# Union of countable sets is countable.

Let’s say we have a possibly infinite countable collection of countable sets $A_1, A_2, \cdots, A_n, \cdots$, prove that $\bigcup A_i$ is countable too. We are going to do this using Axiom of Choice.

Since every set $A_i$ is countable, there is a set of injections from $N$ to $A_i$, let’s call it $M_i$. Using the axiom of choice we can create a set of maps $\{g_1, g_2, \cdots\}$ from $N$ to $A_i$. Now, we are going to create an injection from $(N, N)$ to $\bigcup A_i$. If $x_k$ is an element of $\bigcup A_i$ then it’s also an element of some set $A_m$. Therefore we can construct $f(z, m) = x_k$ where $x_k = g_m(z)$. And since we know that $(N, N)$ is countable, we can map $N$ to $\bigcup A_i$. QED.

## 2 thoughts on “Union of countable sets is countable.”

1. Neural Outlet..

Because it’s countable and what-not, the cardinality of the entire union is Aleph-null isn’t it?

1. oh Post author

Yes. Assuming that we take countably-infinite collection of countable sets, like in the post.