# Real numbers

There are certain situations when the rational numbers are apparently not sufficient: for instance, the function $f(x)=x^2-2$ is negative at $x=0$, positive at $x=2$ but does not take the intermediate value zero: $\forall x\in\mathbb Q\ f(x)\ne 0$. Another situation concerns the possibility to define the notions of supremum and infimum for infinite sets: the set $A=\{x\in\mathbb Q: x^2<2\}$ is bounded from two sides, but among its upper bounds $B=\{b\in\mathbb Q:\ \forall a\in A\ a\leqslant b\}$ there is no minimal one.

The idea is to adjoin to $\mathbb Q$ solutions of infinitely many inequalities.

For any rational number $a\in\mathbb Q$ one can associate two subsets $L,R\subset\mathbb Q$ as follows: $L=\{l\in \mathbb Q: l\le a\}$ and $R=\{r\in\mathbb Q: a\le r\}$. Then the number $a$ is the unique solution to the infinite system of inequalities of the form $l\le x\le r$ for different choices of $l\in L,\ r\in R$. This system has the following two features:

1. it is self-consistent (non-contradictory): any lower bound $l$ is no greater than any upper bound $r$, i.e., $L\le R$, and
2. it is maximal: together the two sets give $\mathbb Q=L\cup R$, and none of the sets can be enlarged without violating the first condition.

Definition.
A (Dedekind) cut is any pair of subsets $L,R\subseteq\mathbb Q$ satisfying the two conditions above.

If a rational number $a\in\mathbb Q$ satisfies all the inqualities $l\le a,\ a\le r$ for all $l\in L,\ r\in R$, then we call it a root (or a solution) of the cut. Every rational number is the solution to some cut $\alpha=(L,R)$ as above, and this happens if and only if $L\cap R=\{a\}$. Yet not all cuts have rational solutions (give an example!).

We can associate cuts without rational solutions with “missing” numbers which we want to adjoin to $\mathbb Q$. For this purpose we have to show how cuts can be ordered (in a way compatible with the order on $\mathbb Q$) and how arithmetic operations can be performed on cuts.

## Order on cuts

Let $\alpha=(L,R),\ \beta=(L',R')$ be two different cuts. We declare that $\alpha\triangleleft\beta$, if $L\cap R'\ne\varnothing$, i.e., if there is a rational number $a\in\mathbb Q$ that is at the same time an upper bound for the cut $\alpha$ and a lower bound for the cut $\beta$. If both cuts have rational solutions, this number would be squeezed between these solutions. In the similar way we define the opposite order $\alpha\triangleright\beta$ if and only if $L'\cap R\ne\varnothing$.

To see that this definition is indeed a complete order, we need to check that for any two cuts $\alpha,\beta$ one and only one of the three possibilities holds: $\alpha\triangleleft\beta,\ \alpha\triangleright\beta$ or $\alpha=\beta$ (meaning that $L=L',R=R'$). This is a routine check: if the first two possibilities are excluded, then $L\cap R'=L'\cap R=\varnothing$, and therefore $(L\cup L', R\cup R')$ is a self-consistent cut. But because of the maximality condition, this means that $L\cup L'=L=L'$ and $R\cup R'=R=R'$, that is, $\alpha=\beta$.

## Arithmetic operations on cuts

If $\alpha=(L,R),\ \beta=(L',R')$ are two cuts which have rational solutions $a,b$, then these solutions satisfy inequalities $L\le a\le R,\ L'\le b\le R'$ (check that you understand the meaning of this inequality between sets and numbers ;-)!) Adding these inequalities together means that $c=a+b$ satisfies the infinite system of inequalities $L+L'\le c\le R+R'$, where $L+L'$ stands for the so called Minkowski sum $L+L'=\{l+l':\ l\in L,\ l'\in L'\}$ (the same for $R+R'$). This allows to define the summation on cuts.

Definition.
The sum of two cuts $\alpha=(L,R),\beta=(L',R')$ is the cut $\gamma=(L+L',R+R')$ with the Minkowski sum in the right hand side.

To define the difference, we first define the cut $-\alpha$ as follows, $-\alpha=(-R,-L)$, where (of course!) $-L=\{-l: l\in L\},\ -R=\{-r: r\in R\}$. Note that the upper and lower bounds exchanged their roles, since multiplication by $-1$ changes the direction (sense) of the inequalities. Then we can safely define $\alpha-\beta$ as $\alpha + (-\beta)$. Again, one has to check that this definition is well-behaving and all arithmetic properties are preserved.

To define multiplication, one has to exercise additional care and start with multiplication between positive cuts $\alpha,\beta\triangleright 0$ (do it yourselves!) and then extend it for negative cuts and the zero cut. After introducing this definition, one has to make a lot of trivial checks:

1. that for cuts having rational solutions, we get precisely what we expected, that is, the new operation agrees with the old one on the rational numbers,
2. that they have the same algebraic properties (associativity, distributivity, commutativity etc) as we had for the rational numbers,
3. that they agree with the order that we introduced earlier exactly as this was the case with the rational numbers,
4. … … …. …. …

Of course, nobody ever wrote the formal proofs of these endless properties! (Life is short and one should not waste it for nothing). Yet every mathematician can certainly provide a formal proof for any of them, and nobody of countless students who passed through this ordeal ever voiced any concern about validity of these endless nanotheorems. So wouldn’t we.

## Achievement of the stated goals

Once we constructed the extension of the rational numbers by all cuts and denote the result $\mathbb R$ and call it the set of real numbers, one has to verify that all the problems we started with, were actually resolved. There is a number of theorems about the real numbers that look dull and self-evident unless we know that a heavy price had to be paid for that. Namely, we can guarantee that:

1. Any subset $A\subset\mathbb R$ which admits at least one upper bound, admits the minimal upper bound called $\sup A=\sup_{a\in A}a$ (and, of course, the analogous statement holds for $\inf A$).
2. If $\varnothing\ne I_k=[a_k,b_k]\subseteq\mathbb R$ is a family of nested nonempty closed intervals, $I_1\supseteq I_2\supseteq I_3\supseteq\cdots$, then the intersection $I_\infty=\bigcap_{k=1}^\infty I_k$ is also nonempty.
3. Any function $f:[a,b]\to\mathbb R$ continuous on the closed segment $[a,b]$, takes any intermediate value between $f(a)$ and $f(b)$.

For more detailed exposition, read the lecture notes here.