# Alexandroff extension

(Redirected from One-point compactification)

In the mathematical field of topology, the Alexandroff extension is a way to extend a noncompact topological space by adjoining a single point in such a way that the resulting space is compact. It is named after the Russian mathematician Pavel Alexandroff. More precisely, let X be a topological space. Then the Alexandroff extension of X is a certain compact space X* together with an open embedding c : X → X* such that the complement of X in X* consists of a single point, typically denoted ∞. The map c is a Hausdorff compactification if and only if X is a locally compact, noncompact Hausdorff space. For such spaces the Alexandroff extension is called the one-point compactification or Alexandroff compactification. The advantages of the Alexandroff compactification lie in its simple, often geometrically meaningful structure and the fact that it is in a precise sense minimal among all compactifications; the disadvantage lies in the fact that it only gives a Hausdorff compactification on the class of locally compact, noncompact Hausdorff spaces, unlike the Stone–Čech compactification which exists for any topological space, a much larger class of spaces.

## Example: inverse stereographic projection

A geometrically appealing example of one-point compactification is given by the inverse stereographic projection. Recall that the stereographic projection S gives an explicit homeomorphism from the unit sphere minus the north pole (0,0,1) to the Euclidean plane. The inverse stereographic projection $S^{-1}:\mathbb {R} ^{2}\hookrightarrow S^{2}$  is an open, dense embedding into a compact Hausdorff space obtained by adjoining the additional point $\infty =(0,0,1)$ . Under the stereographic projection latitudinal circles $z=c$  get mapped to planar circles $r={\sqrt {(1+c)/(1-c)}}$ . It follows that the deleted neighborhood basis of $(0,0,1)$  given by the punctured spherical caps $c\leq z<1$  corresponds to the complements of closed planar disks $r\geq {\sqrt {(1+c)/(1-c)}}$ . More qualitatively, a neighborhood basis at $\infty$  is furnished by the sets $S^{-1}(\mathbb {R} ^{2}\setminus K)\cup \{\infty \}$  as K ranges through the compact subsets of $\mathbb {R} ^{2}$ . This example already contains the key concepts of the general case.

## Motivation

Let $c:X\hookrightarrow Y$  be an embedding from a topological space X to a compact Hausdorff topological space Y, with dense image and one-point remainder $\{\infty \}=Y\setminus c(X)$ . Then c(X) is open in a compact Hausdorff space so is locally compact Hausdorff, hence its homeomorphic preimage X is also locally compact Hausdorff. Moreover, if X were compact then c(X) would be closed in Y and hence not dense. Thus a space can only admit a Hausdorff one-point compactification if it is locally compact, noncompact and Hausdorff. Moreover, in such a one-point compactification the image of a neighborhood basis for x in X gives a neighborhood basis for c(x) in c(X), and—because a subset of a compact Hausdorff space is compact if and only if it is closed—the open neighborhoods of $\infty$  must be all sets obtained by adjoining $\infty$  to the image under c of a subset of X with compact complement.

## The Alexandroff extension

Put $X^{*}=X\cup \{\infty \}$ , and topologize $X^{*}$  by taking as open sets all the open subsets U of X together with all sets $V=(X\setminus C)\cup \{\infty \}$  where C is closed and compact in X. Here, $X\setminus C$  denotes setminus. Note that $V$  is an open neighborhood of $\{\infty \}$ , and thus, any open cover of $\{\infty \}$  will contain all except a compact subset $C$  of $X^{*}$ , implying that $X^{*}$  is compact (Kelley 1975, p. 150).

The inclusion map $c:X\rightarrow X^{*}$  is called the Alexandroff extension of X (Willard, 19A).

The properties below all follow from the above discussion:

• The map c is continuous and open: it embeds X as an open subset of $X^{*}$ .
• The space $X^{*}$  is compact.
• The image c(X) is dense in $X^{*}$ , if X is noncompact.
• The space $X^{*}$  is Hausdorff if and only if X is Hausdorff and locally compact.
• The space $X^{*}$  is T1 if and only if X is T1.

## The one-point compactification

In particular, the Alexandroff extension $c:X\rightarrow X^{*}$  is a Hausdorff compactification of X if and only if X is Hausdorff, noncompact and locally compact. In this case it is called the one-point compactification or Alexandroff compactification of X.

Recall from the above discussion that any Hausdorff compactification with one point remainder is necessarily (isomorphic to) the Alexandroff compactification. In particular, if $X$  is a compact Hausdorff space and $p$  is a limit point of $X$  (i.e. not an isolated point of $X$ ), $X$  is the Alexandroff compactification of $X\setminus \{p\}$ .

Let X be any noncompact Tychonoff space. Under the natural partial ordering on the set ${\mathcal {C}}(X)$  of equivalence classes of compactifications, any minimal element is equivalent to the Alexandroff extension (Engelking, Theorem 3.5.12). It follows that a noncompact Tychonoff space admits a minimal compactification if and only if it is locally compact.

## Non-Hausdorff one-point compactifications

Let $(X,\tau )$  be an arbitrary noncompact topological space. One may want to determine all the compactifications (not necessarily Hausdorff) of $X$  obtained by adding a single point, which could also be called one-point compactifications in this context. So one wants to determine all possible ways to give $X^{*}=X\cup \{\infty \}$  a compact topology such that $X$  is dense in it and the subspace topology on $X$  induced from $X^{*}$  is the same as the original topology. The last compatibility condition on the topology automatically implies that $X$  is dense in $X^{*}$ , because $X$  is not compact, so it cannot be closed in a compact space. Also, it is a fact that the inclusion map $c:X\to X^{*}$  is necessarily an open embedding, that is, $X$  must be open in $X^{*}$  and the topology on $X^{*}$  must contain every member of $\tau$ . So the topology on $X^{*}$  is determined by the neighbourhoods of $\infty$ . Any neighborhood of $\infty$  is necessarily the complement in $X^{*}$  of a closed compact subset of $X$ , as previously discussed.

The topologies on $X^{*}$  that make it a compactification of $X$  are as follows:

• The Alexandroff extension of $X$  defined above. Here we take the complements of all closed compact subsets of $X$  as neighborhoods of $\infty$ . This is the largest topology that makes $X^{*}$  a one-point compactification of $X$ .
• The open extension topology. Here we add a single neighborhood of $\infty$ , namely the whole space $X^{*}$ . This is the smallest topology that makes $X^{*}$  a one-point compactification of $X$ .
• Any topology intermediate between the two topologies above. For neighborhoods of $\infty$  one has to pick a suitable subfamily of the complements of all closed compact subsets of $X$ ; for example, the complements of all finite closed compact subsets, or the complements of all countable closed compact subsets.

## Further examples

### Compactifications of discrete spaces

• The one-point compactification of the set of positive integers is homeomorphic to the space consisting of K = {0} U {1/n | n is a positive integer} with the order topology.
• A sequence $\{a_{n}\}$  in a topological space $X$  converges to a point $a$  in $X$ , if and only if the map $f\colon \mathbb {N} ^{*}\to X$  given by $f(n)=a_{n}$  for $n$  in $\mathbb {N}$  and $f(\infty )=a$  is continuous. Here $\mathbb {N}$  has the discrete topology.
• Polyadic spaces are defined as topological spaces that are the continuous image of the power of a one-point compactification of a discrete, locally compact Hausdorff space.

### Compactifications of continuous spaces

• The one-point compactification of n-dimensional Euclidean space Rn is homeomorphic to the n-sphere Sn. As above, the map can be given explicitly as an n-dimensional inverse stereographic projection.
• The one-point compactification of the product of $\kappa$  copies of the half-closed interval [0,1), that is, of $[0,1)^{\kappa }$ , is (homeomorphic to) $[0,1]^{\kappa }$ .
• Since the closure of a connected subset is connected, the Alexandroff extension of a noncompact connected space is connected. However a one-point compactification may "connect" a disconnected space: for instance the one-point compactification of the disjoint union of a finite number $n$  of copies of the interval (0,1) is a wedge of $n$  circles.
• The one-point compactification of the disjoint union of a countable number of copies of the interval (0,1) is the Hawaiian earring. This is different from the wedge of countably many circles, which is not compact.
• Given $X$  compact Hausdorff and $C$  any closed subset of $X$ , the one-point compactification of $X\setminus C$  is $X/C$ , where the forward slash denotes the quotient space.
• If $X$  and $Y$  are locally compact Hausdorff, then $(X\times Y)^{*}=X^{*}\wedge Y^{*}$  where $\wedge$  is the smash product. Recall that the definition of the smash product:$A\wedge B=(A\times B)/(A\vee B)$  where $A\vee B$  is the wedge sum, and again, / denotes the quotient space.

### As a functor

The Alexandroff extension can be viewed as a functor from the category of topological spaces with proper continuous maps as morphisms to the category whose objects are continuous maps $c\colon X\rightarrow Y$  and for which the morphisms from $c_{1}\colon X_{1}\rightarrow Y_{1}$  to $c_{2}\colon X_{2}\rightarrow Y_{2}$  are pairs of continuous maps $f_{X}\colon X_{1}\rightarrow X_{2},\ f_{Y}\colon Y_{1}\rightarrow Y_{2}$  such that $f_{Y}\circ c_{1}=c_{2}\circ f_{X}$ . In particular, homeomorphic spaces have isomorphic Alexandroff extensions.