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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.

JohnK
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  • Misleading title... What if you used a contradiction? – abiessu Nov 04 '13 at 15:03
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    Maybe you can use the well-order of the natural numbers. – Stefan Hamcke Nov 04 '13 at 15:03
  • Thank you for your responses but since I am a begginer, could you please be a little more specific? – JohnK Nov 04 '13 at 15:04
  • "countable finite or empty" There is something wrong in your question. – Michael Greinecker Nov 04 '13 at 15:06
  • @MichaelGreinecker It is taken exactly as it is from the book "Understanding Analysis" by S. Abbott, nothing wrong about it. – JohnK Nov 04 '13 at 15:07
  • When we talk about countability infinite, we mean that we can "count" a set - in other words, we can assign the natural numbers to the elements of B in such a way that we know that a) each element in B is counted at least once, and b) each element in B is counted at most once. This is, formally, called a bijection between the naturals and B, where condition a) is called surjectivity and condition b) is called injectivity. Can you use this? – HJ32 Nov 04 '13 at 15:07
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    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 $f: 1↦\min i[A],\ 2↦\min i[A]\setminus{f(1)},\ 3↦\min i[A]\setminus{f(1),f(2)},...$ – Stefan Hamcke Nov 04 '13 at 15:08
  • @Ioannis : of course, the empty set is finite. – Stefan Smith Nov 04 '13 at 15:09
  • Then the book is wrong. – Michael Greinecker Nov 04 '13 at 15:09
  • @Ioannis : you probably mean "$A$ is countably infinite, finite, or empty". Some would say I should edit the question, but I think OP's should fix their questions. This is redundant because the empty set is finite. This goal is equivalent to $A$ being countable, so it may be easier just to prove $A$ is countable, depending on what definition of "countable" you're using (I have seen definitions in good books that make this problem difficult). – Stefan Smith Nov 04 '13 at 15:13
  • @Ioannis : you improved your question, but it is still less than ideal, since all finite sets are countable, and the empty set is finite. Also, the phrase "one-to-one correspondence" is generally taken to mean that there is a bijection, which is inappropriate here because $A$ might be finite. – Stefan Smith Nov 04 '13 at 15:14
  • The question makes sense now with the edit by Lord_Farin. – Michael Greinecker Nov 04 '13 at 15:41
  • @StefanH Thanks for your response. If I got your response right then I should define $$ n_1=min{n \in \mathbb{N}: f(n)\in A} $$ and then move on excluding the given values of $$f(n)$$, yes? – JohnK Nov 04 '13 at 20:11
  • It should be $f(n+1)=\min{k\in i[A]\mid k>f(n)}$ for $n\ge1$ and $f(1)=\min i[A]$. – Stefan Hamcke Nov 04 '13 at 22:03
  • @StefanH Thank you Stefan. Stick around ;) – JohnK Nov 04 '13 at 22:05
  • You can prove by induction that for each $n$ we have ${f(1)<f(2)<...<f(n)}\subset i[A]$ and contains each $i(a)$ less than $f(n)$, and that for each $a\in A$ there is some $n$ such that $f(n)=i(a)$. Hence you get a bijection from $\Bbb N$ to $i[A]$, which is itself in bijection to $A$. – Stefan Hamcke Nov 04 '13 at 22:10
  • @StefanH Okay, how would you go ahead and prove that? Any help is greatly appreciated, me being a beginner. – JohnK Nov 04 '13 at 22:16
  • I have made the comments into an answer :) – Stefan Hamcke Nov 04 '13 at 22:34

5 Answers5

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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, $|A|\leq\aleph_0$.

1233dfv
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    Your response is very intuitive and thank you for it but I fear that I am required to produce a bisection for that exercise. – JohnK Nov 04 '13 at 20:17
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    To me, it seems that the question is precisely: Why does $|A|\leq|B|$ follow from $A\subset B$ ? – Loic Jul 17 '22 at 11:35
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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 $f(1),f(2),...,f(n-1)$.

Stefan Hamcke
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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 $A$?

GAM
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We start with a bijective mapping

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

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.

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

hhsaffar
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