In our reading we had and proved the following theorem concerning the compactification with respect to a family of bounded functions:
Theorem Let $E$ be a discrete countable infinite space and let $\mathcal{F}$ be a denumerable familiy of bounded functions on $E$. There exists an unique (up to homeomorphism) compactification $\hat{E}=\hat{E}_{\mathcal{F}}$ of $E$ such that (i) every function $f\in\mathcal{F}$ extends to a continuous function on $\hat{E}$ and (ii) the family $\mathcal{F}$ separates boundary points, that is, if $\xi,\eta\in\hat{E}\setminus E$ are distinct, then there exists a function $f\in\mathcal{F}$ so that $f(\xi)\neq f(\eta)$.
Now the following is to show:
If $E$ is a discrete set and if $\mathcal{F}$ only contains constant functions, then the compactification $\hat{E}_{\mathcal{F}}$ of $E$ is the Alexandroff (one point) compactification of $E$.
I think it is to show that the Alexandroff compactification $E^*:=E\cup\left\{\infty\right\}$ fullfills the properties (i) and (ii) of the theorem.
Obviously, property (ii) is fullfilled, because there are no distinct boundary points, because the boundary only consists of the point $\infty$.
For (i), it is to show that each $f=\text{const}$ can be extended continuously to the boundary $E^*\setminus E=\left\{\infty\right\}$. But I think this is very easy:
Since $\infty$ is a boundary point, there is a sequence $(x_n)\subset E$ that tends to $\infty$ as $n\to\infty$. Since $f\equiv\text{const}$ is continuous on $E$, it is $$ \text{const}=\lim_n f(x_n)=f(\lim_n x_n)=f(\infty), $$ thus setting $f(\infty):=\text{const}$ gives the desired continuous extension of $f$ on $E^*$.
So (i) and (ii) are fullfilled. Therefore by the theorem above the compactification $\hat{E}_{\mathcal{F}}$ is the Alexandroff-compactification (up to homeomorphism).
That's it?
Yes, (ii) is vacuously true since there are no distinct elements in $E^*\setminus E$.
And (i) is satisfied because a constant function has the obvious extension by the constant function $E^*\to\Bbb R$ which is continuous as constant functions always are.
So the theorem then implies that the Alexandroff compactification is the compactification whose existence is implied by the theorem.