The Metric Topology

Definition

A metric on a set X is a function

d \colon X \times X \rightarrow R

such that the following hold:

(1) d(x,y) \geq 0 for all x,y \in X, with equality only at x = y.

(2) d(x,y) = d(y,x) for all x,y \in X.

(3) d(x,y) + d(y,z) \geq d(x,z) for all x,y,z\in X (Triangle inequality)

The value of d(x,y) is called the distance between x and y in the metric d.

We can come up with many alternate ways to calculate distance between points.

Definition

Let d be a metric on a set X, and let \varepsilon > 0. The set

B_d(x,\varepsilon) = \{y \mid d(x,y) < \varepsilon\}

of all points y whose distance from x is less than \varepsilon is called the \varepsilon-ball centered at x. We often omit the metric when it is implied, and write simply B(x,\varepsilon).

[Why is this relevant, explain how it relates to topology, why we induce a topology from a metric]

We now have a formal way to express balls in other metrics. We now examine how we can induce a topology from a metric.

Definition

Let d be a metric on the set X. The collection of all balls B_d(x,\varepsilon) for x \in X and \varepsilon > 0, is a basis for a topology on X. This is called the metric topology induced by X.

We must of course prove that taking the collection of balls of any radius centered at any point creates a basis for a topology, but that can wait.

We have seen how a metric can induce a topology, but can we ‘extract’ a metric from a topology? Sometimes.

Definition

Let (X,\mathcal{T}) be a topological space. (X,\mathcal{T}) is metrizable if there exists a metric d on X that induces \mathcal{T} on X.

The following definition is wacky

https://math.stackexchange.com/questions/3327448/equivalence-or-non-equivalence-of-definitions-for-metric-spaces

Definition

Let d be a metric on a set X. A metric space is an ordered pair (X,d).

Though the analysis of metric spaces lies deeper with analysis than topology, we shall still examine some properties.

Properties of Metric Spaces

Definition

Let (X,d) be a metric space. A subset A of X is said to be bounded if there exists some number M such that

d(a_1,a_2) \leq M

for all a_1,a_2 \in A.

We can expand on this definition

Definition

Let (X,d) be a metric space, and A a non-empty bounded subset of X. The diameter of A is defined

\text{diam }A = \sup\{d(a_1,a_2) \mid a_1,a_2 \in A\}.

Definition

Let (X,d) be a metric space, and define {\bar{d} \colon X \times X \rightarrow \mathbb{R}} by

\bar{d}(x,y) = \min\{d(x,y),1\}.

Then \bar{d} is called the standard bounded metric corresponding to d.

We must now prove \bar{d} is a metric.

Theorem

Given a metric d, the standard bounded metric \bar{d} is also a metric.

We have to show that \bar{d} fulfills the conditions for a metric.

(1) \bar{d}(x,y) \geq 0 for all x,y \in X, with equality only at x = y.

Since d is a metric, we have that d(x,y) \geq 0 for all x,y \in X. Because \bar{d}(x,y) is equivalent to either d(x,y) or 1, which are both above 0 for all x,y \in X, we have that \bar{d}(x,y) \geq 0 for all x,y \in X.

Additionally, \bar{d}(x,y) can only equal 0 if d(x,y) does. Since d is a metric, we know this equality is only at x = y, and thus \bar{d}(x,y) = 0 only if x = y.

(2) \bar{d}(x,y) = \bar{d}(y,x) for all x,y \in X.

Because d is a metric

\bar{d}(y,x) = \min\{d(y,x),1\} = \min\{d(x,y),1\} = \bar{d}(x,y)

(3) \bar{d}(x,y) + \bar{d}(y,z) \geq \bar{d}(x,z) for all x,y,z\in X (Triangle inequality)

If either d(x,y) \geq 1 or d(y,z) \geq 1, then the left hand side of the equation is clearly greater than the right, as \bar{d}(x,z) \leq 1. Then we only need check the case where d(x,y) < 1 and d(y,z) < 1.

If d(x,y) + d(y,z) \geq 1, then we are done for the reasons listed above. If d(x,y) + d(y,z) < 1, then because d is a metric, d(x,z) < 1, and \bar{d} = d for all relevant values, the inequality holds.

The boundeness of a set is not a topological property, as it only depends on the metric d that is used for X, and nothing to do with the topology induced by X. To show this consider the following theorem

Theorem

Let (X,d) be a metric space, and let \bar{d} be the standard bounded metric.

Then \bar{d} induces the same topology as d.