# Alexandroff extension

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 (but provides an embedding exactly for Tychonoff 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=√((1+c)/(1−c)){\textstyle 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≥√((1+c)/(1−c)){\textstyle 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

Let $X$ be a topological space. Put $X^{*}=X\cup \{\infty \},$ and topologize $X^{*}$ by taking as open sets all the open sets in X together with all sets of the form $V=(X\setminus C)\cup \{\infty \}$ where C is closed and compact in X. Here, $X\setminus C$ denotes the complement of $C$ in $X.$ 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 space $X^{*}$ is called the Alexandroff extension of X (Willard, 19A). Sometimes the same name is used for the inclusion map $c:X\to X^{*}.$
The properties below 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 T$_1$ if and only if X is T$_1$.

## 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 R$^n$ is homeomorphic to the n-sphere S$^n$. 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.