A set is said to be countable if there exists a one to one correspondence between that set and the set of integers.
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
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 .
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 .
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 .
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 .
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.
If a topological space satisfies the first axiom of countability, then is continuous if and only if whenever converges to , converges to .
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
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 .
If a topological space satisfies the second axiom of countability, then all open covers of have a countable subcover.
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 .
A topological space is separable if it has a countable proper subset such that .
Example: is separable because is a countable subset and .
A topological space is seperable if it has a countable dense subset.
Example: The set of real numbers and complex numbers are both seperable.
If a topological space satisfies the second axiom of countability, then it is separable.
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.
In any topological space, second countability implies seperable and first countable. Prove of this is left for the reader.
If a metric space is separable, then it satisfies the second axiom of countability.
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.
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.
A topological space is countably compact if and only if any infinite subset of that space has at least one limit point.
( )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.
A subset S of a topological space X is relatively countably compact when its closure Cl(S) is countably compact.
A set is an -net of a metric space where if for any within , there is an element such that .
A metric space is totally bounded when it has a finite -net for any .
A countably compact metric space is totally bounded.
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.
A totally bounded set is separable.
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.
A second countable normal T1 topological space is homeomorphic to a metric space.
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
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
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.
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.
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.
- Prove that a separable metric space satisfies the second axiom of countability. Hence, or otherwise, prove that a countably compact metric space is compact.
- 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.