This is a question about $\beta \mathbb{N}$, the Čech-Stone compactification of the natural numbers.
I'm reading a proof of the fact that every zeroset in $\beta X \setminus X$ contains a copy of $\beta \mathbb{N}$ (Where $X$ is an arbitrary Tychonov space). The author implicitly uses the fact that if $$f:\mathbb{N}\to [0,1]$$ goes to zero for $n\to \infty$, then it's unique extension $\beta f$ is zero on the remainder $\beta\mathbb{N}\setminus\mathbb{N}$. Intuitively this makes a lot of sense but I can't translate that intuition into a solid argument. How can one prove this?
$\mathbb{N}$ is dense in $\beta\mathbb{N}$. Thus for any $x\in\beta\mathbb{N}$ there is a net $(x_\alpha)\subseteq\mathbb{N}$ convergent to $x$.
By the continuity of the extension $F:\beta\mathbb{N}\to[0,1]$ we have
$$F(x)=\lim_\alpha f(x_\alpha)$$
Now if $x\not\in\mathbb{N}$ then $f(x_\alpha)\to 0$ by the assumption because in that case $(x_\alpha)$ cannot be bounded.