MTH104 Finite Sets, Classes, & Power Sets Explained

Finite Sets: Definition and Core Concepts

A finite set contains exactly m distinct elements, where m is a non-negative integer (whole number). Unlike infinite sets, finite sets have a known, countable limit. For a more detailed introduction to sets, you can refer to the Comprehensive Overview of Set Theory: Understanding Sets, Operations, and Applications.

Example: {1, 2, 3, ..., 100} is finite because the limit (100) is given. If the '...' had no endpoint, it would be infinite.

Lemma: Counting Elements in Disjoint Finite Sets

A lemma is a proven statement used to prove another statement.

Lemma: If A and B are finite disjoint sets (no common elements), then the number of elements in their union is simply the sum:

n(A ∪ B) = n(A) + n(B)

Example:

A = {1, 2, 3} → n(A) = 3
B = {a, b, c} → n(B) = 3
Since A and B are disjoint, n(A ∪ B) = 3 + 3 = 6 (the set {1, 2, 3, a, b, c} has exactly 6 elements).

Three Key Theorems for Counting Elements

Theorem 1: The Complement Rule

Let A and B be any finite set in a finite universal set U.

Formula: n(Ac) = n(U) – n(A)

Use when: You know the total (universal set) and one subset, and need to find what's left.

Example: In a class of 35 students (U), 20 are male (A). How many are female (Ac)?

n(U) = 35, n(A) = 20
n(Ac) = 35 – 20 = 15 female students

Theorem 2: The Set Difference Rule

Let A and B be finite sets.

Formula: n(A – B) = n(A) – n(A ∩ B)

Use when: You need only the elements in A that are not also in B. The key word is "also" or "common" (intersection).

Example: Math class A has 35 students. 15 of these are also in Biology class B. How many math students are not in Biology?

n(A) = 35, n(A ∩ B) = 15
n(A – B) = 35 – 15 = 20 students are only in Math

Theorem 3: The General Union Rule

Let A and B be finite sets (may have common elements).

Formula: n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

Use when: You need the total number in both sets combined, accounting for double-counting the common elements.

Example: Among 110 college students, 30 are on List A, 35 on List B, and 20 are on both lists. How many students are on either list?

n(A) = 30, n(B) = 35, n(A ∩ B) = 20
n(A ∪ B) = 30 + 35 – 20 = 45 students

Important Distinction:

Lemma (disjoint): No common elements → just add: n(A) + n(B)
Theorem 3 (general): Common elements exist → add and subtract the intersection: n(A) + n(B) – n(A ∩ B)

Classes of Sets and Subclasses

A class of sets is a set whose elements are themselves sets. It is written with an outer set of braces {} containing inner set elements.

Creating a Class

Given S = {1, 2, 3, 4}, create a class A with 3-element subsets: A = { {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4} }

Creating a Subclass

From class A, create a subclass B where the element '2' is fixed in every subset: B = { {1,2,3}, {1,2,4}, {2,3,4} }

B is a subset (subclass) of class A.

Power Sets

The power set of a set A, written P(A), is the set of all possible subsets of A, including the empty set ∅ and A itself. Understanding power sets is foundational for topics like A Level Math S1: Permutations and Combinations Full Introduction Guide, which often rely on counting subsets.

Formula for number of subsets: 2n, where n = number of elements in the original set.

Example: Power Set of {1, 2, 3}

n = 3 → total subsets = 23 = 8

The power set P(A) = {

∅ (empty set),
{1}, {2}, {3} (singletons),
{1,2}, {1,3}, {2,3} (pairs),
{1,2,3} (the full set) }

Tip: Always start with the empty set, then all single-element subsets, then pairs, then triples, etc., until you include the full set.

Rahman Rahim. Assalamu Alaikum. Welcome to my YouTube channel. And today is lecture number four of our MTH 104 Sets

and Logic course: Finite Sets and Classes of Sets. Okay? In this, we'll study finite sets, and then there's a

small topic on classes of sets. But finite sets involve some theorems, okay ? They're very important. So, you have

to understand this carefully. Let's start with a finite set. Finite, as the name suggests, is something that must

have a limit. It has so many whole elements that we know where they end. Okay? And the opposite is infinite,

where we don't know the limit. Okay? So , here's the definition. A set is a solid set, so a set is said to be

finite if it contains exactly m distinct elements. m is shown here as any number. It can also be called n. q

or s can be anything. Okay? The exact, distinct element is that it must contain the exact number. Meaning, we

know that these exact numbers are 100, 20, or 40. So, we know the limit. Okay? Where m is some non-negative number.

Non-negative means integer. Now, we have a number line. On one side of it, there are negative numbers, on the

other side, there are positive numbers. So, what will a finite set always be? It will be positive. Okay? Now, the

example is given: dot-dot means there are numbers in between, and we've already been told the limit. If there

wasn't 100 at the end of dot-dot, it would be an infinite set. Okay? Next, we have a lemma before theorem. What is

a lemma? Now, what happens is that someone says a theorem, so we know that there is a theorem, which we prove.

There is any statement that we prove. A lemma is called a lemma or a lemma. It's a proven statement. Any statement

that can be proven is used to prove another statement. For example, we know that A is a union B. We know that A is

a union B. But let's take the statement A is a union B and tell us how many elements it contains. For example, I've

written its definition here for you to read. Okay? Now, look at the example. Suppose A and B are finite disjoint

sets. Disjoint means they don't have any common elements. Okay? Then A is a union B. Now, we know that A is a

proven statement, right? We do A and B by adding them together. So, here we need to tell what its finite number is

if it is disjoint. So, we need to add it. n means number. The number of A is a union B is equal to the number of

sets A plus the number of sets B. Now, if we want to union A with B, we have a set A and a set B, right? So we put

them together and write them together. This becomes the union. Now, how many elements are there in that set? How do

we find that? There is already a proven statement for A union B. From this, we proved the next statement, which tells

us that if we add the number of N plus the number of B, we get the set A union B. Look at the example, here we have

any one, for example, 1 2 3, and here it is disjoint, so we write A B C. We don't take the same element because it

is disjoint. Now, if we do A union B, we have to write them together. So we wrote 1 2 3 A B C. In this set, the

number of A plus the number of B is equal to the number of N. Now, how many elements of A B are there? 1 2 3, we

wrote three here, and how many elements of P are there? 1 2 3, so we added both of them, six. And if we count this,

there will be six. 1 2 3 4 5 6 So this proves that A union B is equal to the number of A + the number of B. Right?

There's nothing difficult about it. Just add it. We do A union B, right? We write them together. So how many

elements does it have? We can easily add them. 1 2 3 Count all the numbers. So here it's said that A union B is

obtained by adding both A and B. Right? So this is already a proven statement. We proved it again that adding these

two gives us A union B. Next, we start with the theorems. We get theorem number one. Now the important thing

starts from here. Right? These are three theorems, which are slightly different from each other. So you have

to keep this in mind. Right? What was in the first example we used? A disjoint finite set. Disjoint means

that it has no common elements. Okay? Next, we're just going to have a finite set. It's not disjoint. Okay? So, this

could mean that it also has common elements. Okay? From our example, we'll see if it has any common elements. So,

we have the first theorem. Let A B be any set in a finite universal set U. Then we've given it to you. Okay? Okay?

This is the theorem. Using this formula in the example, we'll use this formula in the example. Okay? So, what's our

first formula? We can call it n = a complement. n is a number. Now, in all the theorems, we have to count numbers.

For example, in the previous one, we knew about a union b, but we also saw how many numbers are in it. Similarly,

here we need to see how many ACs there are. So, that means, if we subtract n, the number of universal sets, from the

number of A sets, we get AC. This is the formula. There's nothing to understand. How does this formula apply

to the example? We need to see that. Okay? For example, we have A minutes. If there are 20 male students in a

class of 35 students, okay? There are 35 total students, and 20 of them are male. Then how many female students are

in the class? So, we need to find how many female students there are. So, simply subtract 20 students from the

total, and we get that there are females. So, here you need to see what is the total of a complement? We

subtract a from the universal set, so we have this whole thing. Now, we'll read more details in this later. So,

you have to pay attention to which formula is fixed in which statement. Here, we have the total number of

students, and each one is given to us. Okay? We've given one set. We need to find the second set. So, the formula

would be: Number of C = Number of Unions minus the number of sets A. So, we get 35-20, which is 15. Okay? This

is common sense. How did this come about? You know, subtract from the total. But it will forget which formula

to apply to which example. Now, you've understood the first one. When you understand the second and third, I'll

explain the difference again, what was in this one and what was in this one. Okay? Look at the next theorem we have.

It's Theorem Two. So, suppose a and b are finite sets. Initially, it was a finite set. This one is also finite.

The one before this was disjoint finite . Okay? We have to remember this. So, when the next example comes up, I'll

explain the difference between the two formulas. Now, we have A and B are finite sets. So what do we do? A

difference, this is called difference. Line difference b = a number of A-N number of A intersection B. Okay? We've

already given this. Now, let's look at the question, and you'll understand how this formula comes about. Now, look, A

is the difference B, and we subtract the intersection from the number of A. What is called intersection? It's

called common. Okay? So, we have to subtract the common point between A and B from the set A. Now, let's look at

the example. So, you'll understand how this formula works. Okay? Let's do it this way using the theorem. Okay,

suppose it's in math class A, right? Where is the math class labeled A? It contains 35 students. We'll keep

collecting data. Okay? Just like we write givens, it'll be easier if you keep writing them. Givens, where we

write all the givens. A means we have 35, okay? Students, and 15 of them are also in the bio class. Okay? Second, we

had this math class, and this one was in bio. We have 15 in bio. And these ones are in this one too. They're also

in A, right? The word 'also' is used, right? They were in A and are in B. And then how many students are in class A

that are not in class B? Now, we have a total of 35 students, of which 15 are in bio. So we'll know how many are not

in it. They're not in class A, but in bio. Now, there are 15 in bio, right? So we have to subtract something from

15 to find out how much 35 is. Now let's understand the question: we have 35 students, 15 of whom are also in the

biology class. Let's say a contains 25 students and 15 of them. Out of 35, 15 are also in the biology class. So, how

many students are not in the math class ? The biology class. Okay? Sorry, they are not in the biology class. Class A

is not in class B. 35 is the total. Okay? The total is given. 15 of which are in biology. And how many students

are not in biology is given. So what do we do now? We have to find the difference, not how many are missing

from a. We subtract b. So, we have a difference minus a, which is like this: a-b. So, we have a number of A, total A

, we have to subtract from it. A intersection B common ones. Also means common. So we have to subtract also. We

have to subtract the common student. You'll understand from the question itself. Also means both have it. So, in

which did both do it? They did it in the intersection. Okay? So we subtracted the intersection one. So,

the intersection one is 15, which is in both of them, and the total one is 35. 15 is taken away from 35, so we get 20.

Okay? You have to remember this. Now, the new thing that has come to us in this is also. So, a common thing has

come in both of them. So, that's why the intersection thing has come in this . In the one above this, we were given

only two things. The total class was given and within that class, it was told that these are so many students,

so how many are there. So we used this one above. Okay? By subtracting that from the total, we told ourselves that

so many students are female. And in this one, we have the total, how many of the total are common, and how many

are left behind. So, 20 students are not in class B, but in class A. Okay? Let's come to the next one. The next

theorem, we have Theorem 3. In this, we have Suppose A and B are finite sets. The same thing is repeated. Only at the

beginning, we had disjoint finite sets. All the others are finite sets. And the formula given is a union b = an + nb-n

and intersection. Now, look at this, this formula is becoming the very beginning. When we found the number of

A union B, we got the number of A plus the number of B. But what's the difference between this and that? That

was a disjoint set. It was a finite disjoint set. And what is this? It's just a finite set. So that's why we

wrote minus AN intersection here. Okay? That's the difference between the two. In the first one, when we were finding

the union, what is the number of unions ? A union B, so we did number of A plus number of B. But that was a disjoint

set. Okay? Remember this. A disjoint set only has the full up to this point. And this is a finite set. So, we added

minus total, which means minus the also thing in both, N A intersection B, right? The also thing has come in. Now,

if we look at an example, we have: Suppose 1, 10 students in college, 30 students are on list A, 35 students are

on list B. Okay? One class is A, the other is B, and we have a total of 110. We have the total. One list is also

there, the other list is also there. Students and are both, so students are on both. 20 students are on both. Okay?

The ones in A and B. Next, find the number of students on lists A and B. Both A and B. Now, look, if you want to

find both, you need to add both. But this also includes "on both." Okay? "on both." So "on both," or "also," means "

both." What was in the previous one? It also said "they are also in both." So " also" means "both," and "both" also

means "both." This will remind you of the intersection. And when it says "add both," it means how many students are

there in A and B in total? So, we need to add them. We'll find out by adding A and B. And then we subtract what's in

both. Intersection A = intersection B. So, where did the 30 go? 30 are the students in list A. 35 are the students

in list B. And how much are the two? 20 on both. So, 20 is in both. So, if you subtract this and add it, you get 45.

Okay? Now, it's the same simple thing. You might have been given the data, but you might not know which formula to

apply to which. So, keep in mind that if you were told that this data is like this: A is also a student, you need to

find A and B. And you and both are also students. That means both should be included and add should also be

included. Add this too. So, for both, you would use the intersection formula, and for addition, you would use a union

b, meaning this formula. But if you were just told that there are disjoint sets and you wanted to tell how many

numbers are in a, then you would do an + b n. n b. Okay? And if you have the difference, then you would then add the

intersection formula to the difference, which we did earlier. This is what we did with the difference. You have to

subtract the intersection from the set. Okay? That includes "also." For "also," only "this" appears, and only one set

is given. You're given one class. You're asked for one class. And if you're asked for both, you have to add

them, so you have to use the add formula. Okay? Add and intersection. So , only one class is given, and you're

told to take the total and find the second number. So, you have to do this. It's pretty easy. It's a little

confusing, but you have to use your brain and understand how it came about. Okay? If you still have any problems,

please comment. I'll explain it to you. Okay. What is a set, a class of sets? A class of sets. How do we find the class

of a set? It must be within that set. Is it consisting of a set of sets? Okay ? It must be a set of sets. If we

consider some set in the given class of set, then it is called a subset. Well, this example will clarify things for

you. This might be a bit confusing. We have an example where the value S is equal to the set 1 2 3 4 class which

contains three elements. Now you have to create a class for it which has three elements. This is a total of four

, right? So you have to create a class of three elements. So this big bracket is used for a class. Okay? Then you

have to create sets inside it. You have to create three. You took 1 2 3. There should be three elements. Okay? They

said three elements, so you have to put three. Do it 1 2 3. Do it 1 3 4 or 1 2 3. It's up to you. You can create as

many classes as you want. You have to create classes by moving the numbers forward and backward. You made it 1 2

4. You made it 1 3 4. Then we added 2, 3, 4. Right? You have to do all of them . And all the people who are inside

this bracket, that's all the classes of this big set. Right? Now if you create again from within this class, it will

be called a subclass. Right? What has this become? The first class has been created. Now you've been told again

that you have to create elements from this class in which two is fixed, meaning two is inevitable. The number

two is inevitable. So then see what happens. This is our question: create a class of subsets. Now we have to create

a subset. We have to create a class of sets, a subset. We have to create a subset of the class. Earlier we created

a class of sets. We had the set 1, 2, 3 , 4, and we created a class for that. Right? We created that inside the big

bracket. Now we have to create a subset of that class. So what does the subset contain? It's said that two should be

fixed, and we can take two or any other element. So you've kept two fixed. We took one three. Then we put two fixed.

We took one four. Then we put two fixed . We took three four. Okay? So these are the three we have made. So what

will this be? B is a subclass of which one? Of class a. First we had S. Then we made A. We made a class. So what is

B? A subclass of which one? Of class a. Since B is a subset of class a. This becomes a subset of class a. Okay?

First we have a set. First we have a set. A class is formed from a set, and then if we create something further

from a class, it becomes its subclass. Okay? Hope so got it. Okay, next comes the power set. A power set is very easy

. The power set of a set is a subset of A. A given by PS. Okay, this won't be understood as a normal definition or

something to memorize. So you'll understand it with an example. What example do we have? We have a set 1 2 3

, we have to create power sets of it. Power set means we have to create how many subsets can be formed from this

set. Now you know that every set is bound to have an empty set, right? If an empty set is bound to be formed,

then it will be formed. So, how do you know how many sets you have to create? The formula for this is 2 to the power

n. n is the total number we have. We don't count infinity. We have 1 2 3, and we write three above it. So,

multiplying two by three equals eight. So, we have eight sets. Okay? Now, we've created an empty set first. Then

we'll create each one once. Then we'll create two each. One by one, and then each one separately in curly brackets.

Okay? First, one by one, then two by two, then three by two. Then we'll do two each. One by one. Then we'll do 1

by 3. Okay? Then we'll do 2 by 3. 1 by 3. That's it. Then we turned two into three. And finally, we're all two by

two. Now we have to do three by three, so this is three, so 1 2 3 is left alone at the end. Okay? And finally, we

need to put this inside the big bracket and turn it into one. Okay? These are power sets. Whatever you have inside

the brackets, you need to do it one by one. First, you need to write the empty set, of course. And then write each one

once. Then write each one two by two. Then three by three, then four by four, according to the number you've given.

This way, you can figure out how many you need to make in total. Now, look, this is eight. 1 2 3 4 5 6 7 8 Okay? So

, eight will be its subsets. This concludes our lecture. It was a short lecture. I hope you understood. If

you're confused about anything, please comment. I've clarified your concepts. If you want to give more examples, you

can use the three theorems we've already proposed to further this theorem, for your own understanding.

Okay? Please allow me. Allah Hafiz.

What is a finite set and how is it different from an infinite set?

How do you count elements in the union of two disjoint finite sets?

What is the complement rule for finite sets and when is it used?

How do you calculate the number of elements only in set A but not in set B?

What is the general formula for counting elements in the union of two finite sets that may overlap?

What is a class of sets and a subclass?

What is a power set and how many subsets does it contain?

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