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If $X$ is a set s.t $|X|\geq \aleph _0$ then I want to prove that there exist $A,B\subset X$ s.t $A\cap B=\emptyset$, $A\cup B=X$ and $|A|=|B|=|X|$. How can I construct such subsets?

Vegetal605
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  • You need to appeal to the axiom of choice here. This was asked at least once or twice before. – Asaf Karagila Sep 07 '15 at 12:56
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    Here's one example, http://math.stackexchange.com/questions/1041731/prove-that-if-a-is-an-infinite-set-then-a-times-2-is-equipotent-to-a; also http://math.stackexchange.com/questions/395974/x-ge-aleph-0-wedge-a-subset-x-wedge-a-x-backslash-a and about the relationship with the axiom of choice, see this: http://math.stackexchange.com/questions/393196/the-relationship-of-frak-mm-m-to-ac/ – Asaf Karagila Sep 07 '15 at 12:59
  • @AsafKaragila I don't see how this post answer my question. It only shows that for any infinite cardinal $\alpha$, $2\cdot \alpha=\alpha$. – Vegetal605 Sep 07 '15 at 13:05
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    And what your question is asking, is exactly that. $|X|=|A\cup B|=|A|+|B|=|X|+|X|$. – Asaf Karagila Sep 07 '15 at 13:06
  • @AsafKaragila from this fact I can't know that there exist 2 disjoint subsets of $X$ with the same cardinality of $X$. I just know that if there exist such subsets then the cardinality of their union equals to the cardinality of $X$. – Vegetal605 Sep 07 '15 at 13:08
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    If $|X|=|X|+|X|$, then there is a bijection $f$ between $X$ and $X\times{0,1}$. Use this bijection and define $A=f^{-1}(X\times{0})$ and $B=f^{-1}(X\times{1})$. Is this explicit enough why you are asking the same question? – Asaf Karagila Sep 07 '15 at 13:10
  • yeah this was explicit enough :) thanks. – Vegetal605 Sep 07 '15 at 13:12

1 Answers1

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This is easy to do for $\aleph_0$: take even and odd numbers. You can use this to prove the statement for a more powerful $X$; the trick is to split $X$ into disjoint sets of cardinality $\aleph_0$ (you will need the axiom of choice, naturally). To this end, well order $X$ (which can be done by the Zermelo theorem) and put elements $a\prec b$ into one set if they are within a finite distance, i.e. there exist some $n\ge 1$ and $x_1,\dots,x_n=b$ such that $a$ precedes $x_1$, $x_1$ precedes $x_2$, $\dots$, $x_{n-1}$ precedes $x_n=b$. Now you can "halve" each of your disjoint sets, which are of cardinality $\aleph_0$ (this I leave for you to prove).

You can use the same idea for other things, e.g. to split $X$ into a countable number of equipotent sets.

zhoraster
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  • but my demand was that each subset has the same cardinality of $X$ and this cardinality is not necessarily $\aleph _0$ – Vegetal605 Sep 07 '15 at 12:56
  • @dorsh605, and this is exactly what you get with my construction. You have a union of disjoint countable sets. Then you halve each countable set, obtaining two exactly the same unions of disjoint countable sets. – zhoraster Sep 07 '15 at 13:08
  • What if the well order you find using Zermelo's theorem is e.g. $\omega+1$? Then the maximum will be alone in its subset and you cannot halve it. You need a limit ordinal. – Najib Idrissi Sep 07 '15 at 13:38
  • @Najib Idrissi, you are right, there is a small gap: there might be a finite set, but there may be only one. Then we can remove the gap by adding this in front of any of the remaining infinite sets. – zhoraster Sep 07 '15 at 14:14