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This is a Lemma from N.L. Carothers Real Analysis.

Lemma. An infinite subset $A$ of $\mathbb{N}$ is countable.

Proof. Since $A\ne\emptyset$, there is a smallest element $x_1\in A$. Then $A\backslash\{x_1\}\ne\emptyset$, and there must be a smallest $x_2\in A\backslash\{x_1\}$. By induction we can find $x_1,x_2,\dotsc,x_n,\dotsc\in A$, where $x_n=\min(A\backslash\{x_1,\dotsc,x_{n-1}\})$. Now, suppose that $x\in A\backslash\{x_1,x_2,\dotsc\}\ne\emptyset$. Then the set $\{k:x_k>x\}$ must be nonempty, and hence it has a least element. That is, there is some $n$ with $x_1<x_2<\cdots<x_{n-1}<x<x_n$. But this contradicts the choice of $x_n$.

The part of the proof that I don't understand is that the set $\{k:x_k>x\}$ must be nonempty. Carothers claims that otherwise we would have $x\in A$ and $x<x_1=\min A$. I can't prove that if $\{k:x_k>x\}=\emptyset$ then $x<x_1$.

Help appreciated please.

Asaf Karagila
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user156720
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1 Answers1

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The answer is that $\{k\mid x_k>x\}$ is non-empty, because $A$ is infinite. So there is no $x\in\Bbb N$ such that for all $a\in A$, $a\leq x$.

In other words, this means that whenever $x\in\Bbb N$ there is some $a\in A$ such that $x<a$. So the set $\{k\mid x_k>x\}$ cannot be empty.

Asaf Karagila
  • 393,674