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Could anyone help me solve the below problem?

Use the axiom of choice and the generalized principle of recursive definition to show that each infinite set $X$ contains a countably infinite sebest? (H.L Royden, Real Anaysis, 3rd edition, 1988)

Also, can one prove it without using the axiom of choice or the principle of recursive definition?

Thank you!

Sunghee Yun
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2 Answers2

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It is impossible to completely avoid AC, but it is possible to use a weaken version of it:

Let $X$ be infinite, then let $[X]^{<\omega}$ be the set of finite subsets of $X$.

Define the partition, $P$, of $[X]^{<\omega}$ by the equivalence relation of $|\cdot |=|\cdot |$, that is:

For all $n∈\Bbb N$, $A_n=\{x∈[X]^{<\omega}\mid |x|=n\}$

$P=\{A_i\}_{i∈\Bbb N}$

Now by the axiom of countable choice we can find a sequence $B_n$ such that for each $n$, we have $B_n\in A_{2^n}$, that it: $|B_n|=\boldsymbol{2^n}$.

Define the sequence $C_0=B_0$, $C_n=B_n\setminus\bigcup_{i<n}B_i$, because $|B_n|>\sum_{i<n}|B_i|$, $C_n$ is never empty, also notice that for each $m≠k$ we must have $C_m∩C_k=\emptyset$.

Again, by the axiom of countable choice, we can find a sequence $a_n$ such that for each $n∈\Bbb N$, we have $a_n∈C_n$, the set $\{a_i\}_{i∈\Bbb N}$ is countable infinity subset of $X$.

ℋolo
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  • Awesome answer! thx! – Sunghee Yun Apr 18 '20 at 02:28
  • The below answer seems to be much simpler. Do you think there's any flaw in it? – Sunghee Yun Apr 18 '20 at 02:30
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    @SungheeYun it is not wrong per se, but it doesn't explain where exactly you use the axiom of choice. The bellow answer uses dependent choice which is stronger than countable choice, and because you asked about how to avoid choice, it is better to use the weakest choice principal possible – ℋolo Apr 18 '20 at 02:35
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Let S be an infinite set

$S\ne \emptyset \Rightarrow \exists x_1 \in S$

Because S is not finite,

$\Rightarrow\exists x_2 \in S - \{x_1\}$

$\Rightarrow\exists x_3 \in S - \{x_1, x_2\}$

...

$\Rightarrow\exists x_{n+1} \in S - \{x_1, x_2, ..., x_n\}$

A function $f: \mathbb{N} \rightarrow S$

$f(\mathbb{N}) = \{x_1, x_2, x_3, ...\}$

$\Rightarrow f$ is one-to-one and onto

$\Rightarrow f(\mathbb{N}) \subseteq S$ and is countable

cole_theman
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  • Thanks for your answer. I think I understand you answer. I'm not sure of this since I'm not an expert, but why do you think many others say we need some kind of axioms, e.g., axiom of choice, to prove this? Do you think there's any possibility for your answer missing something? – Sunghee Yun Apr 18 '20 at 02:32