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I try to give a more constructive proof of the following lemma

$\qquad$ Let $S$ be countably infinite and $A$ an infinite subset of $S.$ Then $A$ is countable.

Here the "constructive proof" means that to prove that $A$ is countable is to construct a proper bijection from the set of positive integers onto $A.$ And I prefer to a proof of without using Zorn's Lemma.

I have tried and already got something. Since $S$ is countably infinite, there exists a bijection $ f: \mathbb{N} \to S.$ Here we denote the set of positive integers by $\mathbb{N}.$ Then I constructed the following mapping by the method given in Apostol's book Mathematical Analysis, Page 39:

$k(1)=\inf\{i\in\mathbb{N}\mid f(i)\in A\}$. Assume $k(1), k(2), \dots,k(n-1)$ has been constructed, let $k(n)=\inf\{i\in\mathbb{N}\mid f(i)\in A, i>k(n-1)\}, n=2,3,\dots.$ Continue this process on. Then let $h(n)=f(k(n)).$ I have proved that $h: \mathbb{N}\to A$ is injective.

Because I have only used the continuity axiom of real numbers and the principle of induction, I think this proof is constructive.

But at present I do not know how to prove $h$ is surjective. Can anyone help me to prove that the above $h$ is a surjection from $\mathbb{N} $ to $A?$

nuage
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  • The inclusion of $A$ into $S$ is an injection, so $|A| \leq |S|$. But $A$ is infinite, and $|S|$ is the smallest infinite cardinal. So they must be equally large. Would that work? – Arthur Apr 30 '14 at 09:37

2 Answers2

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Let's do it for special case $A\subset S=\mathbb{N}$. Starting with $f=id_{\mathbb{N}}$ it comes to defining $h\left(n\right)$ inductively as the smallest integer in $A-\left\{ h\left(1\right),\dots,h\left(n-1\right)\right\} $. Since $A$ is infinite this set will not be empty. If $a\in A-h\left(\mathbb{N}\right)$ then $h\left(n\right)>a$ for $n$ large enough. This however contradicts that $h\left(n\right)$ is the smallest integer in $A-\left\{ h\left(1\right),\dots,h\left(n-1\right)\right\} $. The conclusion is that $A-h\left(\mathbb{N}\right)=\emptyset$ or equivalently that $h$ is surjective. This special case can be applied to prove the general case. It shows that your function $k$ is surjective if its codomain is defined as $\{i\mid f(i)\in A\}$. If $\iota$ denotes the inclusion of this set into $\mathbb N$ then composition $h=f\circ\iota\circ k$ is consequently surjective.

drhab
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Let $S$ be countable, with elements $a_{1}$, $a_{2}$, . . . , and let $A$ be a subset of $S$. Among the elements $a_{1}$, $a_{2}$,... let $a_{n_{1}}$,$a_{n_{2}}$,...be those in the set $A$. If the set of numbers $n_{1}$, $n_{2}$, . . . has a largest number, then $A$ is finite. Otherwise A is countably infinite (consider the correspondence $i$ $\leftarrow\rightarrow$ $a_{n_{i}}$ ).

Remark: Credits should go to Kolmogorov and Fomin. Elements of Theory of Functions and functional analysis.

Airbag
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  • The proof you mentioned seems that it is based on Axiom of Choice. And I want to give a more constructive proof, better of without using Axiom of Choice. – nuage May 01 '14 at 01:27
  • @azhi: Where do you think the axiom of choice is being used here? – Asaf Karagila May 01 '14 at 04:24