proof that countable unions are countable

Let $C$ be a countable set of countable sets. We will show that $\cup C$ is countable.

Let $P$ be the set of positive primes (http://planetmath.org/Prime). $P$ is countably infinite, so there is a bijection between $P$ and $\mathbb{N}$. Since there is a bijection between $C$ and a subset of $\mathbb{N}$, there must in turn be a one-to-one function $f:C\rightarrow P$.

Each $S\in C$ is countable, so there exists a bijection between $S$ and some subset of $\mathbb{N}$. Call this function $g$, and define a new function $h_{S}:S\rightarrow\mathbb{N}$ such that for all $x\in S$,

 $h_{S}(x)=f(S)^{g(x)}$

Note that $h_{S}$ is one-to-one. Also note that for any distinct pair $S,T\in C$, the range of $h_{S}$ and the range of $h_{T}$ are disjoint due to the fundamental theorem of arithmetic.

We may now define a one-to-one function $h:\cup C\rightarrow\mathbb{N}$, where, for each $x\in\cup C$, $h(x)=h_{S}(x)$ for some $S\in C$ where $x\in S$ (the choice of $S$ is irrelevant, so long as it contains $x$). Since the range of $h$ is a subset of $\mathbb{N}$, $h$ is a bijection into that set and hence $\cup C$ is countable.

Title proof that countable unions are countable ProofThatCountableUnionsAreCountable 2013-03-22 12:00:01 2013-03-22 12:00:01 Koro (127) Koro (127) 9 Koro (127) Proof msc 03E10