Topology: One-Point Compactification and Locally Compact Spaces

Compactifications

There’re lots of similarities between completeness and compactness, beyond the superficial resemblance of the words. For example, a closed subset of a compact (resp. complete) space is also compact (resp. complete). Two differences though:

compactness is a topological concept while completeness depends on the underlying metric;
compact metric spaces are complete (proposition 4 here) but not vice versa.

[ Intuitively, complete metric spaces can become unbounded and thus fail to be compact. But merely boundedness is not enough to ensure compactness, since any metric can be massaged to a bounded one without affecting the underlying topology. Indeed, it turns out a metric space is compact if and only if it’s complete and totally bounded. We won’t prove that here, but the interested reader may attempt it (it’s not too hard). ]

Also, any metric space can be “filled out” to give a completion. Likewise, can we compactify a general topological space?

Definition. A compactification of a topological space X is a compact space Y containing X, such that $X\hookrightarrow Y$ is a dense subspace of Y.

Upon doing that, we immediately run into the problem of uniqueness, i.e. the completion is unique while the compactification isn’t. E.g. (0, 1) can be compactified to give both [0, 1] and the unit circle $S^1 := \{(x,y)\in\mathbf{R}^2 : x^2 + y^2 = 1\}.$

Nonetheless, let’s continue exploring compactifications. The easiest way is to add just one point.

One-Point Compactifications

Task. Given a topological space X, we wish to construct a compact space Y by appending one point: $Y = X \cup \{\infty\}.$ This is called a one-point compactification of X.

Examples.

A one-point compactification of (0, 1) (or R) is given by the circle $S^1 := \{ (x,y)\in\mathbf{R}^2 : x^2+y^2=1\}$ in the plane.
A one-point compactification of [0, 1) is given by [0, 1].
A one-point compactification of $(0, 1) \cup (2, 3)$ is given by the union of two circles which are tangent to each other.
A one-point compactification of the plane R² is given by the 2-sphere $\{(x,y,z)\in\mathbf{R}^3 : x^2+y^2+z^2=1\}$ in 3-space.

Describing the Topology of Y

What would the topology of Y look like? We’ll start by making a reasonable assumption: that X is open in Y (i.e. {∞} is closed in Y). Now, given any open subset V of Y.

If $\infty\not\in V,$ then V is an open subset of Y contained in X, so V is open in X.
If $\infty\in V$ , then C := Y–V is closed in X. Also C is compact since it’s a closed subset of the compact space Y.

This inspires us to define open subsets of Y as follows:

not containing ∞: open subsets U of X, or
containing ∞: sets Y–C, where C is a closed subset of X which is compact.

Checking that this gives us a topology.

Indeed, a union of sets of the first type is of the first type.
A union of sets of the second type $\cup_i (Y-C_i) = Y-\cap_i C_i$ is of the second type (it’s compact since ∩C_i is closed is closed in each C_i).
Finally, $U\cup (Y-C) = Y-(C\cap (X-U))$ and C ∩ (X–U) is closed in X and compact since it’s a closed subset of C.

Obviously, X is open in Y and has the subspace topology.

Showing that Y is compact.

Let $Y = (\cup_i U_i) \cup (\cup_j (Y-C_j))$ be an open cover. Then there’s at least one C_j, and

$C_j \subseteq (\cup_i U_i) \cup (\cup_{j'\ne j} (Y-C_{j'})).$

Since C_j is compact, it has a finite subcover. Together with Y–C_j, this forms a finite subcover of Y.

Definition. The above construction is called the Alexandroff extension of the space X. It extends X by one point to give a compact space Y.

Examples and Properties

Suppose X = (0, 1). We map its Alexandroff extension $f:Y\to S^1$ by $t\mapsto (\cos(2\pi t), \sin(2\pi t))$ and $\infty \mapsto (1, 0).$ Clearly f is continuous outside ∞. For continuity at ∞, an open subset containing (1, 0) must contain the set V of $(\cos(2\pi t), \sin(2\pi t)),$ $-\epsilon<t<\epsilon,$ for some ε>0. Then $f^{-1}(V)$ is the set $Y - [\epsilon, 1-\epsilon]$ which is open in the Alexandroff extension. Now we’ll finish the proof by a trick.

Lemma. If X is a compact space, Y is Hausdorff, and $f:X\to Y$ is a bijective continuous map, then f is a homeomorphism.

Proof.

Continuity means $f^{-1}$ maps closed subsets to closed subsets. It suffices to show f maps closed subsets to closed subsets. But if C is closed in X, then C is compact and thus f(C) is compact. Since Y is Hausdorff, f(C) is closed in Y. Done. ♦

Let’s take the second example $X=[0,1)$ . Map the Alexandroff extension $f:Y \to [0,1]$ by taking t to itself and ∞ to 1. Once again f is continuous outside ∞. For continuity at ∞, an open subset containing 1 in [0, 1] must also contain V = (1-ε, 1] for some ε>0. But then $f^{-1}(V) = Y - [0, 1-\epsilon]$ which is open in the Alexandroff extension. Hence f is continuous and we invoke the above lemma to show that it is a homeomorphism.
Let X = Q. Then Y is not Hausdorff since there’re no disjoint open subsets U and V separating 0 and ∞. To see why, replacing an open subset by a smaller one, we may take $U=(-\epsilon, +\epsilon)\cap \mathbf{Q}$ containing 0. For ∞, we have V = Y–C, for a compact subset C of Q containing U. This means $[-\epsilon, +\epsilon] \cap\mathbf{Q}$ is closed in C and hence also compact, which is impossible by this lemma.

Lemma. The subspace $C:=[a, b] \cap \mathbf{Q}\subset \mathbf{R}$ for a<b is not compact.

Proof.

Indeed, C is not closed in R since its complement Y := R–C contains an irrational r, a<r<b, and no open neighbourhood of r is contained in Y. ♦

Now we’ll state some properties of the Alexandroff extension Y of X.

Property 1. X is dense in Y if and only if X is not compact.

Proof.

The closure of X in Y is either X or Y. Thus, X is not dense in Y iff X is closed in Y iff {∞} is open in Y iff X = Y-{∞} is compact. ♦

Property 2. Y is Hausdorff if and only if X is Hausdorff, and every $x\in X$ is contained in an open subset U such that $\text{cl}_X(U)$ is compact.

Proof.

(→) : every subspace of a Hausdorff space is Hausdorff. To show X is locally compact, let $x\in X.$ Pick open subsets U, V of Y such that $x\in U, \infty\in V, U\cap V=\emptyset.$ Now V = Y–C for some closed compact subset C of X. Then $x\in U\subseteq C.$ Since cl_X(U) is closed in C, it’s compact.

(←) : since X is open in Y, any two points in $X\subseteq Y$ can be separated by open subsets. The only problem is $x\in X, y = \infty\in Y.$ Pick open subset U of X containing x such that C := cl_X(U) is compact. Then U and Y–C are open subsets of Y which separate x and y. ♦

Finally, we end this section with an exercise.

Exercise.

Prove that if Y, Y’ are the Alexandroff extensions of X, X’ respectively, and f : X → X’ is continuous, then the map g : Y → Y’ which takes an element x of X to f(x) and ∞ to ∞ is also continuous.

Answer.

Since X is open in Y and f is continuous, g is also continuous on X. It remains to check ∞. Let V = Y’ – D be an open subset of Y’ where D is a closed compact subset of X’. Now f-1(V) = X’ – f-1(D), where C := f-1(D) is closed in X.

Locally Compact Spaces

The following definition is inspired by the above results.

Definition. A topological space X is said to be locally compact if for each $x\in X,$ there’s an open subset U containing x such that cl(U) is compact.

For example, compact spaces are clearly locally compact. Our discussion above tells us:

One-Point Compactification Theorem. If X is a Hausdorff and locally compact space which is not compact, then the Alexandroff extension Y is the unique one-point compactification of X which is Hausdorff.

The following alternative definition of local compactness is rather common.

Theorem 1. If X is Hausdorff, then it is locally compact if and only if for any $x\in X$ and open subset U containing x, there’s an open subset V containing x such that

$x\in V\subseteq \text{cl}(V) \subseteq U$

and cl(V) is compact.

Proof.

(←) is obvious. For (→), let U be an open subset of X containing x. Pick open V containing x such that cl(V) is compact. Let C = cl(V)-U = cl(V) ∩ (X–U), which is closed in cl(V) and hence compact. For each $y\in C,$ separate x and y by open subsets $W_y, W_y'$ such that

$x\in W_y,\ y\in W_y', \ W_y \cap W_y'=\emptyset.$

Since $\cup_{y\in C} W_y' \supseteq C$ and C is compact, there’s a finite set of points $y_1, y_2, \ldots, y_n$ such that $W' := \cup_{i=1}^n W_{y_i}' \supseteq C.$ Let $W = U\cap V\cap (\cap_{i=1}^n W_{y_i})$ which gives an open subset containing x which doesn’t intersect W’. Thus, cl(W) also doesn’t intersect W’, and $\text{cl}(W) \subseteq \text{cl}(V) - C \subseteq U$ :

$x\in W \subseteq \text{cl}(W) \subseteq U$ , with cl(W) closed in cl(V) and hence compact. ♦

If X is not Hausdorff, then this theorem fails. E.g. example 3 above (Alexandroff extension Y of Q) is compact, and hence locally compact by our definition. Yet the open subset $(-1, 1)\cap\mathbf{Q}$ of Y doesn’t contain an open subset whose closure in Y is compact. This is because any such open subset contains $(a,b)\cap\mathbf{Q}$ and the closure thus contains $[a,b]\cap \mathbf{Q}$ so this $[a,b]\cap \mathbf{Q}$ must be compact (impossible).

Hence, we’ll only consider locally compact spaces which are also Hausdorff. Note that a locally compact metric space is not necessarily complete, e.g. (0, 1) is a counter-example.

Properties of Locally Compact Spaces

The following result explains why the property is “local”.

Theorem 2. Let X be a Hausdorff space.

If X is locally compact, then so is any open subset U.

If $X = \cup_i U_i$ is an open cover of X and each U_i is locally compact, then so is X. In particular, a disjoint union of locally compact spaces is locally compact.

Proof.

[ Note: throughout the proof, be careful when talking about open and closed subsets, since we’ve to specify the ambient space too. ]

1st statement: if U’ is an open subset of U containing x, then it’s also an open subset of X containing x. By theorem 1, there’s an open subset V of X such that $x\in V\subseteq \text{cl}_X(V) \subseteq U',$ where cl_X(V) is compact. Since V is an open subset of X contained in U, it’s also open in U; also $\text{cl}_U(V) =\text{cl}_X(V) \cap U = \text{cl}_X(V).$

2nd statement: if U is an open subset of X containing x, pick any i such that $x\in U_i.$ By theorem 1, there exists V (open in U_i and hence in X) containing x such that $x\in V\subseteq \text{cl}_{U_i}(V) \subseteq U\cap U_i$ and $\text{cl}_{U_i}(V)$ is compact and hence closed in X. Thus $\text{cl}_X(V) \subseteq\text{cl}_{U_i}(V)$ and conversely, $\text{cl}_{U_i}(V) = \text{cl}_X(V) \cap U_i \subseteq \text{cl}_X(V).$ So $x\in V\subseteq \text{cl}_X(V) \subseteq U$ and cl_X(V) is compact. ♦

Exercise.

Prove that if X is locally compact and Hausdorff, then so is any closed subset $C\subseteq X.$

Answer (Highlight to read).

Let U be an open subset of C containing x; write U = U’ ∩ C for some open subset U’ of X. By theorem 1, there’s an open subset V’ of X, containing x, such that its closure cl_X(V’) is compact and contained in U’. The open subset V = V’ ∩ C of C then contains x, and its closure cl_C(V’) = cl_X(V’) ∩ C is closed in cl_X(V’) and hence compact. ♦

Theorem 3. If X and Y are locally compact and Hausdorff, so is X × Y.

Proof.

Let $(x, y) \in X\times Y$ . By local compactness of X, there’s an open subset U of X containing x such that $\text{cl}_X(U)$ is compact. Likewise, there’s an open subset V of Y containing y such that $\text{cl}_Y(V)$ is compact. Thus, (x, y) is contained in U × V, whose closure $\text{cl}(U\times V) = \text{cl}(U)\times \text{cl}(V)$ is compact. ♦

The product of infinitely many locally compact spaces is not locally compact in general (to get a locally compact space, one uses the restricted product which leads to adeles and ideles in algebraic number theory). E.g., let X = R^I, where I is infinite. Then any non-empty open subset U of X contains a basic open set of the form $\prod_{i\in I} U_i,$ where each $U_i\subseteq \mathbf{R}$ is open and equality holds for all but finitely many i. The same must hold for any open $V\subseteq U$ and cl(V) can’t be compact since projecting it onto one of the coordinates covers the whole R.

Examples

Any discrete topological space is locally compact and Hausdorff.
R is locally compact since if U contains x, then it must contain some (x-2ε, x+2ε) for some ε>0. Thus U contains (x-ε, x+ε) and its closure [x-ε, x+ε] which is compact.
Thus, Rⁿ is locally compact, and so is any open or closed subset of Rⁿ.
The set X = {1/n : n = 1, 2, 3, … } is locally compact since it’s discrete. Its closure, given by $X\cup \{0\}$ , is in fact compact.
The set of rationals Q is not locally compact since it’s Hausdorff but its Alexandroff extension is not.

Exercise

Is it true that if f : X → Y is continuous and X is locally compact, then so is Y? [ Recall that this is true if we replace “locally compact” by “compact”. ]

Answer (Highlight to read)

No, take X = Q with the discrete topology and Y = Q as a subspace of R. Then the identity map f : X → Y is continuous but Y = f(X) is not locally compact. ♦

6 Responses to Topology: One-Point Compactification and Locally Compact Spaces

Paulina says:

August 12, 2013 at 12:21 pm

That’s an excellent blog. It’s a great supplement to an ordinary topology handbook. There’s not much information here, but enough to make what I’ve read in Engelking clear. Thank you a lot for posting this.

- limsup says:
  
  August 12, 2013 at 2:45 pm
  
  Thank you for your kind words. I try to motivate every definition I make, to the best of my ability.
  
Kirtan says:

April 25, 2020 at 9:05 am

This blog is far better than Munkres topology. Amazing work. Though I like to point out a error. Image of complete metric space under uniformly continious map is not complete. For e.g. take identity map form (0,1) to (0,1). Let domain be endowed with discrete metric and co domain be endowed with standard metric. identity map is uniformly continious. Domain is complete while co domain is not complete

- limsup says:
  
  April 25, 2020 at 3:59 pm
  
  Thanks – very good catch there. I’ll remove that line.
  
Vanya says:

February 27, 2021 at 10:31 am

Is there any use of compactifying a space , for example to solve a problems that might not be possible by working with the original space?

- limsup says:
  
  March 1, 2021 at 10:26 am
  
  Roughly speaking, because compact Hausdorff spaces are nice. For example, a bijective continuous map of such spaces is a homeomorphism.
  
  Another example: a compact Hausdorff space X can be recovered from $C(X)$ (space of continuous $X\to \mathbb R$ ) by the Stone-Weierstrass theorem. By compactifying a locally compact Hausdorff space, we also obtain the locally compact version of the Stone-Weierstrass theorem.