How To Find The Cardinality Of A Set

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1.2.3 Cardinality: Countable and Uncountable Sets

Here we demand to talk virtually cardinality of a set, which is basically the size of the set. The cardinality of a set is denoted by $|A|$. We first discuss cardinality for finite sets and then talk near infinite sets.

Finite Sets:

Consider a fix $A$. If $A$ has only a finite number of elements, its cardinality is simply the number of elements in $A$. For example, if $A=\{ii,4,6,8,10\}$, and then $|A|=v$. Earlier discussing infinite sets, which is the main give-and-take of this section, we would like to talk nearly a very useful rule: the inclusion-exclusion principle. For two finite sets $A$ and $B$, we have $$|A \loving cup B |=|A|+|B|-|A \cap B|.$$ To see this, note that when we add $|A|$ and $|B|$, we are counting the elements in $|A \cap B|$ twice, thus by subtracting it from $|A|+|B|$, nosotros obtain the number of elements in $|A \cup B |$, (y'all can refer to Figure one.16 in Problem 2 to run across this pictorially). We can extend the same thought to three or more sets.

Inclusion-exclusion principle:

1. $|A \loving cup B |= |A|+|B|-|A \cap B|$,


2. $|A \loving cup B \loving cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$.

Generally, for $n$ finite sets $A_1, A_2, A_3,\cdots, A_n$, we can write

$$\biggl|\bigcup_{i=1}^n A_i\biggr|=\sum_{i=1}^north\left|A_i\correct|-\sum_{i < j}\left|A_i\cap A_j\correct|$$ $$\>\>\>\>\>\>\>+\sum_{i < j < k}\left|A_i\cap A_j\cap A_k\correct|-\ \cdots\ + \left(-1\right)^{due north+1} \left|A_1\cap\cdots\cap A_n\right|.$$

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Example

In a party,

• there are $10$ people with white shirts and $8$ people with red shirts;
• $iv$ people have black shoes and white shirts;
• $3$ people accept blackness shoes and reddish shirts;
• the full number of people with white or red shirts or black shoes is $21$.

How many people have black shoes?

• Solution

  • Let $W$, $R$, and $B$, be the number of people with white shirts, cherry shirts, and blackness shoes respectively. Then, here is the summary of the available information: $$|West|=10$$ $$|R|=8$$ $$|W \cap B|=iv$$ $$|R \cap B|=iii$$ $$|W \cup B \cup R|=21.$$ Also, it is reasonable to assume that $W$ and $R$ are disjoint, $|Due west \cap R|=0$. Thus by applying the inclusion-exclusion principle we obtain

    $|W \cup R \loving cup B|$	$ = 21$
    	$= |W| + |R| + |B|- |Westward \cap R| - |Due west \cap B| - |R \cap B| + |W \cap R \cap B|$
    	$=10+8+|B|-0-4-3+0$.

    
    Thus $$|B|=10.$$

    Notation that another style to solve this problem is using a Venn diagram every bit shown in Figure 1.11.

    

    Fig.ane.11 - Inclusion-exclusion Venn diagram.

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Space Sets:

What if $A$ is an infinite set? It turns out we need to distinguish between two types of infinite sets, where one type is significantly "larger" than the other. In particular, one blazon is chosen countable, while the other is chosen uncountable. Sets such as $\mathbb{N}$ and $\mathbb{Z}$ are called countable, but "bigger" sets such as $\mathbb{R}$ are called uncountable. The difference between the 2 types is that you can list the elements of a countable set $A$, i.due east., you lot tin can write $A=\{a_1, a_2,\cdots\}$, but y'all cannot list the elements in an uncountable set. For example, you tin can write

• $\mathbb{N}=\{ane,ii,3,\cdots\}$,
• $\mathbb{Z}=\{0,i,-1,ii,-ii,three,-3,\cdots\}$.

The fact that yous can list the elements of a countably space set means that the set tin be put in i-to-one correspondence with natural numbers $\mathbb{N}$. On the other hand, yous cannot listing the elements in $\mathbb{R}$, so it is an uncountable fix. To exist precise, here is the definition.

Definition

Prepare $A$ is chosen countable if i of the following is true

1. if it is a finite set, $\mid A \mid < \infty$; or
2. it can be put in one-to-one correspondence with natural numbers $\mathbb{North}$, in which case the set is said to be countably space.
A set up is called uncountable if it is not countable.

Hither is a simple guideline for deciding whether a set is countable or not. Equally far every bit applied probability is concerned, this guideline should be sufficient for virtually cases.

• $\mathbb{Northward}, \mathbb{Z}, \mathbb{Q}$, and any of their subsets are countable.
• Any set containing an interval on the real line such equally $[a,b], (a,b], [a,b),$ or $(a,b)$, where $a < b$ is uncountable.

The above rule is usually sufficient for the purpose of this book. However, to brand the statement more than concrete, hither we provide some useful results that help u.s.a. prove if a set is countable or non. If yous are less interested in proofs, you lot may make up one's mind to skip them.

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Theorem

Any subset of a countable prepare is countable.
Any superset of an uncountable set is uncountable.

Proof

The intuition behind this theorem is the following: If a gear up is countable, and so whatever "smaller" set up should also be countable, so a subset of a countable gear up should be countable as well. To provide a proof, we tin can argue in the following fashion.

Let $A$ be a countable set up and $B \subset A$. If $A$ is a finite set, then $|B|\leq |A| < \infty$, thus $B$ is countable. If $A$ is countably infinite, and then nosotros can listing the elements in $A$, then by removing the elements in the list that are not in $B$, we tin obtain a list for $B$, thus $B$ is countable.

The second part of the theorem tin be proved using the first office. Assume $B$ is uncountable. If $B \subset A$ and $A$ is countable, by the first part of the theorem $B$ is also a countable set which is a contradiction.

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Theorem

If $A_1, A_2,\cdots$ is a list of countable sets, so the fix $\bigcup_{i} A_i=A_1 \cup A_2 \cup A_3\cdots$ is likewise countable.

Proof

It suffices to create a list of elements in $\bigcup_{i} A_i$. Since each $A_i$ is countable we can listing its elements: $A_i=\{a_{i1},a_{i2},\cdots\}$. Thus, we have

• $A_1=\{a_{11},a_{12},\cdots\}$,
• $A_2=\{a_{21},a_{22},\cdots\}$,
• $A_3=\{a_{31},a_{32},\cdots\}$,
• ...

At present we need to make a list that contains all the above lists. This tin can be done in different ways. One way to practice this is to use the ordering shown in Effigy 1.12 to brand a list. Here, we can write $$ \bigcup_{i} A_i=\{a_{11}, a_{12},a_{21}, a_{31}, a_{22}, a_{13}, a_{fourteen}, \cdots \} \hspace{100pt} (1.1)$$

Fig.1.12 - Ordering to make a list.

We accept been able to create a list that contains all the elements in $\bigcup_{i} A_i$, and then this set is countable.

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Theorem

If $A$ and $B$ are countable, and so $A \times B$ is also countable.

Proof

The proof of this theorem is very like to the previous theorem. Since $A$ and $B$ are countable, we can write $$A = \{a_1, a_2, a_3, \cdots \},$$ $$B = \{b_1, b_2, b_3, \cdots \}.$$ Now, we create a list containing all elements in $A \times B = \{(a_i,b_j) | i,j=ane,2,3,\cdots \}$. The idea is exactly the same every bit before. Figure 1.13 shows ane possible ordering.

Fig.one.13 - Ordering to brand a list.

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The in a higher place arguments can be repeated for any set up $C$ in the class of $$C=\bigcup_i \bigcup_j \{ a_{ij} \},$$ where indices $i$ and $j$ vest to some countable sets. Thus, any set in this grade is countable. For example, a issue of this is that the fix of rational numbers $\mathbb{Q}$ is countable. This is because we tin can write $$\mathbb{Q}=\bigcup_{i \in \mathbb{Z}} \bigcup_{j \in \mathbb{N}} \{ \frac{i}{j} \}.$$

The higher up theorems confirm that sets such as $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ and their subsets are countable. Yet, every bit we mentioned, intervals in $\mathbb{R}$ are uncountable. Thus, you lot can never provide a list in the course of $\{a_1, a_2, a_3,\cdots\}$ that contains all the elements in, say, $[0,1]$. This fact tin can be proved using a so-called diagonal argument, and nosotros omit the proof hither as information technology is not instrumental for the remainder of the volume.

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The impress version of the book is available through Amazon hither.

Source: https://www.probabilitycourse.com/chapter1/1_2_3_cardinality.php

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