. First method is not really a mathematically rigorous one. However, it will pave the way to a more rigorous proof using methods of mathematical induction. Let's think about the following. If you have a set of n elements and let's consider it has k subsets. I don't know what k is. I mean, I presume that k is two to the power of n, but let's just not think about this. k subsets. Now let's consider what happens if I add one new element. So from x two to the xn are all old elements and then I will add one more element y. Well, it will definitely increase the number of subsets, but how will it increase the number of subsets? Well, first of all, all these k subsets which used to be obtained from first n elements will still be there. So I will definitely have the same k subsets. Now, introducing a new element y, how will it change the number of subsets? Well, let's think about this way. This new element y can either belong to one of these old subsets or it will not belong then. So if you have these old subsets and there are k of them, then I can either add y to them or not add y to them. So if I do not add y to them, that's the k subsets which I used to have before. And if I will add y, I will have k new subsets. So basically what I'm saying is adding one element actually doubles the number of subsets. This is a relatively, it's kind of an explanation of how the real proof using the method of mathematical induction will be conducted. And now the real proof. I will just comment on mathematical induction. Yes, you can go to another lecture which contains the explanation of this method, but basically in just a couple of words, if you would like to prove some formula for any positive integer n, you can check if this formula is true for some initial number n, let's say number one. And then you can prove the following theorem. Let's consider that the formula is true for certain number. If assuming it's true for certain number, it will be true for the next number. That actually proves that the formula is true for any number. Why? Because if I have this theorem proven, which means from n equals k, it follows that it will be true for n equals k plus one. I will apply it to n is equal to one. Okay, if the theorem or formula is true for n equals one, using this logic, I can actually say that it's true. Now this is true because it's checked for. So using this theorem, I can derive that this is true for n equals two as well. So this is true. Again, using the same formula for k equals two, I have that for n is equal to three is true, et cetera, et cetera, et cetera. So this process goes to infinity and for any integer n, sooner or later using this logic, I come up with a statement that this is true. All right, so basically, let's do exactly this type of thing in this case. So I would like to prove that the number of subsets is two to the power of n. Okay, let's check it for some beginning for some initial number n. Well, actually I did it already for n equals zero and one and two and three. So this is checked. Now, let's assume that formula is true for some number n equals k. So number of subsets equals two to the power of k. This is assumption. Now, based on this assumption, step number three in the method of mathematical induction, based on these assumption, based on this assumption, I would like to prove that this formula is true for the next k, for the k equals to k plus one. But this is exactly what I did before. I said, let's assume that we have a new element y and what happens with the number of subsets? Well, I have explained that the number of subsets will be doubled because this element y can either be included in one of the old subsets or excluded. And that actually from each old subset, it makes two new subsets with or without y. So for n equals to k plus one, all I can say that the number of subsets is equal to old number of subsets, which was in the previous k, and I have assumed it's true in the power of k, doubled. But this is exactly two to the power of k plus one, right? So I doubled the previous number, get this, and this is exactly the same formula for n is equal to k plus one. That basically concludes the proof using the method of mathematical induction that this formula is correct. So basically what you can say is given any set of n elements, the question of how many subsets does this particular set have can be answered and the formula is this. So regardless of what kind of elements as long as they're all different obviously, we can say that any set of n elements contains two to the n's degree of subsets. Would be interesting if you can come up with some other proof of the same theorem. It looks like a kind of very interesting thing and a nice formula. I like when the formula is actually simple. It's very easy to remember it. So in this particular case, this actually corresponds to some criteria of beauty. n elements, two to the power of n subsets. Well, that concludes this particular problem. Thank you very much.