# Proving : Every infinite subset of countable set is countable [duplicate]

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An infinite subset of a countable set is countable

I was going through rudin when I encountered the proof for "Every infinite subset of a countable set is countable".

What he states is that we take a set $A$. Let $E \subset A$ be an infinite subset. We need to prove that $E$ is countable.

The book states that we should pick the smallest number of A and then proceed picking smaller numbers and simultaneously indexing them with $J_n = \{1,2,3,\cdots,n\}$.

What I don't understand is how can we pick a minimum out of countably infinite numbers?

Suppose I define $A = \{ y | y =\dfrac{1}{n} \forall n\in\mathbb{N}\}$, we can never get a minimum. 0 can't be the minimum since the domain is assumed to be $[1,\infty)$.

The logic of maximum also won't work because we can define the set $\mathbb{N}$ itself. We can't pick a maximum.

Am I missing something?

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## marked as duplicate by Asaf Karagila, Marvis, MJD, Martin Sleziak, t.b.Jul 10 '12 at 10:57

The duplicate suggestion aside, there are a dozen more questions on this same question under different guises here. – Asaf Karagila Jul 9 '12 at 22:00
Here it is again – MJD Jul 9 '12 at 22:48
You wrote: What I don't understand is how can we pick a minimum out of countably infinite numbers? Probably in the book this is done for subset of $\mathbb N$, and this can be done since the set of positive integers is well-ordered. This fact is closely related to the mathematical induction on the set $\mathbb N$, see e.g. this question Proving that the set of natural numbers is well-ordered. – Martin Sleziak Jul 10 '12 at 10:53

The minimum he means may be referring to the index. Since the index is the natural numbers the minimum makes sense. More precisely.

Suppose $A$ is countable. Then there exists a sequence $(a_n)$ of all the elements of $A$. Let $B \subset A$ be infinite. Then there is a smallest $n_0 \in \mathbb{N}$ such that $a_{n_0} \in B$. Since $B$ is infinite, there exists a next smallest $n_1 > n_0$ such that $a_{n_1} \in B$. And so forth. The sequence $(a_{n_i})_{i \in \mathbb{N}}$ is the desired bijection between $\mathbb{N}$ and $B$.

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You’ve badly misunderstood what he wrote. I’ll expand his proof, emphasizing where it differs from your understanding of it.

We start with the countably infinite set $A$. Since it’s countably infinite, there is a bijection $g:\Bbb Z^+\to A$. For each $n\in\Bbb Z^+$ let $x_n=g(n)$; then $A=\{x_n:n\in\Bbb Z^+\}$. These are the elements $x_n$ of Rudin’s proof; $\{x_n:n\in\Bbb Z^+\}$ is called an enumeration of $A$. The elements of $A$ can be anything; in particular, they need not be numbers of any kind. But they are now numbered, or indexed, by the positive integers, and we can compare those indices.

We also have the infinite set $E\subseteq A$. Imagine going through the listing $\langle x_1,x_2,\dots\rangle$ of $A$ one element at a time; eventually you must reach an element of $E$, since $E\ne\varnothing$. Let $I_1=\{n\in\Bbb Z^+:x_n\in E\}$. $I_1$ is a set of positive integers, so it has a smallest member; call that smallest member $n_1$. Now let $I_2=I_1\setminus\{n_1\}$, all of $I_1$ except the smallest member $n_1$. $I_2$ is another set of positive integers, so it also has a smallest element; call that element $n_2$.

In general, suppose that you’ve chosen $n_1,\dots,n_{k-1}$ so that $x_{n_1},\dots,x_{n_{k-1}}\in E$. Let $I_k=I_1\setminus\{n_1,\dots,n_{k-1}\}$. $E$ is infinite, so $E\setminus\{x_{n_1},\dots,x_{n_{k-1}}\}\ne\varnothing$, and therefore $I_k\ne\varnothing$. Thus, we may let $n_k$ be the smallest member of $I_k$ and so continue constructing the sequence $\langle n_1,n_2,n_3,\dots\rangle$.

All we’re really doing here is going through $A$ in the order $x_1,x_2,x_3,\dots$ until we come to the first element of $E$; that element is $x_{n_1}$ for some positive integer $n_1$. Then we continue through $A$ until we come to another element of $E$; that element is $x_{n_2}$ for some positive integer $n_2>n_1$. At each stage we’ve found only finitely many members of $E$, and $E$ is infinite, so we can keep going. It’s intuitively clear that this procedure cannot miss any elements of $E$, because at each stage we take the first one in the listing that we haven’t already taken.

Thus, at the end of all this the map $f:\Bbb Z^+\to E:k\mapsto x_{n_k}$ is a bijection between $\Bbb Z^+$ and $E$, so $E$ is countably infinite.

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A very nice and clear answer. Thank you. – BU982T May 1 '13 at 19:22
You're very welcome. – Brian M. Scott May 1 '13 at 23:17
I have been studying the same subject and I found your answer very helpful as well. It's an example of clarity and conciseness. – JohnK Nov 5 '13 at 10:34
@Ioannis: Thank you. – Brian M. Scott Nov 5 '13 at 10:38
@BrianM.Scott . I have a small question. It seems that this listing 'process' is done at our own will. If that is the case, why does it have to be an infinite countable subset? Why not a finite countable subset? I could apply the process 3 times for example and leave myself with a subset containing 3 countable elements. – Gustavo Montano Mar 7 '14 at 13:50

$A$ is assumed countable, which means that there exists a bijection $f:A\to\mathbb N$. When the proof speaks of an ordering on $A$, that is implicitly (?) assumed to be the ordering induced by $f$, such that $a_1$ counts as "smaller tham" $a_2$ if and only if $f(a_1)<f(a_2)$.

In your example with $A=\{\tfrac 1n\mid n\in\mathbb N\}$ the relevant ordering of the elements of $A$ is not the ordering they inherit as members of $\mathbb R$, but the ordering mediated by the map $n\mapsto \frac 1n$. So in that case you need to consider $1/2$ as coming "before" $1/10$, for example.

If you find this confusing, you can instead start by proving the property only for $\mathbb N$ itself -- that is, that every infinite subset of $\mathbb N$ is countable. In the general case where you have $E\subseteq A$, consider this to $f(E)$, which is a subset of $\mathbb N$. Once you know (from the version you have already proved) that $f(E)$ is countable -- that is, that there exists a bijection $g:f(E)\to\mathbb N$ -- you can see that the composition $g\circ f$ is a bijection $E\to\mathbb N$.

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Rudin doesn't pick the minimum of $E$ in his proof. He picks the smallest index that corresponds to an element in $E$. Let me illustrate:

Let $E \subset A$ and $A$ is countable. Since $A$ is countable, there is a 1:1 mapping $f$ of $J$ onto $A$. ($J$ is the set of all positive integers)

Write the elements of $A$ in the order they are mapped to $J$. In other words, write:

$$f(1), f(2), f(3), \cdots$$

We're going to use the list above to build a 1:1 mapping from $J$ onto $E$. Scan the list and pick the first element that belongs to $E$. This is the image of $1$ in the 1:1 mapping we're building. Continue scanning and pick the second element that belongs to $E$. This is the image of $2$.

Continue ad infinitum and you'll have the mapping required to prove that $E$ is countable.

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I think Rudin meant the element with the smallest index. Countability has to do with an ordering of the set, not with how its elements might be ordered with respect to some other metric.

Suppose you have a non-empty subset $E$ of a countable set $A=\{a_k\}_{k=1}^\infty$. Here is a finite duration procedure that will pick the first element of $E$ in $A$.

1. Start with $k=1$.

2. If $a_k\in E$, that is your pick; otherwise, increase $k$ by $1$ and go back to Step 2.

Since $E$ is non-empty, it must contain some element $a_n$. The procedure above will terminate after at most $n$ cycles.

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Let your set S be countable. Then, let S' be an infinite subset of that set. Let $\phi:\mathbb{N}\rightarrow S$. Then, select the first member of the image of the list that is in your set. Define that to be the first member of your new sequence. Repeat this again and again. Each time you will have more to list, as S is infinite. Note that by doing this, you obtain a bijection, as $\phi$ is 1-1. Moreover, it's clear that you can get every member of your subset this way.

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