Proof:"Infinite subset of $\mathbb N$ is countable [duplicate]

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I've read a proof of the statement:"An infinite subset of $\mathbb N$ is countable; that is, if $A \subset \mathbb N$ and if $A$ is infinite, then $A$ is equivalent to $\mathbb N$." in Carothers' textbook and there is one part of the proof I don't understand.

Proof

Recall that $\mathbb N$ is well ordered. That is, each nonempty subset of $\mathbb N$ has a smallest element. Thus, since $A \ne \emptyset$, there is a smallest element $x_1 \in A$. Then $A \setminus \{x_1\} \ne \emptyset$, and there must be a smallest $x_2 \in A \setminus \{x_1\}$. But now $A \setminus \{x_1,x_2\} \ne \emptyset$, and so we continue, setting $x_3=\min(A \setminus \{x_1,x_2\})$. By induction, we can find $x_1,x_2,x_3,...,x_n,... \in A$, where $x_n=\min(A \setminus \{x_1,...,x_{n-1}\})$.

How do we know that this process exhausts $A$? Well, suppose that $x \in A \setminus \{x_1,x_2,...\} \ne \emptyset$. Then the set $\{k : x_k>x\}$ must be nonempty (otherwise we would have $x \in A$ and $x<x_1=\min A$), and hence it has a least element. That is, there is some $n$ with $x_1<...<x_{n-1}<x<x_n$. But this contradicts the choice of $x_n$ as the first element in $A \setminus \{x_1,...,x_{n-1}\}$. Consequently, $A$ is countable.

My questions

It is affirmed that the set $\{k : x_k>x\}$ can't be empty (I don't get the reason Carothers gives for this being impossible).

As I see it, if $\{k : x_k>x\}$ is empty, then $x_k \leq x$ for all $k$. So the set $\{x_1,x_2,...\}$ obtained by the method of extracting the smallest element from each remaining nonempty subset of $\mathbb N$ is finite, but we've seen that the set constructed from choosing $x_n=\min(A \setminus \{x_1,...,x_{n-1}\})$ is infinite. So, wouldn't this be the reason why the set $\{k : x_k>x\}$ can't be empty rather than "otherwise we would have $x \in A$ and $x<x_1=\min A$"?.

I couldn't even understand why would $x$ be in $A$ instead of $x \in A \setminus \{x_1,x_2,...\}$ or why $x<x_1$.

I know it is a very small part of the proof but I would like to understand all the steps of it so I would appreciate if someone could clear up my doubt.

marked as duplicate by Adam Hughes, Cookie, Andrés E. Caicedo, user147263, J. W. PerryAug 1 '14 at 6:01

• If you do not want to think about why the process exhaust $A$, you can also observe that the function $f : \mathbb{N} \rightarrow A$ defined by $f(k) = x_k$ is an injection. The inclusion $i : A \rightarrow \mathbb{N}$ is also an injection. Hence the Cantor-Schroder-Bernstein Theorem assert there exists a bijection between $\mathbb{N}$ and $A$. – William Aug 1 '14 at 5:36
• I agree that the paragraph you quote is not perfect. But it says that if the set is empty, then for some $n$, $x<x_n$ but from the choice for $x_n$ this means $x<x_{n-1}$ and so on, and then we get $x<x_1$ which is the contradiction the author points to. – Asaf Karagila Aug 1 '14 at 6:19
I'll try to break things down even further. Suppose that $A \setminus \{x_1, x_2, \ldots\}$ is nonempty so that there exists some $x \in A \setminus \{x_1, x_2, \ldots\}$. Then we know that $x \in A$ and for all $i \in \mathbb N$, we have that $x \neq x_i$. Now we want to show that the set: $$S = \{k \in \mathbb N : x_k > x\}$$ must be nonempty. To see this, we argue by contradiction.
Suppose instead that $S = \emptyset$. Then for all $k \in \mathbb N$, we know that $x_k < x$. But recall that $x \in A \subseteq \mathbb N$. Hence, we may take $k = x + 1 \in \mathbb N$. Now recall that for any $m,n \in N$, we know that: $$m < n \iff m + 1 \leq n$$ Thus, since: $$x_1 < x_2 < \cdots < x_{x-1} < x_x < x_{x+1} < x$$ it is possible to prove by induction that: $$x > x_{x+1} \geq x_x + 1 \geq x_{x-1} + 2 \geq \cdots \geq x_2 + (x - 1) \geq x_1 + x$$ so that $x_1 < 0$, a contradiction. Thus, we conclude that $S$ must be nonempty, as desired.