Topology/Compactness

Topology
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The notion of Compactness appears in a wide variety of contexts. In particular, compactness is a "tameness property" that tells you that the objects you are dealing with are in some sense well-behaved.

Definition

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Let   be a topological space and let  

A collection   of open sets   is said to be an Open Cover of   if  

  is said to be Compact if and only if every open cover of   has a finite subcover. More formally,   is compact iff for every open cover   of  , there exists a finite subset   of   that is also an open cover of  .

If the set   itself is compact, we say that   is a Compact Topological Space.

Compactness of topological spaces can also be expressed by one of the following equivalent characterisations:

  • Every filter on   containing a filter basis of closed sets has a nonempty intersection.
  • Every ultrafilter on   converges.

Important Properties

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  • Every closed subset of a compact set is compact
    Proof:
    Let   be a compact set, and let   be a closed subset of  . Consider any open cover   of  . Observe that   being open, the collection of open sets  is an open cover of  . As   is compact, this open cover has a finite subcover  .
    Now, consider the collection  . This collection is obviously finite and is also a subcover of  . Hence, it is a finite subcover of  


  • Every compact subset of a Hausdorff space is closed.
    Proof:
    Let   be compact. If the complement   is empty, then   is the same as the space; thus closed. Suppose not; that is, there is a point  . Then for each  , by the Hausdorff separation axiom we can find   and   disjoint, open and such that   and  . Since   is compact and the collection   covers  , we can find a finite number of points   in   such that:
     
    It then follows that:
     . Hence, every   has an open neighbourhood  .
    As   can be represented as the union of open sets  ,   is open and   is closed.


  • Every compact set in a metric space is bounded.
    Proof:
    Let   be a metric space and let   be compact.
    Consider the collection of open balls   for some (fixed)  . We see that   is an open cover of  . As   is compact, it has a finite subcover, say  . Let  . We see that  , and hence,   is bounded.


  • Heine-Borel Theorem: For any interval  , and for any open cover   of that interval, there exists a finite subcover of  .
    Proof:
    Let   be the set of all   such that   has a finite subcover of  .   is non-empty because   is within the set. Define  .
    Assume if possible,  . Then there is a finite cover of sets within   for  .   is within a set   within the cover  . Thus, there exists a   such that  . Then   is also within  , contradicting the definition of  . Thus,  . Therefore,   has a finite subcover.

Sources differ as to what exactly should be called the 'Heine-Borel Theorem'. It seems that Emile Borel proved the most relevant result, dealing with compact subsets of a Euclidean Space. However, we provide the simpler case, for reals.


  • Let   be topological spaces. If   is continuous, and   is compact, then the image of  ,  , is compact.
    Proof:
    Let   be any open cover of  . Consider the inverses   where  . These inverses are open because   is continuous. This covers  , and thus there is a finite subcover of  ,  . Then the images   is a finite subcover of  .


  • If a set is compact and Hausdorff, then it is normal.
    Proof:
    Let   be compact and Hausdorff. Consider two closed subsets   and   which are themselves compact by theorem 1 above. For every   and  , there exist two disjoint sets   and   such that   and  . The union of all such   for a fixed   is a cover for  , and thus it has a finite subcover, say,   and let   be the union of its members.
    Let  , and let  . Observe that   being finite,   is open. The union   covers  , and therefore it has a finite subcover  . Let   be the union of all members of this subcover.
    Let   denote the set of all elements   such that  . Take the intersection  , which is open.
    Then   is an open superset of  ,   is an open superset of  , and they are disjoint. Thus,   is normal.
  • In a compact metric space X, a function from X to Y is uniformly continuous if and only if it is continuous.
    Proof:
  • If two topological spaces are compact, then their product space is also compact.
    Proof:
    Let X1 and X2 be two compact spaces. Let S be a cover of X1×X2. Let x be an element of X1. Consider the sets Ax,y within S that contain (x,y) for each y in X2.   forms a cover for X2, with a finite subcover {Ax,yi}. Let Bx be the intersection of   within {Ayi}, which is open. Thus, {Bx} forms an open cover, which has a finite subcover, {Bxi}. The corresponding sets {Axi,yi} is finite, and forms an open subcover of the set.
  • All closed and bounded sets in the Euclidean Space are compact.
    Proof:
    Let S by any bounded closed set in  . Then since S is bounded, it is contained in some "box" of the products of closed intervals of R. Since those closed intervals are compact, their product is also compact. Therefore, S is a closed set in a compact set, and is therefore also compact.

Tychonoff's Theorem

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The more general result on the compactness of product spaces is called Tychonoff's Theorem. Unlike the compactness of the product of two spaces, however, Tychonoff's Theorem requires Zorn's Lemma. (In fact, it is equivalent to Axiom of Choice.)

Theorem: Let  , and let each   be compact. Then the X is also compact.

Proof: The proof is in terms of nets. Recall the following facts:

Lemma 1 - A net   in   converges to   if and only if each coordinate   converges to  .

Lemma 2 - A topological space   is compact if and only if every net in   has a convergent subnet.

Lemma 3 - Every net has a universal subnet.

Lemma 4 - A universal net   in a compact space   is convergent.

We now prove Tychonoff's theorem.

Let   be a net in  .

Using Lemma 3 we can find a universal subnet   of  .

It is easily seen that each coordinate net   is a universal net in  .

Using Lemma 4 we see that each coordinate net converges, because   is compact.

Using Lemma 1 we see that the whole net   converges in  .

We conclude that every net in   has a convergent subnet, so, by Lemma 2,   must be compact.  

Relative Compactness

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Relative compactness is another property of interest.

Definition: A subset S of a topological space X is relative compact when the closure Cl(x) is compact.

Note that relative compactness does not carry over to topological subspaces. For example, the open interval (0,1) is relatively compact in R with the usual topology, but is not relatively compact in itself.

Local Compactness

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The idea of local compactness is based on the idea of relative compactness.

If, in a topological space X, every element has a neighborhood that is relatively compact, then X is locally compact.

It can be shown that all compact sets are locally compact, but not conversely.

Exercises

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  1. It is not true in general for a metric space that a closed and bounded set is compact. Take the following metric on a set X:

 

a) Show that this is a metric

b) Which subspaces of X are compact

c) Show that if Y is a subspace of X and Y is compact, then Y is closed and bounded

d) Show that for any metric space, compact sets are always closed and bounded

e) Show that with this particular metric, closed and bounded sets need not be compact


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