$\mid{B}\mid=\aleph_0 , A\cap{B}=\emptyset$. Prove $\mid{A} \cup B \mid\,=\,\mid{A} \mid$

Question

Let $A$ be an infinite set , and a set $B$. $\mid{B}\mid=\aleph_0 , A\cap{B}=\emptyset$. Prove $\mid{A} \cup B \mid\,=\,\mid{A} \mid$ .

What I thought to do is to take a set $C\subseteq A\, ,\, \mid{C} \mid=\aleph_0$ and define an invertible function $f : A\cup B \rightarrow A$ that sends every element in $A$ and not in $C$ to themselves , and sends the elements in $B\cup{C}$ somewhere (I can't think where). Maybe I should use the Axiom of Choice? $\,$Thanks.

Answer 1

This isn't true if $A$ is finite.

If $B= \{ b_1, b_2, ....\}$ is the innumeration of $B$ and $A= \{a_1, ....a_n\}$ is an enumeration of $A$ that $f:\mathbb N\to A\cup B$ via $f(m) = a_m$ if $m \le n$ and $f(m) = b_{m-n}$ if $m > n$ is a bijection so $|A\cup B| = \aleph_0=|B|$.

If $|A| = \aleph_0$ then if $A= \{a_1, a_2, ...\}$ is an enumeration of $A$ then $f:\mathbb N \to A\cup B$ via $f(m) = a_{\frac m2}$ if $m$ is even and $f(m) = b_{\frac {m+1}2}$ if $m$ is odd is a bijection. So $|A\cup B| = \aleph_0 = |A| = |B|$.

If $A$ is uncountable we need an axiom of countable choice to declare that it is possible to find an $A' \subset A$ where $|A'| = \aleph_0$. That is that there is an injection from $h:\mathbb N\to A$ and $A' = h(\mathbb N)$.

With that we can define a bijection $f:A \to A\cup B = (A\setminus A')\cup (A'\cup B)$ via: $f(x) =x$ if $x\in A\setminus A'$ and $f(x) = g(x)$ where $g: A' \to A'\cup B$ is a bijection (which is possible as we proved above $|A'\cup B| =|A'|$). So $|A\cup B| = |A|$.

Answer 2

You will need to use (at the very least) the fact that being "Dedekind-infinite" is equivalent to being "infinite" somewhere in your argument. This statement is independent of $ZF$, and requires some weak form of the axiom of choice to prove.

To see why: assume that $A$ is a Dedekind-finite, but infinite set (and without loss of generality we can assume that $A \cap \mathbb{N} = \emptyset$). Then clearly $\aleph_0 \leq |A \cup \mathbb{N}|$, yet by definition $\aleph_0 \not\leq |A|$. Therefore $|A \cup \mathbb{N}| \neq |A|$.

On the other hand, if every infinite set is Dedekind-infinite, then $A$ being infinite implies the existence of an injection $f : \mathbb{N} \rightarrow A$. You also must have a bijection $g : \mathbb{N} \rightarrow B$ by assumption that $|B| = \aleph_0$. Now can you use $f$ and $g$ to build a bijection from $A \cup B$ into $A$? (Hint: what happens if you send the elements of $B$ to $f[2 \mathbb{N}]$?)