# 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 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.
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