Topology/Countability

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Bijection[edit]

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

Examples[edit]

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

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

Proof: let f:\bold{Z}\rightarrow\bold{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[edit]

First Axiom of Countability[edit]

Definition[edit]

A topological space Xis said to satisfy the First Axiom of Countability if, for every x\in X there exists a countable collection \mathcal{U}of neighbourhoods of x, such that if N is any neighbourhood of x, there exists U\in\mathcal{U} with U\subseteq N.

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 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\in\mathbb{N}, has the neighborhood B_{1/n}(x) where \tfrac{1}{n}<r.

Theorem[edit]

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

Proof[edit]

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\subset N. Define
B_n=\bigcap_{i=1}^{n} A_n.

Then form a sequence \{ a_i\} such that a_i\in B_i. Then obviously \{ a_i\} converges to x.

Theorem[edit]

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

Proof[edit]

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[edit]

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

Proof[edit]

Let X satisfy the first axiom of countability, and let f:X\to Y be continuous. Let \{ x_n\} be a sequence which converges to x. Let B be any open neighborhood of f(x). As f is continuous, there exists an open neighbourhood A\subset f^{-1}(B) of x. Since \{ x_n\} to x, then there must exist an N\in\mathbb{N} such that A must contain x_n when n>N. Thus, f(A) 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 of Y. Let x_n\in 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[edit]

Definition[edit]

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

Theorem[edit]

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

Proof[edit]

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

Separable Spaces[edit]

Definition[edit]

A topological space X is separable if it has a countable proper subset A such that \mathrm{Cl}(A)=X.

Example: \mathbb{R}^n is separable because \mathbb{Q}^n is a countable subset and \mathrm{Cl}(\mathbb{Q}^n)=\mathbb{R}^n.

Definition[edit]

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

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

Theorem[edit]

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

Proof[edit]

Consider a countable base of a space X. 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 X since any neighborhood of any element of X 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.

Corollary[edit]

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

Theorem[edit]

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

Proof[edit]

Let X be a metric space, and let A be a countable set such that \mathrm{Cl}(A)=X. Consider the countable set B of open balls \{ B_{1/k}(p)|k\in N, p\in 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. Let r' be a number of the form 1/n that is less than r. Because \mathrm{Cl}(A)=X, there is an element x'\in A such that d(x',x)<\tfrac{r'}{4}. Then the ball B_{r'/2}(x') is within B and is a subset of O because if y\in B_{r'/2}(x'), then d(y,x)\leq d(y,x')+d(x',x)<\tfrac{3}{4}r'<r. Thus B_{r'/2}\subseteq O that contains x. The union of all such neighborhoods containing an element of O is O. Thus B is a base for X.

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

Countable Compactness[edit]

Definition[edit]

A subset A of a topological space X is said to be Countably Compact if and only if all countable covers of A 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[edit]

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

Proof[edit]

(\Rightarrow)Let \{ x_i\}, (i=1,2,3,...) be a set within X without any limit point. Then this sequence is closed, since they are all isolated points within the sequence. Let S_n=\{ x_i\} for (i=n, n+1, n+2, ...). The X\setminus S_n are all open sets, and so is a countable cover of the set, but any finite subcover \{ X\setminus S_{n_i}\} of this cover does not cover X because it does not contain S_{n_{max\{ i\} }}. This contradicts the assumption that X is countably compact.

(\Leftarrow)Let \{ S_n\} be open 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\notin B_n. There is a limit point x of this set of points, which must also be a limit point of X\setminus B_n. Since X\setminus B_n is closed, x\in X \setminus B_n. Thus, x\notin B_n and thus is not within any S_n, so S_n is not an open cover of X. Thus, X is countably compact.

Relative Countable Compactness[edit]

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

Definition[edit]

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

Total Boundedness[edit]

Definition[edit]

A set N\subseteq X is an \varepsilon-net of a metric space X where \varepsilon>0 if for any b within X, there is an element x\in N such that d(b,x)<\varepsilon.

Definition[edit]

A metric space X is totally bounded when it has a finite \varepsilon-net for any \varepsilon>0.

Theorem[edit]

A countably compact metric space is totally bounded.

Proof[edit]

Any infinite subset of a countably compact metric space X must have at least one limit point. Thus, selecting x_1,x_2,x_3,\ldots where x_n is at least \varepsilon apart from any x_d where d<n, one must eventually have formed an \varepsilon-net because this process must be finite, because there is no possible infinite set with all elements more than \varepsilon apart.

Theorem[edit]

A totally bounded set is separable.

Proof[edit]

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.

Urysohn's Metrizability Theorem[edit]

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

Theorem[edit]

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

Proof[edit]

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 O_n set of it. Select a point x_n 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 f_n: X \rightarrow [0,1] such that:

f_n(x_n)=0
f_n(X/O_n)=1

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 f_n(O_n)<1, meaning that the function value of any point within the open set is less than 1.

Now define the function g: X \rightarrow H from X to the Hilbert cube to be g(x)=(f_1(x),\frac{f_2(x)}{2},\frac{f_3(x)}{4},...).

To prove that this is continuous, let a_n \rightarrow a be a sequence that converges to a. Consider the open ball B_\epsilon(f(a)) where \epsilon>0. There exists an N such that

\sum_{i=N}^\infty(\frac{1}{2^i})^2<\frac{\epsilon^2}{2}.

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

|f_n(y)-f_n(z)|<\frac{2^n\epsilon}{\sqrt{2N}}

or

(\frac{f_n(y)-f_n(z)}{2^n})^2<\frac{\epsilon^2}{2N}.

Let

S=\bigcap_{i=1}^{N-1} S_i.

In addition, since a_n\rightarrow a, there exists an M_i (i=1,2,3,...,M-1) such that when n>M_i, that a_n\in S_i, and let M be the maximum of M_i so that when n>M, then a_n\in S.

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

\sum_{i=1}^\infty (\frac{f_i(a_n)-f_i(a)}{2^i})^2 = \sum_{i=1}^{N-1}(\frac{f_i(a_n)-f_i(a)}{2^i})^2 + \sum_{i=N}^\infty(\frac{f_i(a_n)-f_i(a)}{2^i})^2 \le \frac{N-1}{2N}\epsilon^2+\sum_{i=N}^\infty(\frac{f_n(y)-f_n(z)}{2^n})^2 \le \frac{N-1}{2N}\epsilon^2+\frac{\epsilon^2}{2} < \epsilon^2.

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 a\in U_a and b\in U_b, and select an element of the base O_n that contains a and is within U_a. It follows that f_n(a)<1 whereas f_n(b)=1, proving that the function g is one-to-one, and that there exists an inverse g^{-1}.

To prove that the inverse g^{-1} is continuous, let O_n be an open set within the countable base of X. Consider any point x within O_n. Since f_n(x)<1, indicating that there exists an \epsilon_n>0 such that when

|f_n(z)-f_n(x)|<2^n\epsilon_n

then z\in O_n.

Suppose that g(z)\in g(X)\cap B_{\epsilon_n}(g(y)). Then

(\frac{f_n(z)-f_n(x)}{2^n})^2\le\sum_{i=1}^\infty(\frac{f_i(z)-f_i(x)}{2^i})^2\le \epsilon_n^2

Implying that |f_n(z)-f_n(x)|\le 2^n\epsilon^n indicating that z\in O_n.

Now consider any open set O around x. Then there exists a set of the base x\in O_n\subseteq O and an \epsilon_n>0 such that whenever g(z)\in g(X)\cap B_{\epsilon_n}(g(y)), then z\in O_n, meaning that z\in O. 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 Theorem[edit]

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.

Theorem[edit]

A Hausdorff space is a continuous image of the unit interval [0,1] if and only if it is a compact, connected, locally connected and second-countable space.


Exercise[edit]

  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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