# Transcendental and real numbers

## Constructing the real numbers

### Cauchy sequences

#### Cauchy sequence

A cauchy sequence is a sequence such that for an any arbitrarily small number $$\epsilon$$, there is a point in the sequence where all possible pairs after this are even closer together.

$$(\forall \epsilon >0)(\exists N\in \mathbb{N}: \forall m,n \in \mathbb{N} >N)( |a_m - a_n|<\epsilon)$$

This last term gives a distance between two entries. In addition to the number line, this could be used on vectors, where distances are defined.

As a example, $$\dfrac{1}{n}$$ is a cauchy sequence, $$\sum_i \dfrac{1}{n}$$ is not.

#### Completeness

Cauchy sequences can be defined on some given set. For example given all the numbers between $$0$$ and $$1$$ there are any number of different cauchy sequences converging at some point.

If it is possible to define a cauchy sequence on a set where the limit is not in the set, then the set is incomplete.

For example, the numbers between $$0$$ and $$1$$ but not including $$0$$ and $$1$$ are not complete. It is possible to define sequences which converge to these missing points.

More abstractly, you could have all vectors where $$x^2+y^2<1$$. This is incomplete (or open) as sequences on these vectors can converge to limits not in the set.

Cauchy sequences are important when considering real numbers. We could define a sequence converging on $$\sqrt 2$$, but as this number is not in the set, it is incomplete.

### Incompleteness of the rational numbers

#### The square root of $$2$$ is not a rational number

Let’s prove there are numbers which are not rational. Consider $$\sqrt 2$$ and let’s show that it being rational leads to a contradiction.

$$\sqrt 2=\dfrac{x}{y}$$

$$2=\dfrac{x^2}{y^2}$$

$$2y^2=x^2$$

So we know that $$x^2$$ is even, and can be shown as $$x=2n$$.

$$2y^2=(2n)^2$$

$$y^2=2n^2$$

So $$y$$ is even. But if both $$x$$ and $$y$$ are even, then the fraction was not reduced.

This presents a contraction so the original statement must have been false.

So we know there isn’t a rational solution to $$\sqrt 2$$.

### $$\sigma$$-algebra

#### Review of algebra on a set

An algebra, $$\Sigma$$, on set $$s$$ is a set of subsets of $$s$$ such that:

• Closed under intersection: If $$a$$ and $$b$$ are in $$\Sigma$$ then $$a\land b$$ must also be in $$\Sigma$$

• $$\forall ab [(a \in \Sigma \land b \in \Sigma )\rightarrow (a\land b \in \Sigma)]$$

• Closed under union: If $$a$$ and $$b$$ are in $$\Sigma$$ then $$a\lor b$$ must also be in $$\Sigma$$.

• $$\forall ab [(a \in \Sigma \land b \in \Sigma )\rightarrow (a\lor b \in \Sigma)]$$

If both of these are true, then the following is also true:

• Closed under complement: If $$a$$ is in $$\Sigma$$ then $$s \backslash a$$ must also be in $$\Sigma$$

We also require that the null set (and therefore the original set, null’s complement) is part of the algebra.

#### $$\sigma$$-algebra

A $$\sigma$$-algebra is an algebra with an additional condition:

All countable unions of sets in $$A$$ are also in $$A$$.

This adds a constraint. Consider the real numbers with an algebra of all finite sets.

This contains all finite subsets, and their complements. It does not contain $$\mathbb{N}$$.

However a $$\sigma$$-algebra requires all countable unions to be including, and the natural numbers are a countable union.

The power set is a $$\sigma$$-algebra. All other $$\sigma$$-algebras are subsets of the power set.