Set theory problem

**wannabeguru** · February 9th 2009, 09:43 PM

- edit -

**clic-clac** · February 10th 2009, 01:43 AM

Hi

Of course the proof can depend on how you defined "infinite", but, basically, we can always say that a infinite set minus a finite subset of its elements is still infinite (and by the way non empty) $(\star )$

Then, using the axiom of choice, you can choose an element $x$ in $S,$ and name it (in your head)* $a_0.$ Since $S_0:=S-\{x\}\neq\emptyset$ , you can choose an element $y$ in $S_0$ and name it $a_1$ , again $S_1:=S_0-\{y\}\neq\emptyset$ and you can choose... You can repeat a countable amount of times this operation using the axiom of choice and $(\star )$ .

Finally, an injection $\mathbb{N}\hookrightarrow S$ would be $n\mapsto a_n,$ which proves what you wanted.

*Why in your head? Well we can rename the elements in $S,$ but that may not be very funny : if $S$ has already elements that are named $a_k$ for a $k\in \mathbb{N},$ that would be a problem.
One way to do would be to consider $S'=\{b_\lambda;\ \lambda\in S\}$ and say that $S\rightarrow S':\lambda\mapsto b_\lambda$ is a bijection, so we can use $S'$ instead of $S$ in the proof.

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