$[(|X| \ge \aleph_0) \;\wedge\; (A \subset X) \;\wedge\; (|A| = |X\backslash A|)] \Rightarrow |A| = |X|$

Question

Suppose that $X$ is an infinite set of cardinality $\alpha$. Also, suppose that, for some $A \subseteq X$, we have that $|A| = |X\backslash A|$. I want to show that $|A| = |X|$.

When, for example, $X = \mathbb{N} = \{1, 2, 3, \dots \}$, and $A = \{2n: n \in \mathbb{N}\}$, the requisite bijection $X\rightarrow A$ is about as obvious as it can be (since the definition of $A$ already provides one such bijection; in fact, in this case, the fact $|A| = |X\backslash A|$ is not even used.) But I have a hard time pinning down the proof of the more general case stated above.

Answer 1

It does require some of the axiom of choice, which may explain why you have a bit of a hard time proving that in the general case.

But if you assume that $X$ is a well-ordered set, then it's easy. Note that for infinite well-ordered sets $A,B$ it's true that $|A\cup B|=\max\{|A|,|B|\}$.

Note that the above follows from having $|A|+|A|=|A|$ for every well-ordered $A$ (and the fact that well-ordered cardinalities are all comparable). To prove that $|A|+|A|=|A|$ simply write $A=\{a_\alpha\mid\alpha<\kappa\}$ for some limit ordinal $\kappa$, then note that $$A_1=\{a_\alpha\mid\alpha=\delta+2n\text{ for a non-successor }\delta,\text{ and }n\in\omega\},\text{ and}\\ A_2=\{a_\alpha\mid\alpha=\delta+2n+1\text{ for a non-successor }\delta,\text{ and }n\in\omega\},$$

are both equipotent with $A$.

Answer 2

I work in ZFC. $X=A\cup (X\setminus A)$ with $A$ and $X\setminus A$ disjoint sets. Therefore $\vert X\vert =\vert A\vert +\vert X\setminus A\vert =\vert A\vert +\vert A\vert$. Now, since this equation holds and $\vert X\vert\geq \omega$ also $\vert A\vert \geq \omega$ (otherwise $\vert A\vert$ would be finite and then $\vert A\vert +\vert A\vert$ could not be equal to $\vert X\vert$). Hence $\vert X\vert=\vert A\vert +\vert A\vert=\vert A\vert$, since $\vert A\vert$ is an infinite cardinal.

To prove that if $\kappa$ is an infinite cardinal $\kappa +\kappa =\kappa$, one can argues as follows. It suffices to show that $\kappa\times 2=\kappa$. Take $F:=\{f\colon \alpha\longrightarrow \alpha\times 2:\ 0\neq \alpha\subseteq \kappa\ \text{and f is bijective}\}$ ($0=\emptyset$) with the partial order given by set theoretic inclusion. Notice that $F\neq\emptyset$ since $\omega\leq \kappa\implies \omega \subseteq \kappa$ and $\omega\approx \omega\times 2$ in an obvious manner. It is not difficult to show that $(F,\subseteq)$ satisfies the assumptions of Zorn's Lemma (easy exercise). Therefore we get a maximal element $f\in F$, so that, in particular, $f\colon \alpha\longrightarrow \alpha\times 2$ with $0\neq\alpha\subseteq \kappa$ and $f$ bijective. To conclude it suffices to prove that $\kappa\setminus \alpha$ is a finite set (if $\kappa$ is an infinite cardinal and $n\in\omega$ then $\kappa +n\approx \kappa$). But this is clear: if $\vert\kappa\setminus \alpha\vert\geq\omega$, there is a bijection between $\omega$ and a subset $X\neq\emptyset$ of $\kappa\setminus \alpha$, i.e there is a bijection $g\colon X\longrightarrow X\times 2$ and therefore $f\cup g\colon \alpha\cup X\longrightarrow (\alpha\cup X)\times 2$ is a bijection such that $f\subsetneq f\cup g$, contradicting maximality of $f$.