Prove that $|AUC| = |A|$, where $A$ is an uncountable set and $C$ is a countable set.

Question

Let $A$ be an uncountable set and let $C$ be a countable set with $A \cap C = \{\}$. Show that $|A\cup C| = |A|$. I'm quite lost as to how to approach solving this problem. I know that CSB theorem is involved to conclude there is a bijection, but still confused. Any help is appreciated!

Answer 1

We construct a bijection between $A\cup C$ and $A$.

Pick and remove an element from $A$ repeatedly, to obtain a sequence of distinct elements of $A$. (Note this requires the axiom of choice; I don’t think your statement can be proven without this axiom.) Call this sequence $B$, and everything else $D$. So $A=B\cup D$, where $B$ is countably infinite.

We can then construct a bijection from two countable sets to one (this is straightforward, and I’ll leave it as an exercise).

So suppose $f$ is a bijection from $B\cup C$ to $B$.

Then our bijection $g$ from $A\cup C$ to $A$ can be defined as $g(x) = x$ if $x \in D$, and $g(x)=f(x)$ if $x \in B\cup C$. This clearly works, and we are done.

Answer 2

We can use the Axiom of Choice to find a countable subset $D\subset A$. Now since $D$ and $C$ are both countable, then $C\cup D$ is a countable set. Hence we may find a bijective function $g: C\cup D\to D$ since both sets have the same cardinality. Now define the following bijective function

$$ f: A\cup C\to A$$ which is defined by $$ f(x) = \begin{cases}x \hspace{8mm}\text{if} \hspace{4mm}x\not\in C\cup D\\ g(x) \hspace{2mm} \text{if} \hspace{4mm}x\in C\cup D\\ \end{cases}$$

The proof that $f$ is a bijection relies on $g$ being a bijection. Does this help you get started?

Answer 3

The Cantor-Schröder–Bernstein theorem promises us a bijection of $A$ and $C$ if we can produce an injection from $A$ to $A \cup C$ and one from $A \cup C$ to $A$.

An injection from $A$ to $A \cup C$ is trivial, let's call it $\phi$, shown below.

$$ \phi(x) = x $$

The other direction is more difficult and writing down an explicit injection isn't going to work unless we know more about $A$.

Let's imagine a family of sequences $F$ indexed by an ordered set $I$ with no maximum element such that $\cup F = A$ and every $x \in F$ has the same cardinality as $C$, further suppose the sequences in $F$ are disjoint.

Let $f_1$ be a sequence in $F$. We define a function $\tau_{f_1}(x)$ that transfers $x$ from the sequence that contains it to the corresponding position in $f_1$.

Let $w(x)$ return an index in $I$ corresponding to any element of $I$ that is strictly greater than the index of the sequence that contains $x$.

Given all of this, we can define an injection.

$$ \psi(x) = \tau_{F_{w(x)}}(x) $$

Our function $\psi$ sends each element to the corresponding position in a later sequence.

Therefore we have two injections.