# Topology/Countability

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## Bijection

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

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

Proof: let $f:\bold{Z}\rightarrow\bold{Z}$ 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

$\sum_{i=1}^\infty (\frac{f_i(a_n)-f_i(a)}{2^i})^2 =$ $\sum_{i=1}^{N-1}(\frac{f_i(a_n)-f_i(a)}{2^i})^2 +$ $\sum_{i=N}^\infty(\frac{f_i(a_n)-f_i(a)}{2^i})^2 \le$ $\frac{N-1}{2N}\epsilon^2+\sum_{i=N}^\infty(\frac{f_n(y)-f_n(z)}{2^n})^2 \le$ $\frac{N-1}{2N}\epsilon^2+\frac{\epsilon^2}{2} < \epsilon^2.$

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 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 $[0,1]$ 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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