Metric Space

Axioms

Given a set $B$ and a function $d(a, b)$ where $a, b \in B$, we say the set and this function comprise a metric space if the three properties hold

(M1) Positive Definite

Informally, we say a function is positive definite if its value is always greater than or equal to zero. Equality holds only when its arguments are equal.

$$d(a, b) \ge 0$$

In other words, we can measure distance in a metric space.

(M2) Symmetric

Measurement can occur in either direction, and you can expect the same answer.

$$d(a, b) = d(b, a)$$

(M3) Triangle Inequality

Imagine the triangle equality did not hold. Then you could draw a triangle where the sides did not connect -- but that's impossible by definition of a triangle. In general metric spaces, this casual observation has a precise formulation in the triangle equality.

$$d(a, b) = \biggl( \sum(a_i - b_i)^2 \biggr) ^{1/2}$$

Show that the two dimensional plane ${R}^2$ is a metric space using the distance function $d(a, b) = | a - b |$

Properties

Convergence

Given a sequence of points ${ a_n }$ in the metric space, we say that the sequence converges when, as we progress through the entire sequence, the distance between each item and some fixed point $a^{*}$approaches zero.

$$\lim_{n \to \infty } d(a_n, a^{*}) = 0$$

Think about this in terms of positive definitiveness and commutivity. The smallest distance possible is zero, and the order of arguments does not matter. If the sequence always gets closer to zero, then the "last" point in the sequence must be getting closer and closer to the fixed point.

Closure

Given a subset of a metric space, $A \sube B$, check that every convergent sequence in ${a_n} \in A$has it's limit in $A$. If so, then we say the subset is closed.

A typical example is to imagine a "ball" in $B$ with its perimiter a fixed distance $r$ from its center. Pick any two points inside the ball to see that the distance is less than $r$. If so, then the ball is closed. Formally, we'd write

$$A := \{ b \in B : d(a, b) \le r \}, \\ r \gt 0$$

Openness

Openness is the opposite of closure, in the sense that, if you remove a subset $C$ from $B$and the resulting set $B \setminus C$ is closed, then $C$ must be open.

In writing this formally, we simply relax the inequality condition

$$A := \{ b \in B : d(a, b) \lt r \}, \\ r \gt 0$$

Notice we've only changed from "less than to equal to" to "less than"; we have not changed direction of the inequality. It's not possible to change the direction in forming the "ball" due to the positive definiteness condition.

Compactness

To demonstrate that a subset is compact, we need to show that every sequence has a subsequence that converges.

Use the familiar $R^2$ plane to imagine this. Draw a shape on the plane; this will be the subset $K \sub R^2$. Choose any sequence of points from one end of the shape to the other, and take a portion from that sequence. It's easy to see that the subsequence has a limit point that's still inside the shape. Doing this for all possible sequences, we notice that all the limit points are inside the shape. There's no "holes".

Continuity

Continuity is defined regarding a function $\phi$. This function takes an argument $a, b \in B$ and outputs a value $\phi(a), \phi(b) \in B'$. Both $B$and $B'$are metric spaces. If each distance measured in $B'$is less than some positive value $\epsilon$, then the corresponding measurement in $B$must be less than some positive value $\delta$. Formally,

$$\exists{a} : d(a, b) \lt \delta \Rightarrow d'(\phi(a), \phi(b)) \lt \epsilon$$

If the function is not continuous at the point $a$, then this correspondence breaks.

Uniform Continuity

To say a function is uniformly continuous, we strengthen the above definition to apply to every point $a, b \in B$

$$\forall{a, b} : d(a, b) \lt \delta \Rightarrow d'(\phi(a), \phi(b)) \lt \epsilon$$

Theorems

See Metric Spaces