If $L$ is a Dedekind cut, is the set $\{ x^2:x \in L \}$ a Dedekind cut? The Problem(s):
(a) If $L$ is a Dedekind cut, is the set $\{ x^2:x \in L \}$ a Dedekind cut?
(b) What about the set $\{ 0 \} \cup \{ \frac{1}{x}:x \in L \setminus 0 \}$?
Where I Am:
Here's the definition of a Dedekind cut with which I'm working:
A subset $L \subset \mathbb Q$ is a Dedekind cut if
(I) $L$ is proper (i.e. $L \ne \emptyset$ and $L \ne \mathbb Q$);
(II) $L$ has no maximal element; and
(III) $\forall a,b \in \mathbb Q$ such that $a \lt b$, $b \in L \implies a \in L$.
For (a): I don't think this is a Dedekind cut because it can't include any negative numbers, but that's not technically one of the conditions, so perhaps that's not correct? It seems to satisfy the three conditions, though.
For (b): Likewise, this one seems to meet all the criteria, as well. I'm sure one of these must not be a Dedekind cut, though...
If someone could point me in the right direction here, it'd be greatly appreciated. Thanks!
 A: Brian Tung's comment has the answer to both:
(a) Condition (III) prevents $\{x^{2} \mid x \in L\}$ being a Dedekind cut for any Dedekind cut $L$.
Every Dedekind cut is an open interval $(-\infty,b)$ for some $b$. In particular this interval definitely contains some negative rationals. For example, if $b > 0$, we must have, by condition (III), $-b \in L$ as $-b < b$ and if $b \leq 0$, we can similar apply (III) and state $b-1 \in L$.
However $\{x^{2} \mid x \in L\}$, regardless of which cut $L$ is, contains no negative numbers.
(b) Again, we can give a cut for which (III) fails (and again, Brian Tung's suggestion is probably the clearest). Let $L = \{x \in \mathbb{Q} \mid x < 1\}$. In particular consider what happens to the rationals in the interval $(0,1)$ when we invert them: for every $y \in (0,1)$ we have $\frac{1}{y} > 1$. Then the set $L' = \{\frac{1}{x} \mid x\in L\} \cup \{0\}$ is $(-\infty,0) \cup (1,\infty)$. That is, there's a gap between $0$ and $1$, so we again violate (III). Pick any $y \in L'$ where $y >1$. Then by (III) we must have that $\frac{1}{2} \in L'$ as $\frac{1}{2} < y$, giving a contradiction.
