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, $\sigma$-compact, Hausdorff topological space.

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

Let $X$ be a generalized continuum. Because $X$ is $\sigma$-compact, it is an increasing union of compact subspaces $(K_n)$; moreover, by locally compactness, we may suppose that $K_n \subset \mathrm{int} ~ K_{n+1}$ for all $n \geq 1$. Such a family of compact subspaces is called an exhausting sequence.

Definition: Let $X$ be a generalized continuum and $(K_n)$ be an exhausting sequence. An end $\epsilon$ is a decreasing family $(C_n)$ of subspaces such that $C_n$ is an unbounded (ie. not relatively compact) connected component of $X \backslash K_n$.

Let $E(X)$ denote the set of ends of $X$ and $\overline{X} = X \cup E(X)$. If $\epsilon=(C_n)$ is an end and $U \subset X$ is an open subspace, $\epsilon < U$ will mean that $C_n \subset U$ for some $n\geq 1$.

The Freudenthal compactification of $X$ is defined as $\overline{X}$ endowed with the topology generated by

$\{ U \cup \{ \epsilon \in E(X) \mid \epsilon < U \} \mid U \subset X \ \mathrm{open} \}$.

First, notice that $\overline{X}$ does not depend on the sequence of compact subspaces we chose. Indeed, it is possible to consider the increasing functions $f$ associating to any compact $K \subset X$ an unbounded connected component $f(K)$ of $X \backslash K$, and then to define a space $\tilde{X}= X \cup F(X)$, where $F(X)$ is the set of functions we just mentionned, endowed with the topology generated by

$\{U \cup \{ f \in F(X) \mid f(K) \subset U \ \mathrm{for \ some \ compact} \ K \subset X \} \mid U \subset X \ \mathrm{open} \}$.

Clearly, the restriction of a function $f \in F(X)$ to the family $(K_n)$ we fixed to construct $\overline{X}$ defines an end of $X$, so there exists a natural continuous map $\tilde{X} \to \overline{X}$. Conversely, an end $\epsilon = (C_n)$ naturally corresponds to a function of $F(X)$: if $K \subset X$ is compact, there exists some $n \geq 1$ such that $K \subset K_n$, and $\epsilon(K)$ may be defined as the unbounded connected component of $X \backslash K$ containing $C_n$. Therefore, the map $\tilde{X} \to \overline{X}$ turns out to be a homeomorphism. In particular, $\overline{X}$ does not depend on the sequence $(K_n)$ since neither does $\tilde{X}$.

Theorem 1: Let $X$ be a generalized continuum. Then $\overline{X}$ is a compactification of $X$, that is $X$ is a(n) (open) dense subset of $\overline{X}$ and $\overline{X}$ 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, $e(X)= |E(X)|$.

Theorem 2: Let $X,Y$ be two generalized continua. If $\varphi : X \to Y$ is a homeomorphism, then $\varphi$ extends to a homeomorphism $\overline{\varphi} : \overline{X} \to \overline{Y}$; in particular, $\overline{\varphi}$ induces a homeomorphism $E(X) \to E(Y)$.

Proof. Let $(K_n)$ be an exhausting sequence for $X$. Then $(\varphi(K_n))$ is an exhausting sequence for $Y$. Moreover, if $(C_n)$ is an end of $X$ then $(\varphi(C_n))$ is naturally an end of $Y$: it defines our extension $\overline{\varphi} : \overline{X} \to \overline{Y}$. We easily check that $\overline{\varphi}$ is a homeomorphism. $\square$

Corollary: The number of ends is a topological invariant.

Of course, two non-homeomorphic spaces may have the same number of ends (eg. $e(\mathbb{R}^n)=1$ for all $n \geq 2$). 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 $n,m \geq 2$ and $S \subset \mathbb{R}^n$, $R \subset \mathbb{R}^m$ be two finite subsets. If $|S| \neq |R|$ then $\mathbb{R}^n \backslash S$ and $\mathbb{R}^m \backslash R$ are not homeomorphic.

Proof. Just notice that $e(\mathbb{R}^n \backslash S)= |S|+1$ and similarly $e(\mathbb{R}^m \backslash R)= |R|+1$. $\square$

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

Theorem 3: Let $X$ be a generalized continuum and $Y$ be a compactification of $X$ whose remainder is finite. Then $Y$ is obtained from $\overline{X}$ 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 $\overline{X}$ denote the Freudenthal compactification of $X$. Let

$\overline{X} = X \cup \{ p_1, \dots, p_r \}$ and $Y= X \cup \{ q_1,\dots, q_s \}$.

For all $1 \leq i \leq s$, let $U_i \subset Y$ be an open neighborhood of $q_i$; we may suppose that $U_i \cap U_j = \emptyset$ if $i \neq j$. Let

$C= Y \backslash \bigcup\limits_{i=1}^s U_i \subset X$;

notice that $C$ is compact. For all $1 \leq i \leq r$, let $V_i \subset \overline{X}$ be a connected neighborhood of $p_i$ disjoint from $C$; again, we may suppose that $V_i \cap V_j = \emptyset$ if $i\neq j$.

Because $V_i$ is connected, there exists $j$ such that $V_i \subset U_j$; moreover, because $U_i \cap U_j = \emptyset$ if $i \neq j$, such a $j$ is unique. Therefore, a map $f : \overline{X} \to Y$ may be defined by $f(p_i)=q_j$ and $f_{|X}= \mathrm{Id}_{X}$.

Let $I_j = \{ 1 \leq i \leq r \mid f(p_i)=q_j \}$. Noticing that

$f^{-1}(U_j)= \bigcup\limits_{i \in I_j} V_i \cup \{ p_i \}$

is open in $\overline{X}$ since $i \in I_j$ if and only if $p_i < U_j$, we deduce that $f$ is continuous. Moreover, $f$ is onto.

Let $\tilde{X}$ denote the space obtained from $\overline{X}$ by identifying two ends $p_i$ and $p_j$ whenever $f(p_i)=f(p_j)$.

By construction, $\tilde{f} : \tilde{X} \to Y$ is well-defined and into. Then, $\tilde{f}$ is continuous since $\pi$ is open, and onto since $f$ is onto. Finally, $\tilde{f}$ is a continuous bijection between two compact spaces: it is necessarily a homeomorphism. $\square$

We illustrate the classification given above with the following example:

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

Corollary 2: Let $X$ be a generalized continuum and $Y$ be a compactification of $X$ whose remainder is finite. Then $|Y \backslash X | \leq e(X)$.

In particular, we get that the circle $\mathbb{S}^1$ and the segment $[0,1]$ are the only compactifications of $\mathbb{R}$ with finitely many points at infinity. However, there is a lot of possible compactifications of $\mathbb{R}$ with infinitely many points; more precisely, the following result can be found in K. Magill’s article, A note on compactification:

Theorem: Let $K$ be Peano space. Then there exists a compactification $X$ of $\mathbb{R}$ whose remainder is homeomorphic to $K$.

For example, if $\Gamma \subset [0,+ \infty) \times \mathbb{R}$ denotes the graph of the function $x \mapsto \sin(1/x)$, then $\Gamma$ is homeomorphic to $\mathbb{R}$, and $X = \Gamma \cup \{ 0 \} \cup [ -1 , 1 ]$ is a compactification of $\mathbb{R}$ whose remainder is homeomorphic to $[0,1]$.

See also B. Simon’s article Some pictoral compactification of the real line to find a compactification of $\mathbb{R}$ whose remainder is homeomorphic to the torus $\mathbb{T}^2$.

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

Lemma: Let $X$ be a generalized continuum. Then $E(X)$ is a compact subspace of $\overline{X}$.

Proof of Theorem 1: Fistly, the injection $X \hookrightarrow \overline{X}= X \cup E(X)$ is clearly a homeomorphism onto its image; therefore, from now on, $X$ will be confunded with its image in $\overline{X}$. Then $X$ is clearly dense in $\overline{X}$, so to conclude the proof, we only need to prove that $\overline{X}$ is compact.

Let $\{ O_i \mid i \in I \}$ be an open covering of $\overline{X}$. Because $E(X)$ is compact, there exists $J_1 \subset I$ finite such that for all $\epsilon \in E(X)$, there exist $j \in J_1$ and an unbounded connected component $C_j \subset O_j$ of $X \backslash K_{n_j}$ such that $\epsilon < C_j$. If $p= \max\limits_{j \in J_1} n_j$, then $\{ O_j \mid j\in J_1 \}$ is a finite subcovering of the unbounded connected components of $X \backslash K_p$.

Notice that $K_p \subset \mathrm{int} ~ K_{p+1}$ implies that $K_p \cap \partial K_{p+1} = \emptyset$, so a connected component of $X \backslash K_p$ intersects $\partial K_{p+1}$ or is included in $K_{p+1}$. But $\partial K_{p+1}$ is compact, so there are only finitely many components of $X \backslash K_p$ intersecting $\partial K_{p+1}$. Therefore, if $K$ denotes the union of $K_{p+1}$ with the closure of the bounded components of $X \backslash K_p$ intersecting $\partial K_{p+1}$, then $K$ is compact. In particular, there exists $J_2 \subset I$ finite such that $\{ O_j \mid j \in J_2 \}$ is a finite subcovering of $K$.

Finally, we deduce that $\{ O_j \mid j \in J_1 \cup J_2 \}$ is a finite subcovering of $\overline{X}$. Hence $\overline{X}$ is compact. $\square$

Proof of the lemma. Let $(K_n)$ be an exhausting sequence and let $\pi(K_n)$ denote the set of unbounded connected components of $X \backslash K_n$. First notice that $\pi(K_n)$ is finite. Indeed, if $C \in \pi(K_n)$, it cannot be included in $K_{n+1}$ so it intersects $\partial K_{n+1}$; the conclusion follows by compactness of $\partial K_{n+1}$. Therefore, if each $\pi(K_n)$ is endowed with the discrete topology, from Tykhonov theorem the cartesian product $\prod\limits_{n \geq 1} \pi(K_n)$ is compact. Because there is a natural map

$\phi : E(X) \to \prod\limits_{n \geq 1} \pi(K_n)$,

it is sufficient to prove that $\phi$ is a homeomorphism onto its image and that its image is closed. In order to show that $\phi$ is continuous, we take an open set $O \subset \prod\limits_{n \geq 1} \pi(K_n)$ of the form

$\{ (C_n) \mid C_{n_1}= C_1 \subset \cdots \subset C_{n_r} = C_r \}$

for some $r \geq 1$ and $C_1 \in \pi(K_1), \dots, C_r \in \pi(K_r)$ fixed. Now, if

$\Omega = \{ \epsilon \in E(X) \mid \mathrm{for \ all} \ 1 \leq i \leq r, \ \epsilon < C_{n_i} \} = \bigcap\limits_{i=1}^r \{ \epsilon \in E(X) \mid \epsilon < C_{n_i} \}$,

then $\Omega$ is open since

$\Omega = \bigcap\limits_{i=1}^r E(X) \cap \{ C_{n_i} \cap \{ \epsilon \} \mid \epsilon < C_{n_i} \}$,

and $\phi(\Omega) \subset O$. We deduce that $\phi$ is continuous. In order to show that $\phi$ is an open map, we take an open subspace $V \subset E(X)$ of the form $\{ \epsilon \mid \epsilon < U \}$ for some open subspace $U \subset X$, and we notice that

$\phi (U) = \{ (C_n) \mid \exists i \in \mathbb{N}, \ C_i \subset U \} = \bigcup\limits_{ i \in \mathbb{N} } \bigcup\limits_{ C \in \omega_i } \{ (C_n) \mid C_i = C \}$

where $\omega_i$ is the set of elements of $\pi(K_i)$ included in $U$. Therefore, $\phi(U)$ is open, and we deduce that $\phi$ is an open map. We just proved that $\phi$ is a homeomorphism onto its image. Now we want to prove that $\phi(E(X))$ is closed. We have

$\phi (E(X)) = \left\{ (C_n) \in \prod\limits_{n \geq 1} \pi(K_n) \mid \forall i \leq j, C_j \subset C_i \right\}$.

Now, if $i \leq j$, let $\phi_{ij}$ denotes the map that sends a connected component of $X \backslash K_j$ to the connected component of $X \backslash K_i$ containing it. Because $C_j \subset C_i$ is equivalent to $\phi_{ij} (C_j)=C_i$, we deduce that

$\phi(E(X))= \bigcap\limits_{i \leq j} M_{ij}$,

where

$M_{ij} = \left\{ C \in \prod\limits_{n \geq 1} \pi(K_n) \mid \phi_{ij} \circ \mathrm{pr}_j (C)= \mathrm{pr}_i (C) \right\}$.

But $\phi_{ij} : \pi(K_j) \to \pi(K_i)$ is obviously continuous, $\mathrm{pr}_i$ is continuous by definition and $\pi(K_i)$ is Hausdorff, so $M_{ij}$ is closed. We conclude that $\phi(E(X))$ is closed, and finally compact. $\square$