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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.

K. Hoffmann
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    @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 ("$A-B\rightarrow A$"). – Noah Schweber Nov 05 '17 at 20:25
  • 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? – K. Hoffmann Nov 05 '17 at 20:25
  • @JonasMeyer Thanks for pointing that out! English is not my native language – K. Hoffmann Nov 05 '17 at 20:26
  • @JonasMeyer Good point about the title. – Noah Schweber Nov 05 '17 at 20:27
  • @NoahSchweber and thanks for pointing that out too! – K. Hoffmann Nov 05 '17 at 20:28
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    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? – fleablood Nov 05 '17 at 20:34
  • @fleablood: No, we don't. On the contrary: without the Axiom of Choice, there may exist infinite Dedekind-finite sets, such as amorphous sets, which are infinite sets that do not have two disjoint infinite subsets. – hmakholm left over Monica Nov 07 '17 at 00:24

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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.

William Elliot
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  • 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 $A$? – K. Hoffmann Nov 06 '17 at 11:04
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    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). – William Elliot Nov 06 '17 at 19:44
  • 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? – K. Hoffmann Nov 06 '17 at 19:56