# Notes - Metric Spaces MT23, Basic definitions

### Flashcards

What three axioms define a distance function $d : X \times X \to \mathbb R$ on a set $X$?

- $d(x, y) \ge 0$ and $d(x, y) = 0$ iff $x = y$.
- $d(x, y) = d(y, x)$.
- $\forall x, y, z \in X \text{ } d(x, z) \le d(x, y) + d(y, z)$.

What is the reverse triangle inequality in a metric space $X$?

Can you define a norm for a vector space $V$ over $\mathbb R$?

A function $ \vert \vert \cdot \vert \vert : V \to [0, \infty)$ such that

- $ \vert \vert x \vert \vert = 0$ iff $x = 0$.
- $ \vert \vert \lambda x \vert \vert = \vert \lambda \vert \text{ } \vert \vert x \vert \vert $ for all $\lambda \in \mathbb R$, $x \in V$.
- $ \vert \vert x+y \vert \vert \le \vert \vert x \vert \vert + \vert \vert y \vert \vert $ for all $x, y \in V$.

Given a norm $ \vert \vert \cdot \vert \vert $, how can you define a metric?

Can you give an example of a metric that is not formed from a norm?

The discrete metric.

If a metric is defined by a norm (or vice versa), what additional properties do you get that are not true for a general metric?

- Translational invariance: $d(x + z, y + z) = d(x, y)$.
- Homogeneity: $d(\lambda x, \lambda y) = \vert \lambda \vert d(x, y)$.

Suppose you have $f : X \to Y$ bijection and $X$ is a metric space. How can you define a metric on $Y$?

A common exam way of making a new metric from old ones in exam questions is

\[\rho(x, y) = \frac{d(x, y)}{1 + d(x, y)}\]
Quickly prove that this is still a metric.

Symmetry and positive definiteness are immediate.

For the triangle inequality, we need to show that

\[\rho(x, y) \le \rho(x, z) + \rho(y, z) \quad\forall x, y, z\]Or, expanding out:

\[\frac{d(x, y)}{1 + d(x, y)} \le \frac{d(x, z)}{1 + d(x, z)} + \frac{d(y, z)}{1 + d(y, z)}\]Letting $a = d(x, y)$, $b = d(x, z)$ and $c = d(y, z)$, this becomes:

\[\frac{a}{1 + a} \le \frac{b}{1 + b} + \frac{c}{1 + c}\]We first use this inequality (for $b, c \ge 0$):

\[\frac{b + c}{1 + b + c} \le \frac{b}{1 + b} + \frac{c}{1 + c}\]which can be seen by expanding out

\[\frac{b + c}{1 + b + c} - \frac{b}{1 + b} + \frac{c}{1 + c} = -\frac{bc(b + c + 2)}{(1 + b + c)(1 + b + c + bc)} \le 0\]Then, undoing the substiution, we have

\[\frac{d(x, z)}{1 + d(x, z)} + \frac{d(y, z)}{1 + d(y, z)} \ge \frac{d(x, z) + d(y, z)}{1+ d(x, z) + d(y, z)}\]Then, by clearing denominators, it turns out that

\[\frac{d(x, z) + d(y, z)}{1+ d(x, z) + d(y, z)} \ge \frac{d(x, y)}{1 + d(x, y)}\]is equivalent to the triangle inequality. So the result follows.

### Proofs

Prove the reverse triangle inequality in a metric space $X$:

$\forall x, y, z \in X \text{ } \vert d(x, y) - d(x, z) \vert \le d(y, z)$

$\forall x, y, z \in X \text{ } \vert d(x, y) - d(x, z) \vert \le d(y, z)$

Todo.

Prove that, given the Euclidiean norm, we have

\[||x + y||_2 \le ||x||_2 + ||y||_2\]

Todo – use the Cauchy-Schwarz inequality.