# Stone-Čech compactification $\beta\mathbb{N}$ of the integers $\mathbb{N}$ with discrete topology has uncountably many points?

How do I show that the Stone-Čech compactification $\beta\mathbb{N}$ of the integers $\mathbb{N}$ with the discrete topology has uncountably many points? There is a hint that crux is to construct a bounded sequence of real numbers which contains uncountably any subsequences converging to different numbers, but I'm still not sure what to do...

Let $f: \mathbb{N} \to [0, 1]$ such that the image of $f$ is dense, $\overline{f(\mathbb{N})} = [0, 1]$ (i.e. an enumeration of the rationals).
There exists an extension of $f$ to the Stone–Čech compactification, $\overline{f}: \beta(\mathbb{N}) \to [0, 1]$. The image of $\overline{f}$ is compact and hence closed, yielding$$\overline{f}(\beta(\mathbb{N})) = \overline{f(\mathbb{N})},$$and so $\overline{f}: \beta(\mathbb{N}) \to [0, 1]$ surjects.
• Note that the same would work with $2^c$ instead of $[0, 1]$, and thus showing that $β\mathbb{N}$ is even bigger. – user87690 Jul 22 '15 at 16:44
Let $(q_n)_n$ be an indexing of all rational numbers in the unit interval $I$, and let $f:\Bbb N\to I$ be a map sending $n$ to $q_n$. Then there is a unique map $f':\beta\Bbb N\to I$ such that $f'\iota = f$, where $\iota:\Bbb N\to\beta\Bbb N$ is the canonical embedding. Can you show that $f'$ is surjective?