Metrics

Definition of metric

1. Metric

   Let $X$ be a set. A metric on $X$ is a function $d\colon X\times X\to \mathbb{R}$ such that:

   • $d(x,y)\geq 0$ for $x,y\in X$, $d(x,y)=0$ if and only if $x=y$.
   • $d(x,y)=d(y,x)$ for all $x,y\in X$.
   • $d(x,y)+d(y,z)\geq d(x,z)$ for all $x,y,z\in X$.

2. Metric space

   If $d$ is a metric on $X$, we say that the pair $(X,d)$ is a metric space.

Metric topology

1. Ball

   If $(X,d)$ is a metric space and $\epsilon>0$, we call [ B_{d}(x,\epsilon)={y\in X\mid d(x,y)<\epsilon}, ] the $\epsilon$-ball centered on $x$.

2. Teorema

   If $(X,d)$ is a metric space, the collection [ \mathcal{B}={B_{d}(x,\epsilon)\mid x\in X,\epsilon>0} ] is a basis for a topology on $X$

Metric topology

1. Metric topology

   The topology generated by the balls, is called the metric topology on $X$. We also say that the metric $d$ induces the topology $\tau_{d}$.

2. Examples
   • Discrete metric
   • Usual metric on $\mathbb{R}^{n}$.

3. Remark

   One can prove that $U\subseteq X$ is open in $\tau_{d}$ if and only if for all $x\in U$ there is $\epsilon>0$ such that $B_{d}(x,\epsilon)\subseteq U$ (that is, the ball can be chosen with $x$ as center)

Metrizable spaces

Metrizable spaces

1. Metrizable space

   We say that the topological space $(X,\tau)$ is metrizable if there is a metric $d$ on $X$ such that $\tau_{d}=\tau$.

Bounded metrics

1. Bounded set

   Let $(X,d)$ be a metric space. We say that $A\subseteq X$ is bounded if there is $M$ such that $d(x,y)\leq M$ for all $x,y\in A$.

2. Bounded metric

   Let $(X,d)$ be a metric space. If we define $\overline{d}$ as: [ \overline{d}(x,y)=\min{d(x,y),1}, ] then $\overline{d}$ is a metric on $X$, and $\tau_{d}=\tau_{\overline{d}}$.

$\mathbb{R}^{n}$ is metrizable

Metrics in $R^{n}$

1. Two metrics in $\mathbb{R}^{n}$

   There are two metrics that we can define on $\mathbb{R}^{n}$:

   • $d(x,y)=\lVert x-y\rVert$,
   • $\rho(x,y)=\max\{\vert x_{1}-y_{1}\vert ,\ldots,\vert x_{n}-y_{n}\vert \}$.

2. Teorema

   The product topology on $\mathbb{R}^{n}$ is the same as $\tau_{d}$ and the same as $\tau_{\rho}$.

Metrics in $\mathbb{R}^{\mathbb{N}}$

Uniform metric

1. Uniform metric

   If $I$ is a set, and $\overline{d}(x,y)=\min\{\vert x-y\vert ,1\}$ for $x,y\in\mathbb{R}$, we can define the metric $\overline{\rho}$ on $\mathbb{R}^{I}=\prod_{i\in I}\mathbb{R}$ by: [ \overline{\rho}(x,y)=\sup{\overline{d}(x_{i},y_{i})\mid i\in I}. ]

   This is called the uniform metric, and the induced topology is the uniform topology.

2. Exercise

   Show that if $I$ is infinite, the uniform topology is different from both the product and the box topology on $\mathbb{R}^{I}$

Metric in $\mathbb{R}^{\mathbb{N}}$

1. The function $D$

   For $x,y\in \mathbb{R}^{\mathbb{N}}$, we define: [ D(x,y)=\sup{\frac{\overline{d}(x_{i},y_{i})}{i}\mid i\in\mathbb{N}} ]

2. $D$ is metric

   $D$ defines a metric on $\mathbb{R}^{\mathbb{N}}$.

Theorem: $\mathbb{R}^{\mathbb{N}}$ is metrizable

1. Teorema

   $\tau_{D}$ is the same as the product topology on $\mathbb{R}^{\mathbb{N}}$.

2. End of statement

   • We know that $D$ is a metric. Let $\tau$ the product topology on $X=\mathbb{R}^{\mathbb{N}}$. We prove that $\tau_{D}\subseteq\tau$.
   • Let $U\in\tau_D$, $x\in U$, we will show that there is $V\in\tau$ with $x\in V\subseteq U$.
   • Let $\epsilon>0$ be such that $B_{D}(x,\epsilon)\subseteq U$. Let $N$ such that $\frac{1}{N}<\epsilon$.
   • Let [ V=(x_1-\epsilon,x_1+\epsilon)\times\cdots\times (x_N-\epsilon,x_N+\epsilon) \times\mathbb{R}\times\mathbb{R}\times\cdots ]
   • We have $x\in V$ and $V\in\tau$, we will show $V\subseteq U$.

Proof (continuation)

• To prove that $V\subseteq U$, it is enough to show that $V\subseteq B_{D}(x,\epsilon)$.
• First, note that if $i\geq N, y\in\mathbb{R}^{\mathbb{N}}$, then $\overline{d}(x_{i},y_{i})\leq 1$ and $\frac{1}{i}\leq \frac{1}{N}$, hence $\frac{\overline{d}(x_{i},y_{i})}{i}\leq \frac{1}{N}$.
• Now, let $y\in V$. We have that $\vert x_i-y_i\vert <\epsilon$ for $i=1,2,\ldots,N$, and so for such values of $i$ we have: [ \frac{\overline{d}(x_{i},y_{i})}{i}\leq \overline{d}(x_{i},y_{i}) \leq d(x_{i},y_i)=\vert x_i-y_i\vert <\epsilon, ]

• Finally:

  $$\begin{align*} D(x,y)&=\sup\{\frac{\overline{d}(x_{i},y_{i})}{i}\mid i\in\mathbb{N} \}\\ &\leq\max\{\frac{\overline{d}(x_{1},y_{1})}{1},\ldots \frac{\overline{d}(x_{N},y_{N})}{N},\frac{1}{N}\}<\epsilon \end{align*}$$
• Hence: $y\in B_{D}(x,\epsilon)\subseteq U$.

Proof (continuation)

• We will show now that $\tau\subseteq\tau_{D}$.
• Let $U$ an element of the base of $\tau$ and $x\in U$. We will see that there is $V\in\tau_D$ such that $x\in V\subseteq U$.
• Let $U=\prod U_i$, with $U_i$ open in $\mathbb{R}$. Suppose that $U_{i}=\mathbb{R}$ if $i\not\in\{\alpha_{1},\ldots,\alpha_{n} \}$. Let $\epsilon_{i}$ such that $(x_{i}-\epsilon_i,x_{i}+\epsilon_i)\subseteq U_i$ for $i\in \{\alpha_{1},\ldots,\alpha_{n} \}$. Without loss we can assume $\epsilon_i\leq 1$.
• Let $\epsilon=\min\{\frac{\epsilon_i}{i}\mid i=\alpha_{1},\ldots,\alpha_{n}\}$. We claim that $B_{D}(x,\epsilon)\subseteq U$. Let $y\in B_{D}(x,\epsilon)$.
• For $i\in\{\alpha_{1},\ldots,\alpha_{n}\}$ we have that: [ \frac{\overline{d}(x_{i},y_i)}{i}\leq D(x,y)<\epsilon\leq \frac{\epsilon_i}{i} ] and so: $\overline{d}(x_{i},y_{i})<\epsilon_{i}\leq 1$.
• It follows that if $i\in\{\alpha_{1},\ldots,\alpha_{n}\}$, then $d(x_{i},y_{i})=\overline{d}(x_{i},y_i)<\epsilon_{i}$. Hence $y\in\prod U_{i}$. \qed

Exercises

1. Show that $\mathbb{R}\times \mathbb{R}$ with the dictionary order is metrizable.
2. Let $(X,d)$ be a metric space.
   • Show that $d\colon X\times X\to \mathbb{R}$ is continuous, where $X\times X$ has the product topology of metric topologies.
   • Show that if $X$ has a topology $\tau$ such that $d\colon X\times X\to \mathbb{R}$ is continuous, then $\tau_{d}\subseteq \tau$.
3. Let $A\subseteq \mathbb{R}^{\mathbb{N}}$ to consist on all sequences that are eventually zero. What is $\overline{A}$? (where $\mathbb{R}^{\mathbb{N}}$ has the uniform topology).