# Prove the following set is also a Dedekind cut

For Dedekind cuts $$α, β > 0^∗,$$ show that

$$α ⊗ β := \{ p ∈ \mathbb{Q} | p < r · s \text{ for some } r ∈ α, s ∈ β \text{ such that } r,s > 0 \}$$

is also a Dedekind cut.

I know the definition of a Dedekind says that a Dedekind cut is a partition of the rationals $$\mathbb {Q} }$$ into two subsets $$A$$ and $$B$$ such that

$$A$$ is non-empty

$$A \ne \mathbb{Q}$$

If $$x,y \in \mathbb{Q}, x and $$y\in A$$ then $$x \in A$$

If $$x \in A$$, then there exists a $$y \in A$$ such that $$y>x$$

To prove that $$α ⊗ β$$ is a Dedekind set, do I only need to show that: if $$α ⊗ β$$ contains some rational $$\gamma$$, it contains every rational to the left of $$\gamma$$ as well; there must be some rational number $$\delta$$ such that every member of $$α ⊗ β$$ is at or to the left of $$\delta$$ and that $$α ⊗ β$$ must not have a largest element? I believe I do not have an exhaustive foundation of a Dedekind set/cut yet so how could I prove $$α ⊗ β$$ is a Dedekind set/cut thoroughly?

• You are given that $\alpha, \beta$ are Dedekind cuts and both contain some positive rationals as well. Use this information to show that the set $\alpha \otimes \beta$ fulfills all the properties of a Dedekind cut given in your question. – Paramanand Singh May 8 '20 at 12:46

Snoop Dogg let's go through your list thoroughly.

𝐴 is non-empty

If $$\alpha = (L_\alpha, R_{\alpha} ),$$ $$\beta = (L_\beta, R_{\beta} )$$, and $$\alpha ,\beta > 0^* =( \mathbb{Q^{-}} , \mathbb{Q^+})$$ then we have that there is an $$a \in L_\alpha$$ and a $$b \in L_\beta$$ such that $$a \not\in \mathbb{Q^{\leq 0}}$$ and $$b \not\in \mathbb{Q^{-}}$$ and by furthermore by the last rule, i.e.

If $$𝑥\in A$$, then there exists a $$y\in A$$ such that $$y>x$$,

we have that there is some $$a' \in L_\alpha$$ and $$b' \in L_\beta$$ such that $$a' > a \geq 0$$ and $$b' > b \geq 0$$; therefore $$\alpha \otimes \beta$$ is not empty since $$0 ,a,b$$ are all in $$\alpha \otimes \beta$$ (for example $$0 < a'\cdot b'$$ and $$a' > 0$$ and $$b'>0$$).

(A,B) is a non-trivial partition, i.e. $$𝐴\neq \mathbb{Q}$$

Since $$\alpha = (L_\alpha, R_{\alpha} ),$$ $$\beta = (L_\beta, R_{\beta} )$$ both satisfy this rule there is some $$r_\alpha$$ and $$r_\beta$$ such that $$r_\alpha > a$$ and $$r_\beta > b$$ for all $$a \in L_\alpha , b \in L_\beta$$; but then for all $$c \in \alpha \otimes \beta$$ we have that $$c < r_\alpha \cdot r_\beta$$ by the definition of $$\alpha \otimes \beta$$ (let $$a'\in L_\alpha ,b' \in L_\beta$$ be witnesses to the definition of $$\alpha \otimes \beta$$; then $$c < a' \cdot b' < r_\alpha \cdot r_\beta$$) so that $$L_{\alpha \otimes \beta} \neq \mathbb{Q}$$.

If $$x,y\in \mathbb{Q}$$, $$x, and $$y \in A$$ then $$x \in A$$

If $$a \in L_\alpha ,b \ \in L_\beta$$ witness the statement $$c < a \cdot b$$, i.e. $$c \in \alpha \otimes \beta$$, and let $$d < c$$ then by definition $$d \in \alpha \otimes \beta$$.

If $$x \in A$$, then there exists a $$y\in A$$ such that $$y>x$$

Let $$a \in L_\alpha ,b \ \in L_\beta$$ be witnesses to the statement $$c < a \cdot b$$ and since $$\alpha, \beta$$ are Dedekind cuts then the rule above, i.e.

If $$x \in A$$, then there exists a $$y\in A$$ such that $$y>x$$

so that there are some $$a' \in L_\alpha ,b' \ \in L_\beta$$ such that $$a< a'$$ and $$b< b'$$. Let $$d = \frac{a \cdot b + a' \cdot b'}{2}$$, then $$c so that If $$c \in L_{\alpha \otimes \beta}$$, then there exists a $$d \in L_{\alpha \otimes \beta}$$ such that $$d>c$$ as was needed. QED