Show that $n+\aleph_0=\aleph_0$

Question

Is there a simple way to show that $n+\aleph_0=\aleph_0$ where $n$ is some positive integer? I can't think of any easy way to prove this.

Answer 1

Make a bijection between the union of a set of size $n$ and $\Bbb N$ and $\Bbb N$. As an example, let the set be $n$ numbered cards. Let card $n$ go to $n\in \Bbb N$ and number $k$ go to $n+k \in \Bbb N$

Answer 2

Let $X=\{x_1,...,x_n\}$ be an $n$-element set. Then

$$f:X\overset.\cup\Bbb N\to\Bbb N, \quad x\mapsto \begin{cases}i&\text{if $x=x_i\in X$}\\x+n&\text{if $x\in\Bbb N$}\end{cases}$$

is a bijection. The inverse is

$$f^{-1}:\Bbb N\to X\overset.\cup\Bbb N,\quad i\mapsto\begin{cases} x_i&\text{if $i\le n$}\\i-n&\text{if $i>n$}\end{cases}$$