Metric space

 

Def 1

Let $X$ be a set and $d : X \times X \to \mathbb{R}$ be a function that satisfies the following $3$ conditions.

 

$\forall x, y, z \in X$

$(\mathrm{i})$  $d(x, y) \geq 0$     &     $d(x, y) = 0$  $\Leftrightarrow$  $x = y$

$(\mathrm{ii})$  $d(x, y) = d(y, x)$ 

$(\mathrm{iii})$  $d(x, z) \leq d(x, y) + d(y, z)$

 

$d$ : metric (or distance) on $X$.

$(X, d)$ : metric space.  

 

 

 

Example

1.  Euclidean metric : $d_{E} : \mathbb{R}^{n} \times \mathbb{R}^{n} \to \mathbb{R}$  by

$d_{E}(\mathbf{x}, \mathbf{y}) = \sqrt{(x_{1} - y_{1})^{2} + \cdots + (x_{n} - y_{n})^{2}}$

 

2.  square metric : $d_{M} : \mathbb{R}^{n} \times \mathbb{R}^{n} \to \mathbb{R}$  by

$d_{M}(\mathbf{x}, \mathbf{y}) = \displaystyle \max_{1 \leq i \leq n} \left|x_{i} - y_{i} \right|$

 

3.  taxi cap metric : $d_{S} : \mathbb{R}^{n} \times \mathbb{R}^{n} \to \mathbb{R}$  by

$d_{S}(\mathbf{x}, \mathbf{y}) =  \left|x_{1} - y_{1} \right| + \cdots + \left|x_{n} - y_{n} \right|$

 

4.  discrete metric : $d : X \times X \to \mathbb{R}$  by

$d(x, y) = \begin{cases}
0 & x = y \\
1 & x \neq y
\end{cases} $

 

 

 

Def 2

Let $X$ be a set  and  $d$ be a metric on $X$.

Let $x_{0} \in X$,  $r>0$. We define

$B_{r}(x_{0}) = \left\{x \in X \; | \; d(x_{0}, x) < r \right\}$  :  open ball with radius $r$

$\overline{B_{r}(x_{0})} = \left\{x \in X \; | \; d(x_{0}, x) \leq r \right\} $  :  closed ball with radius $r$

$S_{r}(x_{0}) = \left\{x \in X \; | \; d(x_{0}, x) = r \right\}$  :  sphere ball with radius $r$

 

 

 

Thm 1

Let $X$ be a set  and  $d$ be a metric on $X$. Then

$\mathscr{B} = \left\{B_{r}(x_{0}) \; | \; x_{0} \in X, \; r>0 \right\}$

is base of some topology on $X$.

 

  Show 1 : $\displaystyle \bigcup_{x_{0} \in X} B_{r}(x_{0}) = X$

  Since $x_{0} \in B_{r}(x_{0}) \subseteq X$,

  $X = \displaystyle \bigcup_{x_{0} \in X} \left\{x_{0} \right\} \subseteq \bigcup_{x_{0} \in X} B_{r}(x_{0}) \subseteq X$

 

  end

 

 

  Show 2 : $\forall x \in B_{r_{1}}(x_{1}) \cap B_{r_{2}}(x_{2})$,  $\exists B_{x} \in \mathscr{B}$  s.t.  $x \in B_{x} \subseteq B_{r_{1}}(x_{1}) \cap B_{r_{2}}(x_{2})$

  $\forall x \in B_{r_{1}}(x_{1}) \cap B_{r_{2}}(x_{2})$

 

  Let $r = \min \left\{r_{1} - d(x_{1}, x), r_{2} - d(x_{2}, x) \right\}$

  Since $d(x_{1}, x) < r_{1}$  and  $d(x_{2}, x) < r_{2}$,

  $r>0$

 

  Let $B_{x} = B_{r}(x) \in \mathscr{B}$

  $\forall y \in B_{x}$  ( i.e. $d(x, y) < r$ )

 

  Therefore

  $d(x_{1}, y) \leq d(x_{1}, x) + d(x, y)$

                   $< d(x_{1}, x) + r$

                   $\leq d(x_{1}, x) + r_{1} - d(x_{1}, x)$

                   $= r_{1}$

 

  $\therefore$  $y \in B_{r_{1}}(x_{1})$

  Similarly we can get $y \in B_{r_{2}}(x_{2})$

 

  $\therefore$  $y \in B_{r}(x_{1}) \cap B_{r_{2}}(x_{2})$

  $\therefore$  $B_{x} \subseteq B_{r_{1}}(x_{1}) \cap B_{r_{2}}(x_{2})$

 

  Since $x \in B_{x}$,

  $x \in B_{x} \subseteq B_{r_{1}}(x_{1}) \cap B_{r_{2}}(x_{2})$

 

  end

 

 

 

Def 3

Let $X$ be a set  and  $d$ be a metric on $X$.

$(X, d)$ is considered a topological space with the topology induced by the base

$\mathscr{B} = \left\{B_{r}(x_{0}) \; | \; x_{0} \in X, \; r>0 \right\}$

, and it is called a metric space.

 

 

 

Thm 2

Let $(X, \mathscr{T})$ be topological space induced by a metric $d$.

T.F.A.E.

$(1)$  $U$ is open set in $X$

$(2)$  $\forall x \in U$, $\exists r > 0$  s.t.  $B_{r}(x) \subseteq U$

 

pf)

$(1) \Rightarrow (2)$

  Let $\mathscr{B} = \left\{B_{r}(x_{0}) \; | \; x_{0} \in X, \; r>0 \right\}$

  So $\mathscr{B}$ is base for $\mathscr{T}$

 

  $\forall x \in U$

 

  Since $U$ is open set in $X$, there exists $B_{r_{1}}(x_{1}) \in \mathscr{B}$  s.t.

  $x \in B_{r_{1}}(x_{1}) \subseteq U$

  # Thm 1

  $\therefore$  $d(x_{1}, x) < r_{1}$

 

  Let $r = r_{1} - d(x_{1}, x)$

 

 

  Claim : $B_{r}(x) \subseteq B_{r_{1}}(x_{1})$

  $\forall t \in B_{r}(x)$

  So $d(x, t) < r$

 

  Therefore

  $d(x_{1}, t) \leq d(x_{1}, x) + d(x, t)$

                   $< (r_{1} - r)+ r$

                   $= r_{1}$

 

  $\therefore$  $t \in B_{r_{1}}(x_{1})$

  $\therefore$  $B_{r}(x) \subseteq B_{r_{1}}(x_{1})$

 

  end

 

  $\therefore$  $B_{r}(x) \subseteq U$

 

$(2) \Rightarrow (1)$

  $\forall x \in U$, there exists $B_{r_{x}}(x)$  s.t.

  $B_{r_{x}}(x) \subseteq U$

 

  Therefore

  $U = \displaystyle \bigcup_{x \in U} \left\{x \right\} \subseteq \bigcup_{x \in U} B_{r_{x}}(x) \subseteq U$

 

  Since $B_{r}(x)$ is open set in $X$,

  $U$ is open set in $X$

 

 

 

Def 4

A topological space $X$ is metrizable space if the topology if $X$ is generated by a metric.

 

 

 

Prop 1

Let $X$ be a set. Then

$(X, \text{discrete topology})$ is metrizable space by discrete metric

 

  Define $d : X \times X \to \mathbb{R}$  by

  $d(x, y) = \begin{cases}
0 & x = y \\
1 & x \neq y
\end{cases} $

 

  Base generated by discrete metric is

  $\mathscr{B} = \left\{B_{r}(x_{0}) \; | \; x_{0} \in X, \; r>0 \right\}  = \left\{\left\{x_{0} \right\} \; | \; x_{0} \in X \right\}$

 

  By Prop 1,

  $\mathscr{B}$ generate discrete topology

 

  $\therefore$  $(X, \text{discrete topology})$ is metrizable space by discrete metric

 

 

 

Lemma 1

Let $d_{M}$, $d_{E}$, and $d_{S}$ be

square metric, Euclidean metric, and taxi cap metric on $\mathbb{R}^{n}$, respectively. Then $\forall \mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}$, 

$d_{M}(\mathbf{x}, \mathbf{y}) \leq d_{E}(\mathbf{x}, \mathbf{y}) \leq d_{S}(\mathbf{x}, \mathbf{y}) \leq nd_{M}(\mathbf{x}, \mathbf{y})$

 

  Let $\mathbf{x} = (x_{1}, x_{2}, \cdots, x_{n})$,  $\mathbf{y} = (y_{1}, y_{2}, \cdots, y_{n})$

 

  $d_{M}(\mathbf{x}, \mathbf{y}) = \displaystyle \max_{1 \leq i \leq n} \left|x_{i} - y_{i} \right| \leq \sqrt{\left|x_{1} - y_{1} \right|^{2} + \cdots + \left|x_{n} - y_{n} \right|^{2}} = d_{E}(\mathbf{x}, \mathbf{y})$

  $d_{E}(\mathbf{x}, \mathbf{y}) = \sqrt{\left|x_{1} - y_{1} \right|^{2} + \cdots + \left|x_{n} - y_{n} \right|^{2}} \leq \left|x_{1} - y_{1} \right| + \cdots + \left|x_{n} - y_{n} \right| = d_{S}(\mathbf{x}, \mathbf{y}) $

  $d_{S}(\mathbf{x}, \mathbf{y}) = \left|x_{1} - y_{1} \right| + \cdots + \left|x_{n} - y_{n} \right| \leq \displaystyle n \max_{1 \leq i \leq n} \left|x_{i} - y_{i} \right| = n d_{M}(\mathbf{x}, \mathbf{y}) $

 

  $\therefore$  $d_{M}(\mathbf{x}, \mathbf{y}) \leq d_{E}(\mathbf{x}, \mathbf{y}) \leq d_{S}(\mathbf{x}, \mathbf{y}) \leq nd_{M}(\mathbf{x}, \mathbf{y})$

 

 

 

Lemma 2

Let $d_{M}$, $d_{E}$, and $d_{S}$ be

square metric, Euclidean metric, and taxi cap metric on $\mathbb{R}^{n}$, respectively. Then $\forall r > 0$, $x \in \mathbb{R}^{n}$, 

$B_{\frac{r}{n}}^{M}(x) \subseteq B_{r}^{S}(x) \subseteq B_{r}^{E}(x) \subseteq B_{r}^{M}(x)$

 

  Show 1 : $B_{\frac{r}{n}}^{M}(x) \subseteq B_{r}^{S}(x) $

  $\forall t \in B_{\frac{r}{n}}^{M}(x)$

  So $d_{M}(x, t) < \dfrac{r}{n}$

 

  By Lemma 1,

  $d_{S}(x, t) \leq n d_{M}(x, t) < r$

 

  $\therefore$  $t \in B_{r}^{S}(x)$

  $\therefore$  $B_{\frac{r}{n}}^{M}(x) \subseteq B_{r}^{S}(x)$

 

  end

 

 

  Show 2 : $B_{r}^{S}(x) \subseteq B_{r}^{E}(x) $

  You can prove it in a similar way as above.

 

  end

 

 

  Show 2 : $B_{r}^{E}(x) \subseteq B_{r}^{M}(x) $

  You can prove it in a similar way as above.

 

  end

 

 

 

Thm 3

Let $\mathscr{T}^{M}$, $\mathscr{T}^{E}$, and $\mathscr{T}^{S}$ be topologies of $\mathbb{R}^{n}$ generated by

square metric, Euclidean metric, and taxi cap metric, respectively. Then

$\mathscr{T}^{M} = \mathscr{T}^{E} = \mathscr{T}^{S}$

 

  Show 1 : $\mathscr{T}^{M} \subseteq \mathscr{T}^{E}$

  $\forall U \in \mathscr{T}^{M}$

 

  For each $x \in U$, there exists $r_{x} > 0$  s.t.

  $B_{r_{x}}^{M}(x) \subseteq U$  

  # Thm 2

 

  By Lemma 2,

  $B_{r_{x}}^{E}(x) \subseteq B_{r_{x}}^{M}(x)$

 

  Therefore

  $U = \displaystyle \bigcup_{x \in U} \left\{x \right\} \subseteq \bigcup_{x \in U} B_{r_{x}}^{E}(x) \subseteq U $

 

  $\therefore$  $U \in \mathscr{T}^{E}$

  $\therefore$  $\mathscr{T}^{M} \subseteq \mathscr{T}^{E} $

 

  end

 

 

  Show 2 : $\mathscr{T}^{E} \subseteq \mathscr{T}^{S}$

  You can prove it in a similar way as above.

 

  end

 

 

  Show 3 : $\mathscr{T}^{S} \subseteq \mathscr{T}^{M}$

  You can prove it in a similar way as above.

 

  end

 

  $\therefore$  $\mathscr{T}^{E} = \mathscr{T}^{M} = \mathscr{T}^{S}$