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 ${\displaystyle \mathbb {Q} } $ into two subsets $A$ and $B$ such that
$A$ is non-empty 
$A \ne \mathbb{Q}$
If $x,y \in \mathbb{Q}, x<y$ 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? 
 A: Snoop Dogg let's go through your list thoroughly. 
  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 $\in A$, 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. $\neq \mathbb{Q}$
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 $x,y\in \mathbb{Q}$, $x<y$, 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 <d < a' \cdot b'$ 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
