# Topology/Countability

 Topology ← Linear Continuum Countability Cantor Space →

## 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 ${\displaystyle f:\mathbf {Z} \rightarrow \mathbf {Z} }$ , where ${\displaystyle f(n)=2n}$ . Hence the even integers are countable.

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

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

• not represented by some ${\displaystyle f(m)}$  for ${\displaystyle m
• ${\displaystyle (x,y)}$  is the lattice point 1 unit from ${\displaystyle f(n-1)}$  nearest to the origin. In the case where there are two such points, an arbitrary choice may be made.

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

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 ${\displaystyle N}$  of a point ${\displaystyle x}$ , there is an open ball ${\displaystyle B_{r}(x)}$  within ${\displaystyle N}$ , and the countable collection of neighborhoods of ${\displaystyle x}$  that are ${\displaystyle B_{1/k}(x)}$  where ${\displaystyle k\in \mathbb {N} }$ , has the neighborhood ${\displaystyle B_{1/n}(x)}$  where ${\displaystyle {\tfrac {1}{n}} .

#### Theorem

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

##### Proof

Let ${\displaystyle \{A_{i}\}}$  be a countable collection of neighborhoods of ${\displaystyle x}$  such that for any neighborhood ${\displaystyle N}$  of ${\displaystyle x}$ , there is an ${\displaystyle A_{i}}$  such that ${\displaystyle A_{i}\subset N}$ . Define
${\displaystyle B_{n}=\bigcap _{i=1}^{n}A_{n}}$ .

Then form a sequence ${\displaystyle \{a_{i}\}}$  such that ${\displaystyle a_{i}\in B_{i}}$ . Then obviously ${\displaystyle \{a_{i}\}}$  converges to ${\displaystyle x}$ .

#### Theorem

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

##### Proof

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

#### Theorem

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

##### Proof

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

### Second Axiom of Countability

#### 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 ${\displaystyle N}$  of that point must contain at least one neighborhood ${\displaystyle A}$  within the collection, and ${\displaystyle A}$  must be a subset of ${\displaystyle N}$ .

#### Theorem

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

##### Proof

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

### Separable Spaces

#### Definition

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

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

#### Definition

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

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

## Countable Compactness

#### Definition

A subset ${\displaystyle A}$  of a topological space ${\displaystyle X}$  is said to be Countably Compact if and only if all countable covers of ${\displaystyle A}$  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 ${\displaystyle X}$  is countably compact if and only if any infinite subset of that space has at least one limit point.

##### Proof

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

(${\displaystyle \Leftarrow }$ )Let ${\displaystyle \{S_{n}\}}$  be open subsets of ${\displaystyle X}$  such that any finite union of those sets does not cover ${\displaystyle X}$ . Define:

${\displaystyle B_{n}=\bigcup _{i=1}^{n}S_{n}}$ ,

which does not cover ${\displaystyle X}$ , and is open. Select ${\displaystyle x_{n}}$  such that ${\displaystyle x_{n}\notin B_{n}}$ . There is a limit point ${\displaystyle x}$  of this set of points, which must also be a limit point of ${\displaystyle X\setminus B_{n}}$ . Since ${\displaystyle X\setminus B_{n}}$  is closed, ${\displaystyle x\in X\setminus B_{n}}$ . Thus, ${\displaystyle x\notin B_{n}}$  and thus is not within any ${\displaystyle S_{n}}$ , so ${\displaystyle S_{n}}$  is not an open cover of X. Thus, ${\displaystyle X}$  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.

## Total Boundedness

#### Definition

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

#### Definition

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

#### Theorem

A countably compact metric space is totally bounded.

##### Proof

Any infinite subset of a countably compact metric space ${\displaystyle X}$  must have at least one limit point. Thus, selecting ${\displaystyle x_{1},x_{2},x_{3},\ldots }$  where ${\displaystyle x_{n}}$  is at least ${\displaystyle \varepsilon }$  apart from any ${\displaystyle x_{d}}$  where ${\displaystyle d , one must eventually have formed an ${\displaystyle \varepsilon }$ -net because this process must be finite, because there is no possible infinite set with all elements more than ${\displaystyle \varepsilon }$  apart.

#### Theorem

A totally bounded set is separable.

##### Proof

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

## 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 ${\displaystyle O_{n}}$  set of it. Select a point ${\displaystyle x_{n}}$  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 ${\displaystyle f_{n}:X\rightarrow [0,1]}$  such that:

${\displaystyle f_{n}(x_{n})=0}$
${\displaystyle f_{n}(X/O_{n})=1}$

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 ${\displaystyle f_{n}(O_{n})<1}$ , meaning that the function value of any point within the open set is less than 1.

Now define the function ${\displaystyle g:X\rightarrow H}$  from X to the Hilbert cube to be ${\displaystyle g(x)=(f_{1}(x),{\frac {f_{2}(x)}{2}},{\frac {f_{3}(x)}{4}},...)}$ .

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

${\displaystyle \sum _{i=N}^{\infty }({\frac {1}{2^{i}}})^{2}<{\frac {\epsilon ^{2}}{2}}}$ .

Moreover, since ${\displaystyle f_{n}}$  is a continuous function from X to [0,1], there exists a neighborhood of ${\displaystyle a}$ , and therefore an open set ${\displaystyle S_{n}}$  of the base within that neighborhood containing a such that if ${\displaystyle y\in S_{n}}$ , then

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

or

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

Let

${\displaystyle S=\bigcap _{i=1}^{N-1}S_{i}}$ .

In addition, since ${\displaystyle a_{n}\rightarrow a}$ , there exists an ${\displaystyle M_{i}}$  (i=1,2,3,...,M-1) such that when ${\displaystyle n>M_{i}}$ , that ${\displaystyle a_{n}\in S_{i}}$ , and let M be the maximum of ${\displaystyle M_{i}}$  so that when n>M, then ${\displaystyle a_{n}\in S}$ .

Let n>M, and then the distance from ${\displaystyle g(a_{n})}$  to g(a) is now

${\displaystyle \sum _{i=1}^{\infty }({\frac {f_{i}(a_{n})-f_{i}(a)}{2^{i}}})^{2}=}$  ${\displaystyle \sum _{i=1}^{N-1}({\frac {f_{i}(a_{n})-f_{i}(a)}{2^{i}}})^{2}+}$  ${\displaystyle \sum _{i=N}^{\infty }({\frac {f_{i}(a_{n})-f_{i}(a)}{2^{i}}})^{2}\leq }$  ${\displaystyle {\frac {N-1}{2N}}\epsilon ^{2}+\sum _{i=N}^{\infty }({\frac {f_{n}(y)-f_{n}(z)}{2^{n}}})^{2}\leq }$  ${\displaystyle {\frac {N-1}{2N}}\epsilon ^{2}+{\frac {\epsilon ^{2}}{2}}<\epsilon ^{2}.}$

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 ${\displaystyle a\in U_{a}}$  and ${\displaystyle b\in U_{b}}$ , and select an element of the base ${\displaystyle O_{n}}$  that contains a and is within ${\displaystyle U_{a}}$ . It follows that ${\displaystyle f_{n}(a)<1}$  whereas ${\displaystyle f_{n}(b)=1}$ , proving that the function g is one-to-one, and that there exists an inverse ${\displaystyle g^{-1}}$ .

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

${\displaystyle |f_{n}(z)-f_{n}(x)|<2^{n}\epsilon _{n}}$

then ${\displaystyle z\in O_{n}}$ .

Suppose that ${\displaystyle g(z)\in g(X)\cap B_{\epsilon _{n}}(g(y))}$ . Then

${\displaystyle ({\frac {f_{n}(z)-f_{n}(x)}{2^{n}}})^{2}\leq \sum _{i=1}^{\infty }({\frac {f_{i}(z)-f_{i}(x)}{2^{i}}})^{2}\leq \epsilon _{n}^{2}}$

Implying that ${\displaystyle |f_{n}(z)-f_{n}(x)|\leq 2^{n}\epsilon ^{n}}$  indicating that ${\displaystyle z\in O_{n}}$ .

Now consider any open set O around x. Then there exists a set of the base ${\displaystyle x\in O_{n}\subseteq O}$  and an ${\displaystyle \epsilon _{n}>0}$  such that whenever ${\displaystyle g(z)\in g(X)\cap B_{\epsilon _{n}}(g(y))}$ , then ${\displaystyle z\in O_{n}}$ , meaning that ${\displaystyle z\in O}$ . 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 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

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

## Exercise

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.
 Topology ← Linear Continuum Countability Cantor Space →