# Subset of a countable set is itself countable

How is it proved that if $$A \subset B\ \text{with}\ B\ \text{countable}$$ then $A$ is either countable, finite, or empty? I think the proof involves a $1-1$ correspondence between $\mathbb{N}$ and $A$ but other than that I do not know how to proceed.

EDIT: I have checked the solution and it advises me to proceed as follows. " As a start to a definition of $$g: \mathbb{N} \rightarrow A$$ set $$g(1)=f(n_1)$$ Then show how to inductively continue this process to produce a $1-1$ function $g$ from $\mathbb{N}$ onto $A$."

So the proof according to my book involves induction.

• Misleading title... What if you used a contradiction? Commented Nov 4, 2013 at 15:03
• Maybe you can use the well-order of the natural numbers. Commented Nov 4, 2013 at 15:03
• Thank you for your responses but since I am a begginer, could you please be a little more specific? Commented Nov 4, 2013 at 15:04
• "countable finite or empty" There is something wrong in your question. Commented Nov 4, 2013 at 15:06
• The inclusion of $A$ in $B$ gives via the bijection $B\equiv\Bbb N$ an injection $i:A\to\Bbb N$. Now you can define a function $f: 1↦\min i[A],\ 2↦\min i[A]\setminus\{f(1)\},\ 3↦\min i[A]\setminus\{f(1),f(2)\},...$ Commented Nov 4, 2013 at 15:08

Via the bijection $\Bbb N\cong B$ we have an injection $i:A\to\Bbb N$.

Define $$f(n+1)=\min\{k∈i[A]∣k>f(n)\}$$ for $n≥1$ and $$f(1)=\min i[A]$$

We claim that for each $n$ we have $\{f(1)<f(2)<...<f(n)\}$ is a subset of $i[A]$ and contains each $i(a)$ less than $f(n)$.
Proof: Induction over $n$:
For $n=1$ clearly $f(1)∈i[A]$, and since there is no $i(a)<f(1)$, the claim is true.
Assume that the statement is true for $n$. Then by definition of $f$, the number $f(n+1)$ is larger than $f(n)$, so the set $\{f(1)<...<f(n)<f(n+1)\}$ is a subset of $i[A]$. Since $f(n+1)$ is the minimal element of $i[A]$ larger than $f(n)$, it must contain each $i(a)$ less than $f(n+1)$.

So we have shown that $f:\Bbb N↦i[A]$ is an injection. Now, for $a\in A$ we have the natural number $l=i(a)$. Since $l$ is less than $f(n)$ for some $n\in\Bbb N$, $f$ being strictly increasing, it must thus be one of $f(1),f(2),...,f(n-1)$.

Any subset of a countable set is countable.

Take $A\subset B$ where $B$ is countable. Then $|A|\leq|B|$ since $A\subset B$. By definition, $|A|\leq|B|$ if there is a one-to-one function from $A$ into $B$. We also see that $|B|\leq\aleph_0$ since $B$ is countable. Thus, $|A|\leq\aleph_0$.

• Your response is very intuitive and thank you for it but I fear that I am required to produce a bisection for that exercise. Commented Nov 4, 2013 at 20:17
• To me, it seems that the question is precisely: Why does $|A|\leq|B|$ follow from $A\subset B$ ?
– Loic
Commented Jul 17, 2022 at 11:35

Hint: Define an injection $f:B \to \mathbb{N}$ (this is possible as B is countable). Define the inclusion mapping $I:A \to B$. Consider $f\circ I:A \to \mathbb{N}.$ What can you say about $I$? What can you then say about $f \circ I$? What can you then conclude about $A$?

$$\tag 1 f: \Bbb N \cong B$$

Using $$f$$ we define another function

$$\tag 2 \Phi: \mathcal P(B) \setminus \emptyset \to B$$ $$\quad \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad X \mapsto f\bigr(\,\text{min}(f^{-1}(X))\,\bigr)$$

If $$A \subset B$$ and $$A$$ is not finite, we can recursively build another function $$g: \Bbb N \to A$$ by setting

$$\quad g(1) = \Phi(A)$$

and with $$g(1), g(2), \dots, g(n)$$ set, defining

$$\quad g(n+1) = \Phi\bigr(A \setminus \{g(1), g(2), \dots, g(n)\}\bigr)$$

Exercise 1: Show that $$g$$ is a well-defined (recursive) function.

Exercise 2: Show that $$g$$ is a bijective correspondence,

$$\tag 3 g: \Bbb N \cong A$$

Here is another proof.

A set $$C$$ is countable if there exist a surjection $$h: \Bbb N \to C$$. In fact, an explicit bijection $$\Bbb N \cong C$$ can be constructed using $$h$$ (see, for example, this).

Addressing the interesting situation, let $$B$$ be an infinite set contained in a countable set $$A$$. Let $$f: \Bbb N \cong A$$ exhibit the $$\text{1:1}$$ correspondence. Choose any element $$b_0 \in B$$ and define

$$F(m) = \left\{\begin{array}{lr} f(m), & \text{for } m \in f^{-1}(B) \\ b_0 , & \text{for } m \in \mathbb N \setminus f^{-1}(B) \end{array}\right\}$$

The mapping $$F: \Bbb N \to B$$ is a surjection.

Hint: Suppose $f$ is the function that maps $B$ to $\mathbb{N}$, sort each item in $B$ by it's $f$ value.

A paper by Hantao Zhang, https://homepage.divms.uiowa.edu/~hzhang/countable.pdf, states that any proof of "a subset of a countable set is countable" has a problem: Suppose N is the set of natural languages. For a Turing-recognizable and undecidable formal language L, if f: N -> L is a bijection, then f must be computable but cannot be increasing. If L is unrecognizable, then f must be uncomputable. Any known proof does not consider these two cases.