# Let the set S be infinite, and the set T countably infinite. Show that S and S U T have the same cardinality

Let the set S be infinite, and the set T countably infinite. S and T are both subsets of R. Show that S and S U T have the same cardinality.

I know we can discuss whether S is countable or uncountable, but is it true that, for example like this, countable set's cardinality basically does not count in the union of itself and an uncountable set?

Let $A$ be a countably infinite subset of $S$.* The first thing we want to do is find a map $g : A \to A \cup T$.
Since $A$ is countable, there's some bijection $\alpha$ from $\mathbb{N}$ to $A$, and same with $T$ (with bijection $\beta$). Define $\gamma : \mathbb{N} \to A \cup T$ as follows: $$\gamma(n) = \begin{cases} \alpha(\frac{n}{2}) & n \textrm{ is even} \\ \beta(\frac{n+1}{2}) & n \textrm{ is odd} \\ \end{cases}$$
Since $\gamma$ is a bijection, this tells us that $A \cup T$ is countable. So there's some bijection $g : A \to A \cup T$. Define $f : S \to S \cup T$ as follows: $$f(s) = \begin{cases} s & s \notin A \\ g(s) & s \in A \\ \end{cases}$$
• Proof of this theorem must requires choice (at least, you assume that every Dedekind-finite set is finite). If $A$ is a Dedekind-finite infinite set and $A\cap\omega=0$, then $|A|<|A\cup\omega|$. – Hanul Jeon Sep 24 '14 at 23:55