# 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.

## Axioms of countability
### First Axiom of Countability
#### 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.

### 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 $N$ 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 ${\mathcal {G}}$.

### Separable Spaces
#### 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.

## Countable Compactness
#### 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.

## Total Boundedness
#### 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
$\sum _{i=1}^{\infty }({\frac {f_{i}(a_{n})-f_{i}(a)}{2^{i}}})^{2}=$ $\sum _{i=1}^{N-1}({\frac {f_{i}(a_{n})-f_{i}(a)}{2^{i}}})^{2}+$ $\sum _{i=N}^{\infty }({\frac {f_{i}(a_{n})-f_{i}(a)}{2^{i}}})^{2}\leq$ ${\frac {N-1}{2N}}\epsilon ^{2}+\sum _{i=N}^{\infty }({\frac {f_{n}(y)-f_{n}(z)}{2^{n}}})^{2}\leq$ ${\frac {N-1}{2N}}\epsilon ^{2}+{\frac {\epsilon ^{2}}{2}}<\epsilon ^{2}.$
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
$({\frac {f_{n}(z)-f_{n}(x)}{2^{n}}})^{2}\leq \sum _{i=1}^{\infty }({\frac {f_{i}(z)-f_{i}(x)}{2^{i}}})^{2}\leq \epsilon _{n}^{2}$
Implying that $|f_{n}(z)-f_{n}(x)|\leq 2^{n}\epsilon ^{n}$ 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

* Prove that a separable metric space satisfies the second axiom of countability. Hence, or otherwise, prove that a countably compact metric space is compact.
* 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.