For every infinite set A there exists infinite B,C such that A=B∪C

Question

Prove for every set $A$, there exists sets $B,C$ such that $A=B∪C$, $B∩C=∅$ and $|A|=|B|=|C|$ when

a. $|A|=ℵ_{0}$

b. $|A|=ℵ$

I'm really not sure of my proof, especially in the part I say A\A' is also infinite.. and if it turns out to be false, how can I do better?

My attempt: a. Let $A$ such that $|A|=ℵ_{0}$. Note that as $A$ is infinite, there exists a bijection from $A$ to a strict subset $A'$ of itself. Also note, that $|A'|=ℵ_{0}$ and that $A=$($A$ \ $A'$)$∪A'$. As A is infinite, ($A$ \ $A'$) is also infinite, therefore setting $B=A'$ and $C=$($A$ \ $A'$) gets us our result.

b. identical.

Answer 1

Michael Barz has shown a counterexample to your assertion that $A\setminus A'$ must be infinite. Here's a hint for "how can I do better?":

(a) Since $|A|=\aleph_0$ by assumption, you know that there is a bijection $f:\mathbb N\to A$. Therefore if you find a solution for $A=\mathbb N$, you can transfer that solution to any other $A$ simply by applying that bijection to each of the two subsets you break $\mathbb N$ into.

In mathematical jargon, this reasoning is usually expressed by saying "without loss of generality we can assume $A=\mathbb N$". (When used alone, this phrasing leaves it to the reader to imagine how you generalize a solution for $\mathbb N$ to a solution for a generic $A$, so in a classroom setting, you'll need also to explain how that works).

Now, can you find two sets $B, C\subseteq \mathbb N$ such that each of them is countably infinite and $\mathbb N$ is their disjoint union?

(b) Since $|A|=\aleph$, without loss of generality we can assume $A=\mathbb R$ ...

Answer 2

This is very wrong. Not every subset of an infinite set is infinite. For instance, if $A = \mathbb{Z}$ and $A' = \mathbb{Z}\setminus\{0\},$ then $A\setminus A' = \{0\}$ is not infinite.

Answer 3

Pick a bijection $f$ from $A \times \{0,1\}$ to $A$ and define $B$ and $C$ to be $f(A \times \{0\})$ and $f(A \times \{1\})$ respectively. Then, $B$ and $C$ are two complementary subsets of $A$ with the same cardinality as $A$.

Answer 4

It's enough to prove this for sets $A$ that are infinite cardinals. We'll use the following

Proposition: Every ordinal can be written uniquely as the ordinal sum of: a limit ordinal or $0$, and an integer.

Thus "the integer part of an ordinal" $\mathsf{int}(\alpha)$ is well-defined, and we can call an ordinal even or odd according as its integer part is even or odd. Given an infinite cardinal $\kappa$, let $B = \{\alpha < \kappa \,|\, \alpha$ is even$\}$, $C = \{\alpha < \kappa \,|\, \alpha$ is odd$\}$. Clearly these are disjoint subsets of $\kappa$, and each is of size $\kappa$ as witnessed by these bijections:

$$\alpha \mapsto \alpha + \mathsf{int}(\alpha) \colon \kappa\to B$$ and $$\alpha \mapsto \alpha + \mathsf{int}(\alpha) + 1 \colon \kappa\to C.$$

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Proof of Proposition:

This is obvious for finite ordinals ($n = 0 + n$), so let $\alpha$ be infinite.

Let $\lambda = \mathsf{sup} \,\{ \gamma \,|\, \gamma < \alpha \,\text{and}\, \gamma \,\text{is a limit ordinal}\}$. Then $\lambda$ is a limit ordinal and $\lambda \le \alpha$. If $\lambda = \alpha$ then $\alpha = \lambda + 0$. Otherwise, $\lambda < \alpha$, so let $\beta = $ the ordertype of the closed-open ordinal interval $[\lambda, \alpha)$. Then $\alpha = \lambda + \beta$. If $\beta \ge \omega$, then $\beta = \omega + \delta$ for some $\delta$, so $\alpha = \lambda + \omega + \delta$; however, this is impossible, because then $\lambda +\omega$ is a limit ordinal $\le\alpha$ that's greater than $\lambda$, but $\lambda$ is the largest limit $\le \alpha$. So $\beta$ must be an integer.

Uniqueness of the $\lambda + n$ decomposition: left as an easy exercise.