Topology/Countability

Bijection
A set is said to be countable if 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 $$f:\mathbf{Z}\rightarrow\mathbf{Z}$$, where $$f(n)=2n$$. Hence the even integers are countable.

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

Proof: let $$f:\mathbf{Z}\rightarrow\mathbf{Z}$$ 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.

Definition
A topological space $$X$$is 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
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
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
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 $$\{ x_n\}\subset A$$ converge 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 $$f:X\to Y$$ is continuous if and only if whenever $$\{ x_n\}$$ converges to $$x$$, $$\{ f(x_n)\}$$ converges to $$f(x)$$.

Proof
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.

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

Proof
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 $$\mathcal{G}$$.

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

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

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

Proof
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)
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.

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

Proof
($$\Rightarrow$$)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
Since there is relative compactness, there is an analogous property called relative countable compactness.

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

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

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, 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
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 $$X$$.

Urysohn's Metrizability Theorem
The following theorem establishes a sufficient condition for a topological space to be metrizable.

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

Proof
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
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
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
If a Hausdorff space is a continuous image of the unit interval, then it is compact, connected, locally-connected and second countable.
 * 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: