NoteNextra-origin/content/Math4201/Math4201_L12.md

Math4201 Topology I (Lecture 12)

Metric spaces

Basic properties and definitions

Definition of metric space

A metric space is a set X with a function d:X\times X\to \mathbb{R} that satisfies the following properties:

1. \forall x,y\in X, d(x,y)\geq 0 and d(x,y)=0 if and only if x=y. (positivity)
2. \forall x,y\in X, d(x,y)=d(y,x). (symmetry)
3. \forall x,y,z\in X, d(x,z)\leq d(x,y)+d(y,z). (triangle inequality)

Example of metric space

Let X=\mathbb{R} and d(x,y)=|x-y|.

Check definition of metric space:

1. Positivity: d(x,y)=|x-y|\geq 0 and d(x,y)=0 if and only if x=y.
2. Symmetry: d(x,y)=|x-y|=|y-x|=d(y,x).
3. Triangle inequality: d(x,z)=|x-z|\leq |x-y|+|y-z|=d(x,y)+d(y,z) since |a+b|\leq |a|+|b| for all a,b\in \mathbb{R}.

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Let X be arbitrary. The trivial metric is $d(x,y)=\begin{cases} 0 & \text{if } x=y \ 1 & \text{if } x\neq y \end{cases}$

Check definition of metric space:

1. Positivity: $d(x,y)=\begin{cases} 0 & \text{if } x=y \ 1 & \text{if } x\neq y \end{cases}\geq 0$ and d(x,y)=0 if and only if x=y.
2. Symmetry: $d(x,y)=\begin{cases} 0 & \text{if } x=y \ 1 & \text{if } x\neq y \end{cases}=d(y,x)$.
3. Triangle inequality use case by case analysis.

Balls of a metric space forms a basis for a topology

Let (X,d) be a metric space. x\in X and r>0, r\in \mathbb{R}. We define the ball of radius r centered at x as B_r(x)=\{y\in X:d(x,y)<r\}.

Goal: Show that the balls of a metric space forms a basis for a topology.

\{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}\text{ is a basis for a topology on }X

Example of balls of a metric space

Let X=\mathbb{R} and $d(x,y)=\begin{cases} 0 & \text{if } x=y \ 1 & \text{if } x\neq y \end{cases}$

The balls of this metric space are:

B_r(x)=\begin{cases}
\{x\} & \text{if } r<1 \\
X & \text{if } r\geq 1
\end{cases}

Note

This basis generate the discrete topology of X.

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Let X=\mathbb{R} and d(x,y)=|x-y|.

The balls of this metric space are:

B_r(x)=\{(x-r,x+r)\}

This basis is the set of all open sets in \mathbb{R}, which generates the standard topology of \mathbb{R}.

Proof

Let's check the two properties of basis:

1. \forall x\in X, \exists B_r(x)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\} such that x\in B_r(x). (Trivial by definition of non-zero radius ball)
2. \forall B_r(x),B_r(y)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\}, \forall z\in B_r(x)\cap B_r(y), \exists B_r(z)\in \{B_r(x)|x\in X,r>0,r\in \mathbb{R}\} such that z\in B_r(z)\subseteq B_r(x)\cap B_r(y).

Observe that for any z\in B_r(x), then there exists \delta>0 such that B_\delta(z)\subseteq B_r(x).

Let \delta=r-d(x,z), then B_\delta(z)\subseteq B_r(x) (by triangle inequality)

Similarly, there exists \delta'>0 such that B_\delta'(z)\subseteq B_r(y).

Take \lambda=min\{\delta,\delta'\}, then B_\lambda(z)\subseteq B_r(x)\cap B_r(y).

Definition of Metric topology

For any metric space (X,d), the topology generated by the balls of the metric space is called metric topology.

Definition of metrizable

A topological space (X,\mathcal{T}) is metrizable if it is the metric topology for some metric d on X.

Q: When is a topological space metrizable?

Lemma: Every metric topology is Hausdorff

If a topology isn't Hausdorff, then it isn't metrizable.

Example of non-metrizable space

Trivial topology with at least two points is not Hausdorff, so it isn't metrizable.

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Finite complement topology on infinite set is not Hausdorff.

Suppose there exists x,y\in X such that x\neq y and x\in U\subseteq X and y\in V\subseteq X such that X-U and X-V are finite.

Since U\cap V=\emptyset, we have V\subseteq X-U, which is finite. So X-V is infinite. (contradiction that X-V is finite)

So X with finite complement topology is not Hausdorff, so it isn't metrizable.

Proof

Let x,y\in (X,d) and x\neq y. To show that X is Hausdorff, it is suffices to show that there exists r,r'>0 such that B_r(x)\cap B_r'(y)=\emptyset.

Take r=r'=\frac{1}{2}d(x,y), then B_r(x)\cap B_r'(y)=\emptyset. (by triangle inequality)

We prove this by contradiction.

Suppose \exists z\in B_r(x)\cap B_r'(y), then d(x,z)<r and d(y,z)<r'.

Then d(x,y)\leq d(x,z)+d(z,y)<r+r'=\frac{1}{2}d(x,y)+\frac{1}{2}d(x,y)=d(x,y). (contradiction d(x,y)<d(x,y))

Therefore, X is Hausdorff.

Other metrics on \mathbb{R}^n

Let \mathbb{R}^n be the set of all $n$-tuples of real numbers with standard topology.

Let d: \mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R} be defined by (the Euclidean distance)

d(u,v)=\sqrt{\sum_{i=1}^n (u_i-v_i)^2}

In \mathbb{R}^2 the ball is a circle.

Let \rho(u,v)=\max_{i=1}^n |u_i-v_i|. (Square metric)

In \mathbb{R}^2 the ball is a square.

Let m(u,v)=\sum_{i=1}^n |u_i-v_i|. (Manhattan metric)

In \mathbb{R}^2 the ball is a diamond.

Lemma: Square metric, Manhattan metric, and Euclidean metric are well defined metrics on \mathbb{R}^n

Proof ignored. Hard part is to show the triangle inequality. May use Cauchy-Schwarz inequality.