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# Contents

## Idea

The one-point compactification of a topological space $X$ is a new compact space $X^* = X \cup \{\infty\}$ obtained by adding a single new point “$\infty$” to the original space and declaring in $X^*$ the complements of the original closed compact subspaces to be open.

One may think of the new point added as the “point at infinity” of the original space. A continuous function on $X$ vanishes at infinity precisely if it extends to a continuous function on $X^*$ and literally takes the value zero at the point “$\infty$”.

This one-point compactification is also known as the Alexandroff compactification after a 1924 paper by Павел Сергеевич Александров (then transliterated ‘P.S. Aleksandroff’).

The one-point compactification is usually applied to a non-compact locally compact Hausdorff space. In the more general situation, it may not really be a compactification and hence is called the one-point extension or Alexandroff extension.

## Definition

### For topological spaces

###### Definition

(one-point extension)

Let $X$ be any topological space. Its one-point extension $X^*$ is the topological space

• whose underlying set is the disjoint union $X \cup \{\infty\}$

• and whose open sets are

1. the open subsets of $X$ (thought of as subsets of $X^*$);

2. the complements $X^\ast \backslash CK = (X \backslash CK) \cup \{\infty\}$ of the closed compact subsets $CK \subset X$.

###### Remark

If $X$ is Hausdorff, then it is sufficient to speak of compact subsets in def. , since compact subspaces of Hausdorff spaces are closed.

###### Lemma

(one-point extension is well-defined)

The topology on the one-point extension in def. is indeed well defined in that the given set of subsets is indeed closed under arbitrary unions and finite intersections.

###### Proof

The unions and finite intersections of the open subsets inherited from $X$ are closed among themselves by the assumption that $X$ is a topological space.

It is hence sufficient to see that

1. the unions and finite intersection of the $(X \backslash CK) \cup \{\infty\}$ are closed among themselves,

2. the union and intersection of a subset of the form $U \underset{\text{open}}{\subset} X \subset X^\ast$ with one of the form $(X \backslash CK) \cup \{\infty\}$ is again of one of the two kinds.

Regarding the first statement: Under de Morgan duality

$\underset{i \in \underset{\text{finite}}{J}}{\bigcap} (X \backslash CK_i \cup \{\infty\}) = \left( X \backslash \left(\underset{i \in \underset{\text{finite}}{J}}{\bigcup} CK_i \right)\right) \cup \{\infty\}$

and

$\underset{i \in I}{\bigcup} ( X \backslash CK_i \cup \{\infty\} ) = \left(X \backslash \left(\underset{i \in I}{\bigcap} CK_i \right)\right) \cup \{\infty\}$

and so the first statement follows from the fact that finite unions of compact subspaces and arbitrary intersections of closed compact subspaces are themselves again compact (this prop.).

Regarding the second statement: That $U \subset X$ is open means that there exists a closed subset $C \subset X$ with $U = X\backslash C$. Now using de Morgan duality we find

1. for intersections:

\begin{aligned} U \cap ( (X\backslash CK) \cup \{\infty\} ) & = (X \backslash C) \cap (X \backslash CK) \\ & = X \backslash (C \cup CK). \end{aligned}

Since finite unions of closed subsets are closed, this is again an open subset of $X$;

2. for unions:

\begin{aligned} U \cup (X \backslash CK) \cup \{\infty\} & = (X \backslash C) \cup (X \backslash CK) \cup \{\infty\} \\ & = (X \backslash (C \cap CK)) \cup \{\infty\} . \end{aligned}

For this to be open in $X^\ast$ we need that $C \cap CK$ is again compact. This follows because subsets are closed in a closed subspace precisely if they are closed in the ambient space and because closed subsets of compact spaces are compact.

### For non-commutative topological spaces ($C^\ast$-algebras)

Dually in non-commutative topology the one-point compactification corresponds to the unitisation of C*-algebras.

## Properties

### Basic properties

We discuss the basic properties of the construction $X^\ast$ in def. , in particular that it always yields a compact topological space (prop. below) and the ingredients needed to see its universal property in the Hausdorff case below.

###### Proposition

(one-point extension is compact)

For $X$ any topological space, then its one-point extension $X^\ast$ (def. ) is a compact topological space.

###### Proof

Let $\{U_i \subset X^\ast\}_{i \in I}$ be an open cover. We need to show that this has a finite subcover.

That we have a cover means that

1. there must exist $i_\infty \in I$ such that $U_{i \infty} \supset \{\infty\}$ is an open neighbourhood of the extra point. But since, by construction, the only open subsets containing that point are of the form $(X \backslash CK) \cup \{\infty\}$, it follows that there is a compact closed subset $CK \subset X$ with $X \backslash CK \subset U_{i \infty}$.

2. $\{U_i \subset X\}_{i \in i}$ is in particular an open cover of that closed compact subset $CK \subset X$. This being compact means that there is a finite subset $J \subset I$ so that $\{U_i \subset X\}_{i \in \J \subset X}$ is still a cover of $CK$.

Together this implies that

$\{U_i \subset X\}_{i \in J \subset I} \cup \{ U_{i_\infty} \}$

is a finite subcover of the original cover.

###### Proposition

(one-point extension of locally compact space is Hausdorff precisely if original space is)

Let $X$ be a locally compact topological space. Then its one-point extension $X^\ast$ (def. ) is a Hausdorff topological space precisely if $X$ is.

###### Proof

It is clear that if $X$ is not Hausdorff then $X^\ast$ is not.

For the converse, assume that $X$ is Hausdorff.

Since $X^\ast = X \cup \{\infty\}$ as underlying sets, we only need to check that for $x \in X$ any point, then there is an open neighbourhood $U_x \subset X \subset X^\ast$ and an open neighbourhood $V_\infty \subset X^\ast$ of the extra point which are disjoint.

That $X$ is locally compact implies by definition that there exists an open neighbourhood $U_k \supset \{x\}$ whose topological closure $CK \coloneqq Cl(U_x)$ is a closed compact neighbourhood $CK \supset \{x\}$. Hence

$V_\infty \coloneqq (X \backslash CK ) \cup \{\infty\} \subset X^\ast$

is an open neighbourhood of $\{\infty\}$ and the two are disjoint

$U_x \cap V_\infty = \emptyset$

by construction.

###### Proposition

(inclusion into one-point extension is open embedding)

Let $X$ be a topological space. Then the evident inclusion function

$i \;\colon\; X \longrightarrow X^\ast$

into its one-point extension (def. ) is

###### Proof

Regarding the first point: For $U \subset X$ open and $CK \subset X$ closed and compact, the preimages of the corresponding open subsets in $X^\ast$ are

$i^{-1}(U) = U \phantom{AAAA} i^{-1}( (X \backslash CK) \cup \infty ) = X \backslash CK$

which are open in $X$.

Regarding the second point: The image of an open subset $U \subset X$ is $i(U) = U \subset X^\ast$, which is open by definition.

Regarding the third point: We need to show that $i \colon X \to i(X) \subset X^\ast$ is a homeomorphism. This is immediate from the definition of $X^\ast$.

###### Remark

For $X$ a compact Hausdorff space and $x_0 \in X$ any point, then $X$ is homeomorphic to the one-point compactification of the complement subspace $X \setminus \{x_0\} \subset X$:

$X \simeq (X \setminus \{x_0\})^\ast \,.$

Observe also that $X \setminus \{x_0\}$, being an open subspace of a compact Hausdorff space is a locally compact topological space, since open subspaces of compact Hausdorff spaces are locally compact, and of course it is Hausdorff, since $X$ is.

###### Proof

Since closed subspaces of compact Hausdorff spaces are equivalently compact subspaces, the open neighbourhoods of $x \in X$ are equivalently the complements of closed, and hence compact closed, subsets in $X \setminus \{x\}$. By def. this means that the function

$\array{ X &\longrightarrow& (X \setminus \{x_0\})^\ast }$

which is the identity on $X \setminus \{x_0\}$ and sends $x_0 \mapsto \infty$ (hence which is just the identity on the underlying sets) is a homeomorphism.

### Universal property

As a pointed compact Hausdorff space, the one-point compactification of $X$ may be described by a universal property:

For every pointed compact Hausdorff space $(Y, y_0)$ and every continuous map $f \colon X \to Y$ such that the pre-image $f^{-1}(K)$ is compact for all compact sets $K$ not containing $y_0$, there is a unique basepoint-preserving continuous map $X^\ast \to Y$ that extends $f$.

This property characterizes $X^\ast$ in an essentially unique manner.

$X$ is dense in $X^*$ precisely if $X$ is not already compact. Note that $X^*$ is technically a compactification of $X$ only in this case.

$X^*$ is Hausdorff (hence a compactum) if and only if $X$ is already both Hausdorff and locally compact (see prop. ).

### Functoriality

The operation of one-point compactification is not a functor on the whole category of topological spaces. But it does extend to a functor on topological spaces with proper maps between them.

## Examples

### Spheres

###### Example

For $n \in \mathbb{N}$ the n-sphere with its standard topology (e.g. as a subspace of the Euclidean space $\mathbb{R}^{n+1}$ with its metric topology) is homeomorphic to the one-point compactification (def. ) of the Euclidean space $\mathbb{R}^n$

$S^n \simeq (\mathbb{R}^n)^\ast \,.$
###### Proof

Pick a point $\infty \in S^n$. By stereographic projection we have a homeomorphism

$S^n \setminus \{\infty\} \simeq \mathbb{R}^n \,.$

With this it only remains to see that for $U_\infty \supset \{\infty\}$ an open neighbourhood of $\infty$ in $S^n$ then the complement $S^n \setminus U_\infty$ is compact closed, and cnversely that the complement of every compact closed subset of $S^n \setminus \{\infty\}$ is an open neighbourhood of $\{\infty\}$.

Observe that under stereographic projection the open subspaces $U_\infty \setminus \{\infty\} \subset S^n \setminus \{\infty\}$ are identified precisely with the closed and bounded subsets of $\mathbb{R}^n$. (Closure is immediate, boundedness follows because an open neighbourhood of $\{\infty\} \in S^n$ needs to contain an open ball around $0 \in \mathbb{R}^n \simeq S^n \setminus \{-\infty\}$ in the other stereographic projection, which under change of chart gives a bounded subset. )

By the Heine-Borel theorem the closed and bounded subsets of $\mathbb{R}^n$ are precisely the compact, and hence the compact closed, subsets of $\mathbb{R}^n \simeq S^n \setminus \{\infty\}$.

Via the presentation of example , the canonical action of the orthogonal group $O(N)$ on $\mathbb{R}^n$ induces an action of $O(n)$ on $S^n$, which preserves the basepoint $\infty$ (the “point at infinity”).

This construction presents the J-homomorphism in stable homotopy theory and is encoded for instance in the definition of orthogonal spectra.

Slightly more generally, for $V$ any real vector space of dimension $n$ one has $S^n \simeq (V)^\ast$. In this context and in view of the previous case, one usually writes

$S^V \coloneqq (V)^\ast$

for the $n$-sphere obtained as the one-point compactification of the vector space $V$.

###### Proposition

For $V,W \in Vect_{\mathbb{R}}$ two real vector spaces, there is a natural homeomorphism

$S^V \wedge S^W \simeq S^{V\oplus W}$

between the smash product of their one-point compactifications and the one-point compactification of the direct sum.

###### Remark

In particular, it follows directly from this that the suspension $\Sigma(-) \simeq S^1 \wedge (-)$ of the $n$-sphere is the $(n+1)$-sphere, up to homeomorphism:

\begin{aligned} \Sigma S^n & \simeq S^{\mathbb{R}^1} \wedge S^{\mathbb{R}^n} \\ & \simeq S^{\mathbb{R}^1 \oplus \mathbb{R}^n} \\ & \simeq S^{\mathbb{R}^{n+1}} \\ & \simeq S^{n+1} \end{aligned} \,.

### Thom spaces

For $X$ a compact topological space and $V \to X$ a vector bundle, then the (homotopy type of the) one-point compactification of the total space $V$ is the Thom space of $V$, equivalent to $D(V)/S(V)$.

For a simple example: the real projective plane $\mathbb{RP}^2$ is the one-point compactification of the ‘open’ Möbius strip, as line bundle over $S^1$. This is a special case of the more general observation that $\mathbb{RP}^{n+1}$ is the Thom space of the tautological line bundle over $\mathbb{RP}^n$.

### Locally compact Hausdorff spaces

###### Example

(every locally compact Hausdorff space is an open subspace of a compact Hausdorff space)

###### Proof

In one direction the statement is that open subspaces of compact Hausdorff spaces are locally compact (see there for the proof). What we need to show is that every locally compact Hausdorff spaces arises this way.

So let $X$ be a locally compact Hausdorff space. By prop. and prop. its one-point extension $X^\ast$ (def. ) is a compact Hausdorff space. By prop. the canonical inclusion $X \to X^\ast$ is an open embedding of topological spaces.

## References

• John Kelly, General Topology (1975)

Last revised on April 18, 2019 at 11:31:45. See the history of this page for a list of all contributions to it.