Intuition for the Stone-Čech compactification via ultrafilters

Since you already know about the Alexandroff one-point compactification, let me begin by saying that the Stone-Cech compactification is at the other extreme, adding as many points at infinity as possible. To see what that could mean, let's consider some other compactifications of $\mathbb R$, starting with the most familiar, the extended real line, obtained by adjoining the two points $+\infty$ and $-\infty$ at the two ends of the line. Compared with the Alexandroff compactification, we're now distinguishing two different ways of "going to infinity". Some sequences that converged to $\infty$ in the Alexandroff compactification fail to converge in the extended real line because part of the sequence goes to the left and part to the right (e.g., $(-1)^nn$).

This idea can be extended, to produce "bigger" compactifications. If you visualize $\mathbb R$ as embedded in the plane as the graph of the sine function and then take its closure in the extended plane $(\mathbb R\cup\{+\infty,-\infty\})^2$, you get a compactification with a whole line segment at $+\infty$ (and another at $-\infty$). Similarly, embedding $\mathbb R$ in $3$-dimensional space as a helix, by $x\mapsto (x,\cos x,\sin x)$, we get a compactification with circles at the ends. Proceeding analogously with all bounded continuous functions $\mathbb R\to\mathbb R$ (in place of $\cos$ and $\sin$), in a very high-dimensional space (in fact, $2^{\aleph_0}$ dimensions), you get one of the standard constructions of the Stone-Cech compactification of $\mathbb R$. Roughly speaking, it separates, into different points at infinity, all of the possible "ways to go to $\infty$" in $\mathbb R$.

The analogous story works for discrete spaces $X$ in place of $\mathbb R$ (except that I don't need to say "continuous" because all functions on a discrete space are continuous). A point in the Stone-Cech remainder of a discrete space $X$ should be thought of as a "way to go to $\infty$" in $X$. But how can such "ways" be described?

Well, in any compactification, each point $p$ at infinity is in the closure of the original space $X$, and so we can describe its location relative to $X$ by the trace on $X$ of its neighborhood filter, i.e. by $\mathcal F=\{U\cap X: p\in\text{interior}(U)\}$ (where $U$ refers to subsets of the compactification). Note that $\mathcal F$ can't contain any finite subsets of $X$, because such subsets are closed in the compactification (as in any $T_1$ space) and thus disjoint from suitable neighborhoods of any point $p$ at infinity.

For the Stone-Cech compactification of a discrete space $X$, this filter $\mathcal F$ must have one additional property, namely that we cannot have two disjoint subsets $A,B$ of $X$ both meeting all the sets in $\mathcal F$. The reason is that then "going to infinity" in $A$ and in $B$ would be two different ways to go to infinity, both leading to the same point $p$.

This additional property of the filter $\mathcal F$ is equivalent to saying that $\mathcal F$ is an ultrafilter. So this is how ultrafilters enter the picture of Stone-Cech compactifications of discrete spaces.

One then ordinarily continues by saying that, (1) since the filter $\mathcal F$ associated with any $p$ at infinity is an ultrafilter, and different $p$'s must correspond to different ultrafilters (in order for the compactification to be Hausdorff), we might as well identify the points $p$ with the ultrafilters $\mathcal F$, and (2) for the sake of uniformity, we might as well identify the principal ultrafilters (which haven't been used yet) with the points of $X$. With these conventions, one can prove (using compactness) that every ultrafilter on $X$ gets identified with a point in the Stone-Cech compactification of $X$. So the Stone-Cech compactification of a discrete space can be identified with the set of ultrafilters on $X$. Finally, one should verify that the topology is necessarily the one you described.

If we have a space $S$ then one can show (it's classical) that $S$ is compact iff every ultrafilter on $S$ has a limit in $S$.

So for an infinite discrete space (which has no ultrafilter limits, except for the fixed ultrafilters) we need to add a point to be a limit for each ultrafilter, and $\beta(S)$ will be the maximal compactification, so we add a different point for each ultrafilter (the one-point compactification adds just one point for all of them).