This note is dedicated to *Freudenthal compactification*, a kind of compactification for some topological spaces. In particular, it will be noticed how such a compactification leads to the *number of ends*, a nice topological invariant, and how it can be used to classify compactifications with finitely many points at infinity. The proof of theorem 1 is mainly based on Baues and Quintero’s book,* Infinite homotopy theory*.

First of all, we introduce the spaces we will work with.

**Definition:** A *generalized continuum* is a locally compact, connected, locally connected, -compact, Hausdorff topological space.

Typically, a generalized continuum may be think of as a locally finite CW complex or as a topological manifold.

Let be a generalized continuum. Because is -compact, it is an increasing union of compact subspaces ; moreover, by locally compactness, we may suppose that for all . Such a family of compact subspaces is called an *exhausting sequence*.

**Definition:** Let be a generalized continuum and be an exhausting sequence. An *end* is a decreasing family of subspaces such that is an *unbounded* (ie. not relatively compact) connected component of .

Let denote the set of ends of and . If is an end and is an open subspace, will mean that for some .

The *Freudenthal compactification* of is defined as endowed with the topology generated by

.

First, notice that does not depend on the sequence of compact subspaces we chose. Indeed, it is possible to consider the increasing functions associating to any compact an unbounded connected component of , and then to define a space , where is the set of functions we just mentionned, endowed with the topology generated by

.

Clearly, the restriction of a function to the family we fixed to construct defines an end of , so there exists a natural continuous map . Conversely, an end naturally corresponds to a function of : if is compact, there exists some such that , and may be defined as the unbounded connected component of containing . Therefore, the map turns out to be a homeomorphism. In particular, does not depend on the sequence since neither does .

**Theorem 1:** Let be a generalized continuum. Then is a *compactification* of , that is is a(n) (open) dense subset of and is compact.

We will prove the theorem only at the end of this note. Before that, we will mention two nice consequences of our construction. The first one is that Freudenthal compactification gives a topological invariant: the *number of ends,* .

**Theorem 2:** Let be two generalized continua. If is a homeomorphism, then extends to a homeomorphism ; in particular, induces a homeomorphism .

**Proof.** Let be an exhausting sequence for . Then is an exhausting sequence for . Moreover, if is an end of then is naturally an end of : it defines our extension . We easily check that is a homeomorphism.

**Corollary:** The number of ends is a topological invariant.

Of course, two non-homeomorphic spaces may have the same number of ends (eg. for all ). However, the number of ends appears to be easy to compute in several situations, so it may be a simple way to prove that two spaces are not homeomorphic. For example:

**Property:** Let and , be two finite subsets. If then and are not homeomorphic.

**Proof.** Just notice that and similarly .

Our second consequence of Freudenthal compactification is the classification of all possible compactifications whose *remainders* are finite (the *remainder* of a compactification of a space is ; sometimes, the points of the remainder are also called *points at infinity*).

**Theorem 3:** Let be a generalized continuum and be a compactification of whose remainder is finite. Then is obtained from by identifying some ends.

Therefore, Freudenthal compactification can be viewed as a maximal compactification among compactifications with finitely many points at infinity.

**Proof.** As above, let denote the Freudenthal compactification of . Let

and .

For all , let be an open neighborhood of ; we may suppose that if . Let

;

notice that is compact. For all , let be a connected neighborhood of disjoint from ; again, we may suppose that if .

Because is connected, there exists such that ; moreover, because if , such a is unique. Therefore, a map may be defined by and .

Let . Noticing that

is open in since if and only if , we deduce that is continuous. Moreover, is onto.

Let denote the space obtained from by identifying two ends and whenever .

By construction, is well-defined and into. Then, is continuous since is open, and onto since is onto. Finally, is a continuous bijection between two compact spaces: it is necessarily a homeomorphism.

We illustrate the classification given above with the following example:

**Corollary 1:** Let be a generalized continuum. Then has only one compactification with one point at infinity: its Alexandroff compactification.

**Corollary 2:** Let be a generalized continuum and be a compactification of whose remainder is finite. Then .

In particular, we get that the circle and the segment are the only compactifications of with finitely many points at infinity. However, there is a lot of possible compactifications of with infinitely many points; more precisely, the following result can be found in K. Magill’s article, *A note on compactification*:

**Theorem:** Let be Peano space. Then there exists a compactification of whose remainder is homeomorphic to .

For example, if denotes the graph of the function , then is homeomorphic to , and is a compactification of whose remainder is homeomorphic to .

See also B. Simon’s article *Some pictoral compactification of the real line* to find a compactification of whose remainder is homeomorphic to the torus .

From now on, we turn to the proof of theorem 1.

**Lemma:** Let be a generalized continuum. Then is a compact subspace of .

**Proof of Theorem 1:** Fistly, the injection is clearly a homeomorphism onto its image; therefore, from now on, will be confunded with its image in . Then is clearly dense in , so to conclude the proof, we only need to prove that is compact.

Let be an open covering of . Because is compact, there exists finite such that for all , there exist and an unbounded connected component of such that . If , then is a finite subcovering of the unbounded connected components of .

Notice that implies that , so a connected component of intersects or is included in . But is compact, so there are only finitely many components of intersecting . Therefore, if denotes the union of with the closure of the bounded components of intersecting , then is compact. In particular, there exists finite such that is a finite subcovering of .

Finally, we deduce that is a finite subcovering of . Hence is compact.

**Proof of the lemma.** Let be an exhausting sequence and let denote the set of unbounded connected components of . First notice that is finite. Indeed, if , it cannot be included in so it intersects ; the conclusion follows by compactness of . Therefore, if each is endowed with the discrete topology, from Tykhonov theorem the cartesian product is compact. Because there is a natural map

,

it is sufficient to prove that is a homeomorphism onto its image and that its image is closed. In order to show that is continuous, we take an open set of the form

for some and fixed. Now, if

,

then is open since

,

and . We deduce that is continuous. In order to show that is an open map, we take an open subspace of the form for some open subspace , and we notice that

where is the set of elements of included in . Therefore, is open, and we deduce that is an open map. We just proved that is a homeomorphism onto its image. Now we want to prove that is closed. We have

.

Now, if , let denotes the map that sends a connected component of to the connected component of containing it. Because is equivalent to , we deduce that

,

where

.

But is obviously continuous, is continuous by definition and is Hausdorff, so is closed. We conclude that is closed, and finally compact.