Metric Space

Definition

Let $X$ be a nonempty set. A metric on $X$ is a function

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

that satisfies the following:

1. (Identity of the Indiscernibles): $\forall p\ne q,d(p,q)>0$ and $d(p,p)=0$
2. (Symmetry): $\forall p,q\in X,d(p,q)=d(q,p)$
3. (Triangle Inequality): $\forall p,q,r\in X,d(p,q)\le d(p,r)+d(r,q)$
   1. The distance between any two points is the shortest distance between them.

We call $(X,d)$ a metric space. Sometimes $X$ is called a space and the elements are called points.

Corollary (Subsets of Metric Spaces)

• Subsets of metric spaces are metric spaces with the same distance function
• every subset of a Euclidean Space is a Metric Space
  • all Euclidean Spaces are Metric Spaces

Segments and Intervals

• A segment $(a,b)$ is a set of real numbers such that $a<x<b$
• An interval $[a,b]$ is the set of real numbers such that $a\le x\le b$

Cells and Boxes

A cell or box in ${\mathbb{R}}^n$ is a Cartesian product of intervals in ${\mathbb{R}}^n$. Open cells come from open intervals and closed cells come from closed intervals.

Convexity

Let $A\subset {\mathbb{R}}^n$ and $A\ne \varnothing$. We say $A$ is convex if the following holds: if $\forall p,q\in A$ the line segment $\overline{pq}\subseteq A$ is entirely contained in $A$. That is, for all $t\in [0,1]$:

$$(1-t)p+tq\in A$$

Open & Closed

A open ball $B$ is a set with center $x$ and radius $r$ where

$$B:=\{y\in {\mathbb{R}}^k\mid |y-x|<r\}$$

A closed ball $C$ is a set with a center $x$ and radius $r$ where

$$C:=\{y\in {\mathbb{R}}^k\mid |y-x|\le r\}$$

The set $E\subset {\mathbb{R}}^k$ is convex if

$$\lambda x+(1-\lambda )y\in E$$

whenever $x,y\in E$ and $0<\lambda <1$. For example, the balls are convex. For if $|y-x|<r,|z-x|<r$ and $\lambda \in (0,1)$, we have

$$\begin{array}{rl}|\lambda x+(1-\lambda )z-x|& =|\lambda (y-x)+(1-\lambda )(z-x)|\\ & \le \lambda |y-x|+(1-\lambda )|z-x|\\ & <\lambda r+(1-\lambda )r\\ & =r\end{array}$$

This is also true for closed balls.

As subsets of a Metric Space $X$

• Let $E\subset X$. Then $E$ is closed if every limit point of $E$ is a point of $E$.
• $E$ is open if every point of $E$ is an interior point of $E$.
  • $E$ is open $⟺E^c$ is closed
• The complement of $E$ (denoted as $E^c$) is the set of all points $p\in X$ such that $p\ne E$.
• $E$ is perfect if $E$ is closed and if every point of $E$ is a limit point of $E$.
  • In other words, $A=A^{\prime }$.
  • In particular, every perfect set is closed.
• $E$ is bounded if there is a real number $M$ and a point $q\in X$ such that $d(p,q)<M$ for all $p\in E$.
  • “You can draw a large circle that encompasses all of $E$.”
• $E$ is dense in $X$ if every point of $X$ is a limit point of $E$, or a point of $E$ (or both).

Neighborhood

A neighborhood of $p$ is a set $N_r(p)$ consisting of all $q$ such that $d(p,q)<r$ for some $r>0$. The number $r$ is called the radius of $N_r(p)$.

Theorem (Open Neighborhood)

Every neighborhood is an open set. Consider a neighborhood $E=N_r(p)$ and let $q$ be any point of $E$. Then there is a positive real number $h$ such that

$$d(p,q)=r-h$$

For all points $s$ such that $d(p,q)<h$, we have then

$$d(p,s)\le d(p,q)+d(q,s)<r-h+h=r$$

so that $s\in E$. Thus $q$ is an interior point of $E$.

• The second inequality is from the Triangle Inequality of it being a Metric Space.

Lemma (Infinite Neighborhoods)

Let $A\subset X$ and $p\in X$ be a limit point of $A$. Then $\forall r>0,N_r(p)\cap A$ is infinite.

Proof:

We prove by contradiction. Suppose there exists $r>0$ such that $N_r(p)\cap A$ is finite. Let $\{q_1,\cdots q_n\}$ be this finite set except if $p\in A$. Now, $\forall i=1,\cdots n$, $q_i\ne p$. So,

$$0<d(p,q_i)<r$$

for $q_i\in N_r(p)$ and $q_i\ne p$. Let $s=\min \{d(p,q_i\mid i=1,\cdots n\}>0$. Now we claim there does not exist $q\in A\cap N_s(p)$ with $q\ne p$. Indeed, $0<s=\min \{d(p,q_i)\mid i=1,\cdots n\}<r$. So $N_s(p)\subset N_r(p)$. However,

$$\{q_1,\cdots q_n\}=A\cap (N_r(p)\backslash \{p\})$$

and $d(p,x)<s⟹x\ne q_i,\forall i$. Therefore $N_s(p)\cap A$ does not contain any $q\ne p$ and contradicts with the fact that $p$ is a limit point.

Corollary (No Limits in Finite Sets)

Let $A\subset X$ be finite. Then $A$ has no limit points.

Limit Point

A point $p$ is a limit point of the set $E$ if every neighborhood of $p$ contains a point $q\ne p$ such that $q\in E$.

If $p\in E$ and $p$ is not a limit point of $E$, then $p$ is called an isolated point of $E$.

Interior Point

A point $p$ is an interior point of $E$ if there is a neighborhood $N$ of $p$ such that $N\subset E$.

Lemma (Open Complement Closed)

Let $(X,d)$ be a metric space and $A\subset X$. $A$ is open $⟺$ $A^c$ is closed

Proof:

Suppose $A$ is open. We want to show $A^c$ is closed. That is, $A^c$ contains all of its limit points.

Since $X=A\cup A^c$, the above is the same as showing no point in $A$ is a limit point of $A^c$.

• Indeed, if there was a limit point $p\in A$, then $A^c$ cannot be closed as $A,A^c$ are disjoint and $p\notin A^c$.
• Precisely, $p\in X$ is a limit point of $A^c$ if $\forall r>0,\exists q\in N_r(p),q=p,q\in A^c$.

If $A=\varnothing$, then nothing to show. Suppose $A$ is nonempty so let $x\in A$. Since $A$ is open, $\exists s>0$ such that $N_s(x)\subset A$. Thus $N_s(x)\cap A^c=\varnothing$ and so $x$ is not a limit point.

Suppose $A^c$ is closed. Let $x\in A$. Then $x\notin A^c⟹x$ is not a limit point of $A^c$. Then $\exists r>0$ such that $N_r(x)\backslash \{x\}\cap A^c=\varnothing$ such that $N_r(x)\subset A$ an $x$ is an interior point of $A$ and $A$ is open as $x$ arbitrary.

Lemma (Openness and Closure)

1. $\varnothing$ and $X$ are both open and closed.
2. An arbitrary union of open sets is open.
3. An arbitrary intersection of closed sets is closed.
4. A finite intersection of open sets is open.
5. A finite union of closed sets is closed.

Proof:

1. Clear from definitions.

2. “Arbitrary” means we can have any kind of union (infinite, uncountably infinite, 3) sets.

   Let $\{{\theta }_{\alpha }\mid \alpha \in I\}$ be a collection of open sets. Put

$$\theta =\bigcup _{\alpha \in I}{\theta }_{\alpha }$$

If $\theta$ is empty, then we are done. Suppose not. Let $p\in \theta$. We want to show that $p$ is an interior point such that $\exists r>0,N_r(p)\subset \theta$. Since $p\in \theta$ then $\exists \alpha \in I$ such that $p\in {\theta }_{\alpha }$. But by assumption, ${\theta }_{\alpha }$ is open, such that $\exists r>0,N_r(p)\subset {\theta }_{\alpha }\subset {\bigcup }_{\alpha \in I}{\theta }_{\alpha }=\theta$. Indeed, $N_r(p)\subset \theta$ and $p$ is an interior point of $\theta$. 3. Consider $\{F_{\alpha }\mid \alpha \in I\}$ a collection of closed sets. We can show that ${\bigcap }_{\alpha \in I}{F}_{\alpha }$ is closed if the union is open. Indeed:

$${\left(\bigcap _{\alpha \in I}{F}_{\alpha }\right)}^c=\bigcup _{\alpha \in I}{F}_{\alpha }^c$$

by DeMorgan’s Law. Note, $\forall \alpha \in I$, since $F_{\alpha }$ is closed, then $F_{\alpha }^c$ is open. Indeed, by $(2)$ then the complement is open and thus the intersection is closed. 4. Suppose $(O_i{)}_{i=1}^n$ is a finite collection of open sets. We want to show:

$$O:=\bigcap _{i=1}^n{O}_i$$

is open. Let $x\in O$. We WTS $x$ is an interior point. So, $x\in O_i$ for all $i$. Hence, $O_i$ is open such that $\exists r_i>0$ such that $N_{r_i}(X)\subset O_i$.

Let $r=\min \{r_i\mid i=1,\cdots ,n\}$. Because the set of radii is finite, $r>0$. Then $\forall i=1,\cdots n$ we have $N_r(x)\subset N_r(x)\subset {\theta }_i$. Thus $N_r(x)\subset {\bigcap }_{i=1}^n{O}_i=O$. 5. Follows from $(4)$ and DeMorgan’s Law.

Definition (Completeness)

From Definition (Completeness)