13.2: Properties
The following facts outline some relationships countability and the set operations. They can be used to more easily prove that a set is countable or uncountable using the already-known countability or uncountability of a related set.
- Every subset of \(\mathbb{N}\) is countable.
- If there exists an injection \(A \hookrightarrow \mathbb{N}\text{,}\) then the set \(A\) is countable.
- Suppose \(A \subseteq B\text{.}\) If \(B\) is countable, then so is \(A\text{.}\)
- Suppose \(A \subseteq B\text{.}\) If \(A\) is un countable, then so is \(B\text{.}\)
- If \(A\) and \(B\) are countable, then \(A\cup B\) and \(A\cap B\) are both countable.
- Proof outline.
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- Assume \(A \subseteq \mathbb{N}\text{.}\) If \(A\) is finite, then it is countable by definition. So assume that \(\vert A \vert = \infty\text{.}\) We can construct a sequence \(\{a_k\}\) that contains each element of \(A\) exactly once as follows.
\begin{align*} a_0 & = \text{ smallest number in } A, \\ a_1 & = \text{ next smallest number in } A, \\ a_2 & = \text{ next smallest number in } A, \\ & \; \vdots \end{align*}
Therefore, \(A\) is countable.- If \(f: A \hookrightarrow \mathbb{N}\) is injective, then \(f: A \rightarrow f(A)\) is a bijection, so that \(A\) and its image \(f(A)\) have the same size. But \(f(A)\) is countable by Statement 1, so using the definition of countable along with Fact 12.3.2 , conclude that \(A\) is countable.
- If \(B\) is countable, then by definition there exists a bijection \(f: B \rightarrow \mathbb{N}\text{.}\) Then \(f\vert _A: A \rightarrow \mathbb{N}\) is an injection. Apply Statement 2.
- This is the contrapositive of Statement 3, under the common assumption \(A \subseteq B\text{.}\)
- For \(A \cap B\text{,}\) consider \(A \cap B \subseteq A\) and apply Statement 3.
Now consider \(A \cup B\text{.}\) For simplicity, we will assume \(A \cap B = \emptyset\text{,}\) so that \(A \cup B = A \sqcup B\text{.}\) Since \(A\) and \(B\) are countable, we can write their elements as sequences:
\begin{align*} A & = \{ \, a_0, \, a_1, \, a_2, \, \ldots \, \}, & B & = \{ \, b_0, \, b_1, \, b_2, \, \ldots \, \} \text{.} \end{align*}
We can then write the elements of \(A \sqcup B\) in a sequence by interleaving these two sequences:\begin{equation*} A \sqcup B = \{ \, a_0, \, b_0, \, a_1, \, b_1, \, a_2, \, b_2, \, \ldots \, \} \text{.} \end{equation*}
Checkpoint \(\PageIndex{1}\)
Prove \(A \cup B\) is countable even in the case \(A \cap B \ne \emptyset\text{.}\)
- Hint.
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Consider the sets
\begin{align*} A' & = A \smallsetminus (A\cap B), & B' & = B \smallsetminus (A\cap B), & C & = A' \sqcup B'. \end{align*}
Then \(A\cup B\) is the disjoint union of \(C\) and \(A\cap B\text{.}\)
The set of prime numbers is countable, since it is a subset of \(\mathbb{N}\text{.}\)
The unit interval \((0,1)\) on the real number line is uncountable because it contains the uncountable subset \(\scr{C}\) from Lemma 13.1.1 .
Set \(\mathbb{R}\) is uncountable.
- Proof
-
This follows from Lemma 13.1.1 and Statement 4 of Proposition \(\PageIndex{1}\).
The Cartesian product set \(\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}\) is uncountable because it has an uncountable subset: the \(x\)-axis has the same size as \(\mathbb{R}\text{.}\)