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.

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.

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 \(\sqrt 2\).

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.

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.