Topology/Countability

Bijection

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

Examples

The Even Integers: There is a simple bijection between the integers and the even integers, namely $f:𝐙\to 𝐙$, where $f(n)=2n$. Hence the even integers are countable.

A 2 - Dimensional Lattice: Let ${𝐙}^2$ represent the usual two dimensional integer lattice, then ${𝐙}^2$ is countable.

Proof: let $f:𝐙\to 𝐙$ represent the function such that $f(0)=(0,)$ and $f(n)=(x,y)$, where $(x,y)$ is whichever point:

• not represented by some $f(m)$ for $m<n$

• $(x,y)$ is the lattice point 1 unit from $f(n-1)$ nearest to the origin. In the case where there are two such points, an arbitrary choice may be made.

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

Axioms of countability

First Axiom of Countability

Definition: a topological space which satisfies the first axiom of countability is a topological space such that for any point, there exists a countable collection of neighborhoods of that set such that for any open neighborhood of that point, there exists a set within that collection that is a subset of that neighborhood.

All metric spaces satisfy the first axiom of countability because for any neighborhood N of a point x, there is an open ball B_r(x) within N, and the countable collection of neighborhoods of x that are B_1/k(x) where k is a natural number has the neighborhood B_1/n(x) where $\frac{1}{n}$<r.

Theorem

Theorem: If a topological space satisfies the first axiom of countability, then for any point of closure x of a set S, there is a sequence {a_i} of points within S which converges to x.

Proof

Let {A_i} be a countable collection of neighborhoods of x such that for any neighborhood N of x, there is an A_i such that A_i⊆N. Define

$B_n=\bigcap _{i=1}^n{A}_n$.

Then form a sequence {a_i} such that a_i∈B_n. Then obviously {a_i} converges to x.

Theorem

A closed subset A of a topological space X is closed if an only if all convergent sequences which converge to an element of X converges to an element of A.

Proof

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

Theorem

If a topological space X satisfies the first axiom of countability, then f:X→Y is continuous if and only if whenever {x_n} converges to x, that {f(x_n)} converges to f(x).

Proof

Let X satisfy the first axiom of countability, and let f:X→Y be continuous. Let {x_n} be a sequence which converges to x. Let B be any open neighborhood of Y. Since {x_n} to x, then there must exist an N such that f^-1(B) must contains x_n when n>N. Thus, f(f^-1(B)) is a subset of B which contains f(x_n) when n>N. Thus, f(x_n) converges to f(x).
Conversely, suppose that whenever {x_n} converges to x, that {f(x_n)} converges to f(x). Let B be a closed subset Y. Let x_n∈f^-1(B) be a sequence which converges onto a limit x. Then f(x_n) converges onto a limit f(x), which is within B. Thus, x is within f^-1(B), implying that it is closed. Thus, f is continuous.

Second Axiom of Countability

Definition: a topological space which satisfies the second axiom of countability is a topological space with a countable base.

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 N of that point must contain at least one neighborhood A within the collection, and A must be a subset of N.

Theorem

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

Proof

Let C be an open cover of X, and let B be a countable base for X. B covers X. For all points x, select an element of C, C_x which contains x, and an element of B_x which contains x and is a subset of C_x (which is possible because B is a base). {B_x} forms a countable open cover for X. For each B_x, select an element of C which contains B_x, and this is a countable subcover of C.

Seperable Spaces

Definition: a topological space T is separable if it has a countable subset A such that Cl(A)=T.

Example: Rⁿ is separable because Qⁿ is a countable subset and Cl(Qⁿ)=Rⁿ.

Theorem

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

Proof

Consider a countable base of a space T. Choose a point from each set within the base. The resulting set A of the chosen points is countable. Moreover, its closure is the whole space T since any neighborhood of any element of T 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 A because A contains at least one point from each base. Thus it is separable.

Theorem

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

Proof

Let T be a metric space, and let A be a countable set such that Cl(A)=T. Consider the countable set B of open balls {B_1/k(p)|k∈N, p∈A}. Let O be any open set, and let x be any element of O, and let N be an open ball of x within O with radius r. Because Cl(A)=T, there is an element x'∈A such that d(x',x)<$\frac{r}{4}$. Let r' be a number of the form 1/n that is less than r. Then the ball B_r'/2(x') is within B and is a subset of O because if y∈B_r'/2(x'), then d(y,x)≤d(y,x')+d(x',x)<$\frac{3}{4}r$. Thus B_r'/2⊆O that contains x. The union of all such neighborhoods containing an element of O is O. Thus B is a base for T.

Corollary (Lindelöf covering theorem)

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 R^{n is a separable metric space, it satisfies the second axiom of countability. This directly implies that any cover a set in Rn has a countable subcover.}

Countable Compactness

Definition: A set X is countably compact if and only if all countable covers of X 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.

Theorem

If a topological space X is countably compact, then any infinite subset of that space must have at least one limit point.

Proof

Let {x_i} (i=1,2,3,...) be a set within X without any limit point. Let S_n={x_i} (i=n, n+1, n+2, ...). The X-S_n is a countable cover of the set, but any finite subcover {X-S_ni} of this cover does not cover X because it does not contain S_nmax{i}.

Theorem

If all infinite subsets of a topological space X contains at least one limit point, then it is countably compact.

Proof

Let {S_n} be subsets of X such that any finite union of those sets does not cover X. Define:

$B_n=\bigcup _{i=1}^n{S}_n$,

which does not cover X, and is open. Select x_n such that x_n∈B_n. There is a limit point x of this set of points, which must also be a limit point of B_n. Since X-B_n is closed, x∈B_n. Thus, x∉B_n and thus is not within any S_n. Thus, X is countably compact.

Total Boundedness

Definition: A set N⊆X is an ε-net of a metric space X where ε>0 if for any b within X, there is an element x∈N such that d(b,x)<ε.

Definition: A metric space X is totally bounded when it has a finite ε-net for any ε>0.

Theorem

A countably compact metric space is totally bounded.

Proof

Any infinite subset of a countably compact metric space X must have at least one limit point. Thus, any infinite subset of X cannot have all elements more than ε apart. Thus, selecting x₁, x₂, x₃, ... where x_n is at least ε apart from any x_d where d<n, one must eventually have formed an ε-net because this process must be finite.

Theorem

A totally bounded set is separable.

Proof

Take the union of all finite 1/n-nets, where n varies over the natural numbers, and that is a countable set such that its closure is the whole space X.

Corollary

A countably compact metric space is separable, satisfies the second axiom of countability, and is thus compact.

Hahn-Mazurkiewicz Theorem

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