12.1: Finite Sets
Recall.
For \(m\in\mathbb{N}\) we have defined the counting set
Clearly, \(\mathbb{N}_{<m}\) contains exactly \(m\) elements. In fact, we have defined the number \(m\) to be the set \(\mathbb{N}_{<m}\text{.}\) (See Example 11.4.2 .)
As the terminology implies, we will use these sets to count the elements of other sets. In particular, given a set \(A\text{,}\) if we can match up the elements of \(A\) with the elements of \(\mathbb{N}_{<m}\text{,}\) one for one, then \(A\) must also contain exactly \(m\) elements.
a set \(A\) for which there exists a bijection \(\mathbb{N}_{<m} \to A\) for some \(m \in \mathbb{N}\text{,}\) \(m \gt 0\)
For finite set \(A\) there exists one unique natural number \(m\) for which a bijection \(\mathbb{N}_{<m} \to A\) exists.
Suppose \(A\) is finite. While there is only one number \(m\) for which a bijection \(\mathbb{N}_{<m} \to A\) exists, there can be many such bijections, and the number of bijections increases as \(m\) increases.
Prove Fact \(\PageIndex{1}\).
the unique natural number \(m\) for with a bijection \(\mathbb{N}_{<m} \to A\) exists
the cardinality of the finite set \(A\)
alternative notation for the cardinality of the finite set \(A\)
alternative notation for the cardinality of the set defined by \(\{\dots\}\)
For \(\Sigma = \{a, b, \ldots, z\} \text{,}\) we have \(\vert \Sigma \vert = 26\text{.}\) Below are two example bijections \(\varphi,\psi: \mathbb{N}_{<26} \rightarrow \Sigma\) that verify this cardinality number.
| \(\sigma\) | \(0\) | \(1\) | \(2\) | \(3\) | \(\cdots\) | \(24\) | \(25\) |
| \(\varphi(\sigma)\) | \(\text{a}\) | \(\text{b}\) | \(\text{c}\) | \(\text{d}\) | \(\cdots\) | \(\text{y}\) | \(\text{z}\) |
| \(\psi(\sigma)\) | \(\text{a}\) | \(\text{z}\) | \(\text{b}\) | \(\text{y}\) | \(\cdots\) | \(\text{m}\) | \(\text{n}\) |
Cardinality of an empty set.
What about the empty set? Clearly we should have \(\vert \varnothing \vert = 0\text{.}\) But is this consistent with our definition of cardinality?
a function with domain \(\varnothing\)
If we accept the existence of empty functions \(\varnothing \to X\) for every set \(X\text{,}\) then the properties of such functions that we need in order to establish \(\vert \varnothing \vert = 0\) will be vacuously true.
- For every set \(X\text{,}\) an empty function \(\varnothing \to X\) is injective.
- An empty function \(\varnothing \to \varnothing\) is a bijection.
- Proof
-
You were asked to verify these statements in Exercise 10.7.12 .
The cardinality of the empty set is \(0\text{.}\)
- Proof.
-
We are required to demonstrate an example of a bijection \(\mathbb{N}_{<0} \to \varnothing\text{.}\) But
\begin{equation*} \mathbb{N}_{<0} = \{n \in \mathbb{N} \vert n \lt 0 \} = \varnothing \text{,} \end{equation*}
so Statement 2 of Proposition \(\PageIndex{1}\) tells that the empty function \(\mathbb{N}_{<0} \to \varnothing\) is indeed a bijection.