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 $r = \sqrt{(1+c)/(1-c)}$. 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
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.

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 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.

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.