Topology/Countability

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BijectionEdit

A set is said to be countable if there exists a one to one correspondence between that set and the set of integers.

ExamplesEdit

The Even Integers: There is a simple bijection between the integers and the even integers, namely  , where  . Hence the even integers are countable.

A 2 - Dimensional Lattice: Let   represent the usual two dimensional integer lattice, then   is countable.

Proof: let   represent the function such that   and  , where   is whichever point:

  • not represented by some   for  
  •   is the lattice point 1 unit from   nearest to the origin. In the case where there are two such points, an arbitrary choice may be made.

Because   exists and is a bijection with the integers, The 2 - dimensional integer lattice is countable.

Axioms of countabilityEdit

First Axiom of CountabilityEdit

DefinitionEdit

A topological space  is said to satisfy the First Axiom of Countability if, for every   there exists a countable collection  of neighbourhoods of  , such that if   is any neighbourhood of  , there exists   with  .

A topological space that satisfies the first axiom of countability is said to be First-Countable.

All metric spaces satisfy the first axiom of countability because for any neighborhood   of a point  , there is an open ball   within  , and the countable collection of neighborhoods of   that are   where  , has the neighborhood   where  .

TheoremEdit

If a topological space satisfies the first axiom of countability, then for any point   of closure of a set  , there is a sequence   of points within   which converges to  .

ProofEdit

Let   be a countable collection of neighborhoods of   such that for any neighborhood   of  , there is an   such that  . Define
 .

Then form a sequence   such that  . Then obviously   converges to  .

TheoremEdit

Let   be a topological space satisfying the first axiom of countability. Then, a subset   of   is closed if an only if all convergent sequences   which converge to an element of   converge to an element of  .

ProofEdit

Suppose that   converges to   within  . The point   is a limit point of   and thus is a limit point of  , and since   is closed, it is contained within  . Conversely, suppose that all convergent sequences within   converge to an element within  , and let   be any point of contact for  . Then by the theorem above, there is a sequence   which converges to  , and so   is within  . Thus,   is closed.

TheoremEdit

If a topological space   satisfies the first axiom of countability, then   is continuous if and only if whenever   converges to  ,   converges to  .

ProofEdit

Let   satisfy the first axiom of countability, and let   be continuous. Let   be a sequence which converges to  . Let   be any open neighborhood of  . As   is continuous, there exists an open neighbourhood   of  . Since   to  , then there must exist an   such that   must contain   when  . Thus,   is a subset of   which contains   when  . Thus,   converges to  .
Conversely, suppose that whenever   converges to  , that   converges to  . Let   be a closed subset of  . Let   be a sequence which converges onto a limit  . Then   converges onto a limit  , which is within  . Thus,   is within  , implying that it is closed. Thus,   is continuous.

Second Axiom of CountabilityEdit

DefinitionEdit

A topological space is said to satisfy the second axiom of countability if it has a countable base.

A topological space that satisfies the second axiom of countability is said to be Second-Countable.

A topological space satisfies the second axiom of countable is first countable, since the countable collection of neighborhoods of a point can be all neighborhoods of the point within the countable base, so that any neighborhood   of that point must contain at least one neighborhood   within the collection, and   must be a subset of  .

TheoremEdit

If a topological space   satisfies the second axiom of countability, then all open covers of   have a countable subcover.

ProofEdit

Let   be an open cover of  , and let   be a countable base for  .   covers  . For all points  , select an element of  ,   which contains  , and an element of the base,   which contains   and is a subset of   (which is possible because   is a base).   forms a countable open cover for  . For each  , select an element of   which contains  , and this is a countable subcover of  .

Separable SpacesEdit

DefinitionEdit

A topological space   is separable if it has a countable proper subset   such that  .

Example:   is separable because   is a countable subset and  .

DefinitionEdit

A topological space   is seperable if it has a countable dense subset.

Example: The set of real numbers and complex numbers are both seperable.

TheoremEdit

If a topological space satisfies the second axiom of countability, then it is separable.

ProofEdit

Consider a countable base of a space  . Choose a point from each set within the base. The resulting set   of the chosen points is countable. Moreover, its closure is the whole space   since any neighborhood of any element of   must be a union of the bases, and thus must contain at least one element within the base, which in turn must contain an element of   because   contains at least one point from each base. Thus it is separable.

CorollaryEdit

In any topological space, second countability implies seperable and first countable. Prove of this is left for the reader.

TheoremEdit

If a metric space is separable, then it satisfies the second axiom of countability.

ProofEdit

Let   be a metric space, and let   be a countable set such that  . Consider the countable set   of open balls  . Let   be any open set, and let   be any element of  , and let   be an open ball of   within   with radius r. Let   be a number of the form   that is less than  . Because  , there is an element   such that  . Then the ball   is within   and is a subset of   because if  , then  . Thus   that contains  . The union of all such neighborhoods containing an element of   is  . Thus   is a base for  .

Corollary (Lindelöf covering theorem)Edit

If a metric space is separable, then it satisfies the second axiom of countability, and thus any cover of a subset of that metric space can be reduced to a countable cover.

Example: Since   is a separable metric space, it satisfies the second axiom of countability. This directly implies that any cover a set in   has a countable subcover.

Countable CompactnessEdit

DefinitionEdit

A subset   of a topological space   is said to be Countably Compact if and only if all countable covers of   have a finite subcover.

Clearly all compact spaces are countably compact.

A countably compact space is compact if it satisfies the second axiom of countability by the theorem above.

TheoremEdit

A topological space   is countably compact if and only if any infinite subset of that space has at least one limit point.

ProofEdit

( )Let  ,   be a set within   without any limit point. Then this sequence is closed, since they are all isolated points within the sequence. Let   for  . The   are all open sets, and so is a countable cover of the set, but any finite subcover   of this cover does not cover   because it does not contain  . This contradicts the assumption that   is countably compact.

( )Let   be open subsets of   such that any finite union of those sets does not cover  . Define:

 ,

which does not cover  , and is open. Select   such that  . There is a limit point   of this set of points, which must also be a limit point of  . Since   is closed,  . Thus,   and thus is not within any  , so   is not an open cover of X. Thus,   is countably compact.

Relative Countable CompactnessEdit

Since there is relative compactness, there is an analogous property called relative countable compactness.

DefinitionEdit

A subset S of a topological space X is relatively countably compact when its closure Cl(S) is countably compact.

Total BoundednessEdit

DefinitionEdit

A set   is an  -net of a metric space   where   if for any   within  , there is an element   such that  .

DefinitionEdit

A metric space   is totally bounded when it has a finite  -net for any  .

TheoremEdit

A countably compact metric space is totally bounded.

ProofEdit

Any infinite subset of a countably compact metric space   must have at least one limit point. Thus, selecting   where   is at least   apart from any   where  , one must eventually have formed an  -net because this process must be finite, because there is no possible infinite set with all elements more than   apart.

TheoremEdit

A totally bounded set is separable.

ProofEdit

Take the union of all finite  -nets, where   varies over the natural numbers, and that is a countable set such that its closure is the whole space  .

Urysohn's Metrizability TheoremEdit

The following theorem establishes a sufficient condition for a topological space to be metrizable.

TheoremEdit

A second countable normal T1 topological space is homeomorphic to a metric space.

ProofEdit

We are going to use the Hilbert cube, which is a metric space, in this proof, to prove that the topological space is homeomorphic to a subset of the Hilbert cube, and is thus a metric space.

First, since all T1 normal spaces are Hausdorff, all single points are closed sets. Therefore, consider any countable base of the topological space X, and any open   set of it. Select a point   within this open set. Since the complement of the open set is closed, and since a point within the open set is also closed, and since these two closed sets are disjoint, we can apply Urysohn's lemma to find a continuous function   such that:

 
 

It is easy to see from the proof of Urysohn's lemma that we have not only constructed a function with such properties, but that such that  , meaning that the function value of any point within the open set is less than 1.

Now define the function   from X to the Hilbert cube to be  .

To prove that this is continuous, let   be a sequence that converges to a. Consider the open ball   where  . There exists an N such that

 .

Moreover, since   is a continuous function from X to [0,1], there exists a neighborhood of  , and therefore an open set   of the base within that neighborhood containing a such that if  , then

 

or

 .

Let

 .

In addition, since  , there exists an   (i=1,2,3,...,M-1) such that when  , that  , and let M be the maximum of   so that when n>M, then  .

Let n>M, and then the distance from   to g(a) is now

         

This proves that it is continuous.

To prove that this is one-to-one, consider two different points, a and b. Since the space is Hausdorff, there exists disjoint open sets   and  , and select an element of the base   that contains a and is within  . It follows that   whereas  , proving that the function g is one-to-one, and that there exists an inverse  .

To prove that the inverse   is continuous, let   be an open set within the countable base of X. Consider any point x within  . Since  , indicating that there exists an   such that when

 

then  .

Suppose that  . Then

 

Implying that   indicating that  .

Now consider any open set O around x. Then there exists a set of the base   and an   such that whenever  , then  , meaning that  . This proves that the inverse is continuous.

Since the function is continuous, is one-to-one, and has a continuous inverse, it is thus a homeomorphism, proving that X is metrizable.

Note that this also proves that the Hilbert cube thus contains any second-countable normal T1 space.

Hahn-Mazurkiewicz TheoremEdit

 
The Hilbert Curve- a space filling curve

The Hahn-Mazurkiewicz theorem is one of the most historically important results of point-set topology, for it completely solves the problem of "space-filling" curves. This theorem provides the necessary and sufficient condition for a space to be 'covered by curve', a property that is widely considered to be counter-intuitive.

Here, we present the theorem without its proof.

TheoremEdit

A Hausdorff space is a continuous image of the unit interval   if and only if it is a compact, connected, locally connected and second-countable space.


ExerciseEdit

  1. Prove that a separable metric space satisfies the second axiom of countability. Hence, or otherwise, prove that a countably compact metric space is compact.
  2. Prove the sufficiency condition of the Hahn-Mazurkiewicz theorem:
    If a Hausdorff space is a continuous image of the unit interval, then it is compact, connected, locally-connected and second countable.
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