I know that $[0,1]$ and a unit circle $\mathbb{S}^1$ are one-point compactifications of $\mathbb{R}$ under some suitable homeomorphism. But how does one construct the Stone–Čech compactification?

  • $\begingroup$ I will appreciate if some can give a material that explained this. $\endgroup$ – Micheal Oguntola May 18 '16 at 10:50
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    $\begingroup$ The one-point compactification is unique and it's a circle. The closed interval is the two-point compactification of $\mathbb R,$ or the one-point compactification of $[0,1).$ $\endgroup$ – bof May 18 '16 at 10:55
  • $\begingroup$ There are a few related questions that might help you, such as math.stackexchange.com/questions/256021/… $\endgroup$ – Arnaud D. May 18 '16 at 10:55
  • $\begingroup$ so can we say $\beta\mathbb{R}=S^1?$ $\endgroup$ – Micheal Oguntola May 18 '16 at 11:02
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    $\begingroup$ Please shed more light on this. $\endgroup$ – Micheal Oguntola May 18 '16 at 11:02

You cannot really construct it, as such. You can define it, and prove its existence (using the Axiom of Choice) but you cannot give a concrete, definable example of a point in the remainder $\beta\mathbb{R}\setminus\mathbb{R}$. It is common to define the half-line $\mathbb{H} = [0,\infty)$ and consider $\beta\mathbb{H} \setminus \mathbb{H}$, because one can show that $\beta \mathbb{R} \setminus \mathbb{R}$ is a disjoint union of two copies of $\beta\mathbb{H} \setminus \mathbb{H}$ (one on the right, and one on the left). This sort of makes sense, because the reals have no holes to fill internally (it's already locally compact and complete), it's compactified at the boundaries, so to say. Adding a single point at infinity yields the (essentially unique) one-point compactification which is homeomorphic to $\mathbb{S}^1$, and because it is ordered we can add two points (at both ends) to get an orderable compactification which is homeomorphic to $[0,1]$.

The Cech-Stone compactification of $X$ is characterised by the function extension property (if we assume for simplicity that $X \subseteq \beta X$): every continuous function $f: X \rightarrow Y$, where $Y$ is any compact Hausdorff space, can be extended to $\beta f: \beta X \rightarrow Y$. This extension $\beta f$ is automatically unique because the co-domain $Y$ is Hausdorff and $X$ is dense in $\beta X$. In fact we could suffice to check this for $Y = [0,1]$ only, it turns out, to have it for all compact Hausdorff $Y$.

This gives unicity: if $\gamma X$ has the same property (where again we assume for simplicity that $X \subseteq \gamma X$), we extend $1_{X,\beta}: X \rightarrow \beta Y$ with $1_X(x) = x$ to $\gamma 1_{X,\beta} : \gamma X \rightarrow \beta X$, and in the same way we have a continuous $\beta 1_{X,\gamma} : \beta X \rightarrow \gamma X$. As these maps are each other's inverses on the dense set $X$ (which we have in both), because the maps are just the identity there, they are each other's inverses on $\gamma X$ and $\beta X$ as well, making these spaces homeomorphic.

This extension property means that we must extend $f(x) = \sin(x)$ from the reals to $[-1,1]$ to all of $\beta \mathbb{R}$, and it's clear we cannot extend it just using one or two points, so that the well-known compactifications are certainly not the Cech-Stone one. E.g. if $f(\infty)$ should equal the limit of $f(2n\pi)$ as well as $f(2n\pi + \frac{1}{2})$, and these are already different. A function can only be extended to the one-point compactification of the reals, if the values outside of $f[-n,n]$ vary less and less for larger $n$, roughly speaking. So this is just a limit class. We need to be able to extend all bounded continuous functions.

Constructions (sketches of them) can be found on Wikipedia or good books like Engelking's "General Topology", or Gillman and Jerrison's "Rings of continuous functions". For the reals (which is normal) the new points of $\beta \mathbb{R}$ correspond to ultrafilters of closed subsets of $\mathbb{R}$ (free ones, i.e. their intersection is empty). But these cannot even be proved to exist without some form of the Axiom of Choice, and no explicit examples of these can be given.

We can e.g. define a filter base $\mathscr{F}$ of closed sets by taking the complements of all open intervals of the form $(a,b)$, where $a < b, a, b \in \mathbb{R}$. These have empty intersection in the reals (as $x \notin \mathbb{R}\setminus (x-1,x+1)$) but any two or finitely many of them intersect (as can easily be checked). Then Zorn's lemma (or some other such principle) tells us there exists some maximal filter of closed sets that contains $\mathscr{F}$, in fact plenty of them, and each of those gives us a point of $\beta \mathbb{R} \setminus \mathbb{R}$. Note that in a compact space any family of closed sets with the finite intersection property has non-empty intersection (so $\mathscr{F}$ witnesses the non-compactness of the reals), so the closure of the sets in $\mathscr{F}$ in $\beta \mathbb{R}$ should have non-empty intersection, and in fact those ultrafilters (considered as points) will be in that intersection in $\beta \mathbb{R}$.

In fact there are $2^\mathfrak{c}$ many new points in $\beta \mathbb{R} \setminus \mathbb{R}$ (as many as points added to $\mathbb{N}$, and as many as there are subsets of $\mathbb{R}$). It's a very large, very non-metrisable space, that still has the reals (and thus $\mathbb{Q} \subseteq \mathbb{R}$ as well) as a dense subspace.

One of the few spaces that we have that the Cech-Stone equals the one-point compactification is $\omega_1$ (the first uncountable ordinal in the order topology) which has $\omega_1+1$ as its unique compactification. So here it is concrete, but this situation is quite rare.


The Stone-Čech compactification of $\mathbb{R}$ can be functorially built as the maximal spectrum of $C_b(\mathbb{R})$, the ring of bounded continuous real functions on $\mathbb{R}$.

The maximal spectrum $\operatorname{Max}(R)$ of a commutative ring is the set of all maximal ideals, with the spectral topology, where a basis of closed sets is given by $$ V(I)=\{\mathfrak{m}\in\operatorname{Max}(R):\mathfrak{m}\supseteq I\} $$ where $I$ is any ideal of $R$. In general this topology is not Hausdorff, but it can be proved that if $X$ is a completely regular Hausdorff space, then the space $\beta X=\operatorname{Max}(C_b(X))$ is compact Hausdorff.

Suppose $f\colon X\to Y$ is a continuous map, where $Y$ is completely regular Hausdorff. Then we have an induced ring homomorphism $f^*\colon C_b(Y)\to C_b(X)$, mapping $\varphi\in C_b(Y)$ to $\varphi\circ f$. If $\mathfrak{m}$ is a maximal ideal of $C_b(X)$, then $\beta f(\mathfrak{m})=(f^*)^{-1}(\mathfrak{m})$ is a maximal ideal of $C_b(Y)$. It's easy to show that $\beta f$ is continuous under the spectral topology, so we get a continuous function $\beta f\colon\beta X\to\beta Y$.

Moreover we get an embedding $X\to \beta X$ by mapping $x\to\mathfrak{m}_x$ defined by $$ \mathfrak{m}_x=\{\varphi\in C_b(X):\varphi(y)=0\} $$

In the special case when $Y$ is compact, the maximal ideals of $C_b(Y)=C(Y)$ are all of the form $\mathfrak{m}_y$, for a unique $y\in Y$, so $Y$ can be identified with $\beta Y$. This shows that $\beta X$ indeed satisfies the universal property of the Stone-Čech compactification, that is, every continuous map $X\to Y$, where $Y$ is compact Hausdorff, can be extended to a continuous map $\beta X\to Y$.

You should be able to find this construction of the Stone-Čech compactification in several books on topology, I believe it is also in Kelley's book.

There's no “explicit” description, because the ideals of $C_b(X)$ can be described in terms of nonprincipal ultrafilters, whose existence depends on the axiom of choice. Only some cases admit an explicit description, because of special properties of $X$.

  • $\begingroup$ Now I got the idea $\endgroup$ – Micheal Oguntola May 18 '16 at 21:29

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