## 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. Acompactificationof a topological space X is a compact space Y containing X, such that 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

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: This is called aone-point compactificationof X.

**Examples.**

- A one-point compactification of (0, 1) (or
**R**) is given by the circle in the plane. - A one-point compactification of [0, 1) is given by [0, 1].
- A one-point compactification of is given by the union of two circles which are tangent to each other.
- A one-point compactification of the plane
**R**^{2}is given by the 2-sphere 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 then
*V*is an open subset of*Y*contained in*X*, so*V*is open in*X*.

- If , 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 is of the second type (it’s compact since ∩
*C*is closed is closed in each_{i}*C*)._{i} - Finally, 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 be an open cover. Then there’s at least one *C _{j}*, and

Since *C _{j}* is compact, it has a finite subcover. Together with

*Y*–

*C*, this forms a finite subcover of

_{j}*Y*.

Definition. The above construction is called theAlexandroff extensionof 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 by and Clearly*f*is continuous outside ∞. For continuity at ∞, an open subset containing (1, 0) must contain the set*V*of for some ε>0. Then is the set 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 is a bijective continuous map, then f is a homeomorphism.

**Proof**.

Continuity means 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 . Map the Alexandroff extension 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 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 containing 0. For ∞, we have*V*=*Y*–*C*, for a compact subset*C*of**Q**containing*U*. This means is closed in*C*and hence also compact, which is impossible by this lemma.

Lemma. The subspace 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 is contained in an open subset U such that is compact.

**Proof**.

(→) : every subspace of a Hausdorff space is Hausdorff. To show *X* is locally compact, let Pick open subsets *U*, *V* of *Y* such that Now *V* = *Y*–*C* for some closed compact subset *C* of *X*. Then Since cl_{X}(*U*) is closed in *C*, it’s compact.

(←) : since *X* is open in *Y*, any two points in can be separated by open subsets. The only problem is 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 belocally compactif for each 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 and open subset U containing x, there’s an open subset V containing x such thatand 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 separate *x* and *y* by open subsets such that

Since and *C* is compact, there’s a finite set of points such that Let which gives an open subset containing *x* which doesn’t intersect *W’*. Thus, cl(*W*) also doesn’t intersect *W’*, and :

, 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 of *Y* doesn’t contain an open subset whose closure in *Y* is compact. This is because any such open subset contains and the closure thus contains so this 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 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 where cl_{X}(*V*) is compact. Since *V* is an open subset of *X* contained in *U*, it’s also open in *U*; also

2nd statement: if *U* is an open subset of *X* containing *x*, pick any *i* such that By theorem 1, there exists *V* (open in *U _{i}* and hence in

*X*) containing

*x*such that and is compact and hence closed in

*X*. Thus and conversely, So and cl

_{X}(

*V*) is compact. ♦

**Exercise.**

Prove that if *X* is locally compact and Hausdorff, then so is any closed subset

**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 . By local compactness of *X*, there’s an open subset *U* of *X* containing *x* such that is compact. Likewise, there’s an open subset *V* of *Y* containing *y* such that is compact. Thus, (*x*, *y*) is contained in *U* × *V*, whose closure 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 where each is open and equality holds for all but finitely many *i*. The same must hold for any open 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,
**R**^{n}is locally compact, and so is any open or closed subset of**R**^{n}. - The set
*X*= {1/*n*:*n*= 1, 2, 3, … } is locally compact since it’s discrete. Its closure, given by , 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. ♦

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.

Thank you for your kind words. I try to motivate every definition I make, to the best of my ability.

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

Thanks – very good catch there. I’ll remove that line.