Let $\mathfrak X$ be a completely regular space. We can then consider its Stone–Čech compactification $\beta \mathfrak X$. Every continuous bounded function $f:\mathfrak X \to \mathbb C$ admits a unique continuous extension $\beta f$ to $\beta\mathfrak X$.
Suppose now we are given a family $\mathcal S\subset C(\mathfrak X; [-1,1])$ which separates points of $\mathfrak X$, namely for any $x,y\in\mathfrak X$ with $x\neq y$ there exists $f\in\mathcal S$ such that $f(x)\neq f(y)$.
Set $\beta\mathcal S=\{\beta f; \;f\in \mathcal S\}\subset BC(\beta\mathfrak X)$.
Question: Is it true that $\beta\mathcal S$ separates points of $\beta\mathfrak X$? If it is false, which functions should I add to $\beta\mathcal S$ in order to get the result?
$\beta\mathcal S$ need not separate $\beta\mathfrak X$. For example, take $\mathfrak X = \mathbb R$ and $\mathcal S$ to be a collection of functions $f_n$ such that each separates the points of $[-n,n]$ and has value $0$ on all points outside $[-(n+1), n+1)]$. The each $\beta f_n$ will be $0$ on all points of the remainder $\beta\mathbb R \setminus \mathbb R$ and so cannot separate any of its points. You could also choose the functions so that each $f_n(0)=0$, then no $\beta f_n$ will be able to separate $0$ from any point of the remainder. A similar construction will be possible in any locally-compact $\mathfrak X$.