# 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 $f:\mathbf {Z} \rightarrow \mathbf {Z}$ , 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
• $(x,y)$  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}} .

#### 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 $\{x_{n}\}\subset A$  which converge to an element of $X$  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' . 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 , 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

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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