Main theorem of Dedekind

Question

Main theorem of Dedekind: It claims, that for every cut $A|A'$ in the set of real numbers there exists a number $b$ that forms the cut. This number $b$ will be $1)$ the largest in the lower class or $2)$ the smallest in the higher class.

I can't get the proof. The proof in the book was given by the contradiction. It says: note by $\mathbf A$ the set of all rational number, that belong to $A$ and by $\mathbf A'$ the set of all rationals, that belong to $A'$. Therefore, $\mathbf A$ and $\mathbf A'$ form a set of all rational numbers. The cut distinguish some real number $b$. Assume, that $b$ is in the lower class $A$. We will prove that it is the maximum in $A$.

contradiction:

If it is not, we can take some number $a_0$ from $A$, that is bigger than $b$. We can also take some rational number $r$ such as $b<r<a_0$. Which means, that $r$ belongs to the $\mathbf A$ class. It is a contradiction, since the rational number $r$, which is in the lower class of the cut, is bigger than the number $b$, which is distinguhed by the lower cut.

I don't see why it can't be so. Does it want to say that the contradiction is in the fact, that if $a_0$ is bigger than $b$, it can't be in the lower class, because the cut was formed by $b$?

Answer 1

This is the original proof by Dedekind (also repeated by Hardy in his A Course of Pure Mathematics) and to understand it properly you need to be very clear of the fact there are are two types of Dedekind cuts being used here. One is the cut $(A, A') $ which involves partitioning the set of reals into two sets $A, A'$. Another is a cut $(\mathbf{A}, \mathbf{A}')$ which involves partition of rationals into two sets. By convention let's assume that bold symbols deal with rationals and usual symbols deal with reals.

Once this distinction is made clear one has to show that the sets $\mathbf{A}, \mathbf{A} '$ do form a Dedekind cut and hence define a real number $b$ and this must be in exactly one of the sets $A, A' $.

Coming back to contradiction, we assume that $b\in A$ and we wish to prove that $b$ is the greatest member of $A$. If not then there is some real number $b'>b$ such that $b'\in A$. Dedekind then uses the fact that between any two distinct real numbers lies a rational number (this needs to be proved and perhaps is given earlier in your textbook). Thus there is a rational $\mathbf{r} $ such that $b<\mathbf{r} <b'$. Now this $\mathbf{r} $ must lie in either $\mathbf{A} $ or $\mathbf{A} '$. Since $b$ is the real number defined by cut $(\mathbf{A}, \mathbf{A}') $ we must have $\mathbf{r} \in \mathbf{A} '$. And since $b' \in A$ and $r<b'$ we must have $r\in A$ and hence $\mathbf{r} \in \mathbf{A} $. Note the usage of normal and bold $A$ and try to understand exactly why these statements are true. Also note the dual role played by $r$ as a rational number and as a real number (this is indicated using normal and bold symbols). And we have reached the contradiction because $\mathbf{r}$ can belong to only one of the sets $\mathbf{A}, \mathbf{A} '$.

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You should also mention the conventions followed by Dedekind. Since I am familiar with his original pamphlet Stetigkeit und irrationale zahlen I could easily understand your post, but it is better to add more context for the benefit of everyone. For Dedekind a section/cut/schnitt is a procedure to partition an ordered set $F$ into two subsets $A, A'$ such that $A\neq\emptyset\neq A', A\cup A'=F, A\cap A'=\emptyset$ and every member of $A$ is less than every member of $A'$. $A$ is called the lower set and $A'$ is called the upper set.

When this procedure is carried out with $F=\mathbb{Q} $ (the set of rationals) then there are three possibilities :

• lower set has a greatest member
• or upper set has a least member
• neither the lower set has a greatest member nor the upper set has a least member.

The great achievement of Dedekind is to show that when the set $F=\mathbb{R} $ (the set of reals) is partitioned in the above manner only the first two possibilities listed above occur. The third possibility does not arise. This is the theorem being discussed here.