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).

ComDiagram

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:

FCompactification

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

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