12.3: Relative Sizes of Sets
We have defined a set \(A\) to be finite when we can count its elements by matching them bijectively with the elements of some counting set \(\mathbb{N}_{<m}\text{.}\) And in this case, by defining \(\vert A \vert = m\text{,}\) we are declaring that \(A\) has the same “size” as \(\mathbb{N}_{<m}\text{.}\)
Expanding on this idea, we can think of every bijection as using the elements of one set to “count” the elements of another.
sets \(A\) and \(B\) for which there exists a bijection \(A \to B\)
If \(B\) has the same size as \(A\text{,}\) then \(A\) has the same size as \(B\text{.}\)
- Proof.
-
If \(f: A \rightarrow B\) is a bijection, then so is \(f^{-1} : B \rightarrow A\text{.}\)
If \(A\) has the same size as \(B\) and \(B\) has the same size as \(C\text{,}\) then \(A\) has the same size as \(C\text{.}\)
- Proof.
-
This is left to you as Exercise 12.6.5 .
We expect our general notion of same size to match with just counting elements of finite sets and getting the same result.
Assume \(A\) and \(B\) are finite sets. Then \(\vert A \vert = \vert B \vert\) if and only if \(A\) and \(B\) have the same size.
- Proof.
-
Assume equal cardinality, show same size.
Assume \(\vert A \vert = \vert B \vert = m\text{.}\) Then by definition there exist bijections \(f: \mathbb{N}_{<m} \rightarrow A\) and \(g: \mathbb{N}_{<m} \rightarrow B\text{.}\) Now \(g \circ f^{-1}\) is a bijection \(A \to B\text{,}\) so \(A\) and \(B\) have the same size according to the technical definition.
Assume same size, show equal cardinality.
Assume \(A\) and \(B\) have the same size. Then by definition there exists a bijection \(f: A \rightarrow B\text{.}\) Now, we have also assumed that \(A\) is finite, so there exists a bijection \(g: \mathbb{N}_{<m} \rightarrow A\text{,}\) where \(m = \vert A \vert\text{.}\) Then \(f \circ g: \mathbb{N}_{<m} \rightarrow B\) is a bijection that demonstrates \(\vert B \vert = m\) as well.
Your intuition may fail you when considering “sizes” of infinite sets. In particular, it is possible to have \(\vert A \vert = \vert B \vert = \infty\text{,}\) where \(A\) and \(B\) do not have the same size.
Even though \(\mathbb{N} \subsetneqq \mathbb{Z}\text{,}\) \(\mathbb{N}\) and \(\mathbb{Z}\) have the same size! The following defines a bijection \(f: \mathbb{N} \rightarrow \mathbb{Z}\text{.}\)
| \(n\) | \(0\) | \(1\) | \(2\) | \(3\) | \(4\) | \(\cdots\) |
| \(f(n)\) | \(0\) | \(-1\) | \(1\) | \(-2\) | \(2\) | \(\cdots\) |
This bijection can be expressed by the formula
In Chapter 13, we will see that even though \(\vert \mathbb{N} \vert = \vert \mathbb{R} \vert = \infty\text{,}\) the sets \(\mathbb{N}\) and \(\mathbb{R}\) do not have the same size!
Recall from first-year calculus that for \(a,b \in \mathbb{R}\) with \(a \lt b\text{,}\) we define the open interval from \(a\) to \(b\) to be the set of all real numbers strictly between \(a\) and \(b\text{:}\)
\begin{equation*} (a,b) = \{x \in \mathbb{R} \vert a \lt x \lt b\} \text{.} \end{equation*}
It turns out that, even though they may have different lengths , the interval \((a,b)\) and the unit interval \((0,1)\) have the same size ! (That is, they somehow contain the same “number” of numbers.)
Construct a bijection \((0,1)\to(a,b)\) in two steps.
-
The map
\begin{align*} f \colon (0,1) & \to (0,b-a), \\ x & \mapsto (b-a)x, \end{align*}is a bijection. (Check!)
-
The map
\begin{align*} g \colon (0,b-a) & \to (a,b), \\ x & \mapsto x+a, \end{align*}is a bijection. (Check!)
Then \(g\circ f: (0,1) \rightarrow (a,b)\) is a bijection.
Define
\begin{align*} S & = \{(x,y) \in \mathbb{R}^2 \vert x^2 + (y-\dfrac{1}{2})^2 = \dfrac{1}{4} \} \text{,} & \hat{S} & = S \setminus \{(0,1)\}\text{.} \end{align*}
Here, \(S\) is a circle in the plane with radius \(\dfrac{1}{2}\) and centre \((0,\dfrac{1}{2})\text{,}\) and \(\hat{S}\) is the circle \(S\) “punctured” at the “north pole”.
We claim that \(\hat{S}\) has the same size as \(\mathbb{R}\text{.}\) Construct a bijection \(\hat{S}\to\mathbb{R}\) in two steps.
- Let \(X\) represent the \(x\)-axis in the plane, i.e.
\begin{equation*} X = \{(x,0) \vert x \in \mathbb{R} \} \subseteq \mathbb{R}^2 \text{.} \end{equation*}
Let \(f: \hat{S} \rightarrow X\) be defined as follows: for \((x,y)\in \hat{S}\text{,}\) let \(f(x,y)\) be the \(x\)-intercept of the line through points \((0,1),(x,y)\text{.}\)
Then \(f\) is a bijection. (Check!)
- We also have a bijection \(g: X \rightarrow \mathbb{R}\) by \(g(x,0) = x\text{.}\)
Therefore, the composition \(g\circ f : \hat{S} \rightarrow \mathbb{R}\) is a bijection.
Example \(\PageIndex{3}\) and Example \(\PageIndex{4}\) can be combined to demonstrate that every finite-length interval \((a,b)\) of real numbers has the same size as the entire set \(\mathbb{R}\) of real numbers. See Exercise 12.6.6 .