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So currently I have a problem on the following question: Let $A$ be an infinite set. Show that there is a countable subset $B\subseteq A$ such that $|A|=|A-B|$ as asked here: Subset of an infinite set with same cardinality.

The construction of a set $B$ seems pretty easy (One thing I have noticed is that $B \neq A$). But showing that there is bijective function with $A-B \rightarrow A$ is where I am currently stuck. We cannot use a slightly altered version of $id_A$ as there is no good way to "hit" the missing elements in $A$ I guess.

So my question is: How can I show that there is such a bijective function or how am I able to construct it?

As this is homework I would appreciate hints over full solutions for now.

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    $\begingroup$ @JonasMeyer I think the first paragraph is meant to say "$\vert A\vert=\vert A-B\vert$; they get it right in their second paragraph ("$A-B\rightarrow A$"). $\endgroup$ – Noah Schweber Nov 5 '17 at 20:25
  • $\begingroup$ Are u sure? The question states that $A$ should be infinite - what e.g is with A being the reals and B being all even integers? As all even integers are of the same Cardinality as the integers itself shouldn't A-B still of the same Cardinality as the reals? $\endgroup$ – K. Hoffmann Nov 5 '17 at 20:25
  • $\begingroup$ @JonasMeyer Thanks for pointing that out! English is not my native language $\endgroup$ – K. Hoffmann Nov 5 '17 at 20:26
  • $\begingroup$ @JonasMeyer Good point about the title. $\endgroup$ – Noah Schweber Nov 5 '17 at 20:27
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    $\begingroup$ Someone better versed than me in the Axiom of Choice answer me this: Without the axiom of choice, Do we know that infinite sets must have countable subsets? $\endgroup$ – fleablood Nov 5 '17 at 20:34
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Let $\{ a_i : i \in N \}$ a denumberable subset of A
with for all distinct j,k in N, $a_j \not = a_k$.
Let $B = \{ a_{2j} : j \in N \}$.

A bijection f from A - B onto A is $a_{2j-1}$ to $a_j$, for j in N
and the identity map for the other points.

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  • $\begingroup$ So it is fine to say that if $B$ is of the same Cardinality as the denumerable subset it follows that $A-B$ is also of the same Cardinality as $A$? $\endgroup$ – K. Hoffmann Nov 6 '17 at 11:04
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    $\begingroup$ Heck no! You have to prove that by constructing a bijection and while you are at it use the axiom of choice to show the existence of the denumberable subset (notice edit). $\endgroup$ – William Elliot Nov 6 '17 at 19:44
  • $\begingroup$ Yeah thats true that is what i did earlier instead of just saying it. I constructed a function between the enumarable set and $B$ called $f$ with $f(a_i)=f(a_{2i})$ which is bijective and then constructed a function which is a bijection between $A$ and the enumarable set. I hope thats the correct reasoning then? $\endgroup$ – K. Hoffmann Nov 6 '17 at 19:56

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