If we define Dedekind-real numbers as Dedekind cuts, i.e. $\sqrt 2 = \{\text{rationals less than }\sqrt2\} \cup \{\text{rationals more than } \sqrt2\}$, can we define addition and multiplication of these real numbers as follows:

These real numbers $\mathbb R$ are a complete lattice in which the rational numbers are embedded. Take the open intervals of $\mathbb R$ and make them the basis for a topology on $\mathbb R$.

Define $+(\mathbb R,\mathbb R)$ and $\times(\mathbb R,\mathbb R)$ as the unique continuous (in this topology) extension of the already known arithmetic functions $+(\mathbb Q,\mathbb Q)$ and $\times(\mathbb Q,\mathbb Q)$?

  • $\begingroup$ "in this topology" -- you mean the product topology on $\mathbb R^2$? $\endgroup$ – oldrinb Jul 31 '15 at 17:26
  • $\begingroup$ I am a little bit rusty on topology but I imagine so yes. $\endgroup$ – John Smith Jul 31 '15 at 17:27
  • $\begingroup$ A function being continuous on a subspace of a topological space does not imply there is a continuous extension to the entire space (simple example: $f(x) = x^{-1}$ on $(0, 1]$ cannot be continuously extended to $[0, 1]$). What you need is uniform continuity, and in order to sensibly talk about that, you need a uniformity (which induces a topology) or a metric (which induces a uniformity). $\endgroup$ – Theo Bendit Jul 31 '15 at 18:18
  • $\begingroup$ So we also need a metric that agrees with Qs metric? Say something like, if we have r1 and r2 (reals) and q1 and q2 (rationals) then q1<r1<r2<q2 implies d(r1,r2)<d(q1,q2) and r1<q1<q2<r2 implies d(r1,r2)>d(q1,q2) ? $\endgroup$ – John Smith Jul 31 '15 at 19:55
  • 2
    $\begingroup$ You need to modify your definition of Dedekind cuts a bit. As you've defined it, $\sqrt{2} = \mathbb Q$, which is not what you want. A better definition would give $\sqrt{2} = \{ \{\text{rationals less than $\sqrt{2}$}\},\,\{\text{rationals greater than $\sqrt{2}$}\}\}$. $\endgroup$ – Jack Lee Jul 31 '15 at 23:13

The idea of Dedikind cuts is that one can define $\mathbb{R}$ and its arithmetic operations and order in an elementary manner, directly in terms of $\mathbb{Q}$ and its the arithmetic operations and order. One doesn't need the concepts of analysis or topology to do this. Of course, after the real numbers and their operations are defined, one can go on to develop calculus and to prove that addition and multiplication are continuous operations, but it is not necessary to bring continuity or other concepts of analysis into the picture simply for the purpose of defining addition and multiplication of real numbers.

For example, given two Dedekind cuts $(A_1,B_1)$ and $(A_2,B_2)$ of $\mathbb{Q}$, their sum is then defined to be $$(A_1,B_1) + (A_2,B_2) = (A_1+A_2,B_1+B_2) $$ where the meaning of addition of subsets is simply $$A_1 + A_2 = \{a_1 + a_2 \, \bigm| \, a_1 \in A_1 \, \text{and} \, a_2 \in A_2 \} $$ Now there is a little work to be done, but it should be pretty elementary: one has to prove that $(A_1+A_2,B_1+B_2)$ is, indeed, a Dedikind cut.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.