If $A$ is infinite set, then there is a subset $B$ of $A$ such that $|A|=|B|=|A-B|$ How to prove that If $A$ is infinite set, then there is subset $B$ of $A$ such that $|A|=|B|=|A-B|$ ? 
If $A$ is countable, it is not hard to show the statement is true.
How about case $A$ is uncountable ?
Thank you. 
 A: As A is infinite, |A| = |A| + |A|.
Let A1 = { (a,1) : a in A }.
Let A2 = { (a,2) : a in A }.
Since |A| = |A1| + |A2| and A1, A2 are disjoint,
there is a bijection f:A -> A1 $\cup$ A2.
Let B = { x in A : f(x) in A2 }.
A: Of course we have to use the Axiom of Choice for this.
First, partition $A$ into countably infinite sets. To do this, take a maximal disjoint family of countably infinite subsets of $A;$ there are at most a finite number of leftover elements, which we can add to one of the countable sets. In this way we get $A=\bigcup_{i\in I}A_i,$ which the sets $A_i$ are countably infinite and pairwise disjoint.
Next, for each $i\in I$ choose a partition $A_i=B_i\cup C_i$ into two disjoint infinite sets. Let $B=\bigcup_{i\in I}B_i$ and $C=\bigcup_{i\in I}C_i.$ Now we have a partition $A=B\cup C.$ To see that $|A|=|B|,$ choose bijections $f_i:A_i\to B_i$ and take the union. To see that $|A|=|C|,$ choose bijections $g_i:A_i\to C_i$ and take the union.
A: You can build the set $B$ and its compliment inductively.
Let $f:|A| \rightarrow A$ be a bijection. Now we build $B_{\alpha}$ and $C_{\alpha}$ inductively. 
$B_{0} = C_{0} = \emptyset$
$B_{\alpha + 1} = B_{\alpha} \cup \{f(2\alpha)\}$ 
$C_{\alpha + 1} = C_{\alpha} \cup \{f(2\alpha + 1)\}$ 
At limit $\alpha$:
$B_{\alpha} = \bigcup_{\beta < \alpha}B_{\beta}$
$C_{\alpha} = \bigcup_{\beta < \alpha}C_{\beta}$
$B = B_{|A|}$ is as required.
