## 414/514 – Cauchy sequences

This is homework 1, due Friday September 9 at the beginning of lecture.

We define absolute value as usual: Given a real $x$, we say that ${}|x|$ is $x$ if $x\ge0$ and is $-x$ otherwise.

Absolute values have useful properties: ${}|x|=|-x|\ge0$ for any $x$. Also, ${}|x|=0$ iff $x=0$. The key property is the triangle inequality: ${}|a+b|\le|a|+|b|$.

Formally, a sequence is a function $s:{\mathbb N}\to{\mathbb R}$. As usual, we write the sequence as $s_1,s_2,\dots$ rather than $s(0),s(1),\dots$

A sequence $s$ is a Cauchy sequence iff for all $\epsilon>0$ there is an $N\in{\mathbb N}$ such that whenever $n,m\in{\mathbb N}$ and $n,m>N$, we have $|s_n-s_m|<\epsilon.$

A sequence $s$ converges iff there is a real $r$ such that for all $\epsilon>0$ there is an $N\in{\mathbb N}$ such that whenever $n\in{\mathbb N}$ and $n>N$, we have $|r-s_n|<\epsilon$.

Note that these concepts also make sense in ${\mathbb Q}$. Now we require all the $s_n$ to be rational, and we require $\epsilon$ and $r$ to be rational as well.

• Show that if a sequence converges, then it is Cauchy.
• Give an example of a Cauchy sequence in ${\mathbb Q}$ that does not converge.
• Show that any Cauchy sequence in ${\mathbb R}$ converges.

Cauchy’s way of defining the reals was to use Cauchy sequences as the basic building blocks rather than cuts. Again, the idea is that we want to have all the limits, and in ${\mathbb Q}$ some of these limits are missing. In the case of cuts, the way of solving the presence of gaps in ${\mathbb Q}$ was by giving names to all the gaps (the cuts), and adding the names.  The easiest repair to the lack of limits here will be the same: We give a name to the limits (the sequences themselves) and the reals will be just the sequences.

There is a problem here that does not occur with the construction using cuts, namely different sequences may have the same limit. We should identify all of them.

Recall that an equivalence relation on a set $X$ is a binary relation $\sim$ that is:

1. Reflexive: For any $x\in X$, $x\sim x$.
2. Symmetric: For any $x,y\in X$, if $x\sim y$ then also $y\sim x$.
3. Transitive: For any $x,y,z\in X$, if $x\sim y$ and $y\sim z$, then $x\sim z$.

If $\sim$ is an equivalence relation, the equivalence classes determined by $\sim$ are the sets ${}[x]=\{y\in X\mid x\sim y\}$. An intuitive way of thinking about this is that we are looking at $X$ from a distance, and so we cannot distinguish points that are close to one another, we just see them smashed together as a single point. Here, two points $x,y$ are close iff $x\sim y$.

Let $s$ and $t$ be two Cauchy sequences of rationals. Say that $s\sim t$ iff $s-t$ converges to $0$. Here, of course,  $s-t$ is the sequence $r_1,r_2,\dots$ with $r_n=s_n-t_n$.

• Show that $\sim$ is an equivalence relation. Check that any Cauchy sequence $s$ is equivalent to infinitely many other sequences.
• Define ${\mathbb R}$ as the set of equivalence classes of the relation $\sim$. The elements of ${\mathbb R}$ are then Cauchy sequences or, more precisely, collections of Cauchy sequences. A typical element of ${\mathbb R}$ is a class ${}[s]$, and we think of ${}[s]$ as the limit of $s$. Of course, we have a copy of ${\mathbb Q}$ inside ${\mathbb R}$: We can identify the rational $q$ with the class $q^*=[s]$ of all sequences $s$ that converge to $q$.
• Define $+,\times,<$ in ${\mathbb R}$ and verify that with these definitions we have an ordered field.
• Verify that ${\mathbb R}$ is complete, meaning that the least upper bound property holds.

This gives a second sense in which ${\mathbb R}$ is complete: It contains the limits of all Cauchy sequences. A small but important point not mentioned above is the following: Given a sequence $s$ of rationals, let $s'$ be its “copy” inside ${\mathbb R}$, i.e., $(s')_n=s_n^*$. Then $s'$ is Cauchy iff $s$ is Cauchy, and $s$ converges to a rational $q$ iff $s'$ converges to $q^*$.