Define a Dedekind cut and show the given set is(n't) a Dedekind cut

[docnet]

Homework Statement

a) What is a Dedekind cut? Give the precise definition.

b) Let $L$ be a Dedekind cut. Is $P=\{x^4|x\in L\}$ a Dedekind cut?

c) Let $L$ be a Dedekind cut. Is $P=\{23x|x\in L\}$ a Dedekind cut?

Relevant Equations

$\mathbb{Q}$

a) a subset $L\subset \mathbb{Q}$ is a Dedekind cut if $L$ is proper, $L$ has no maximal element, and
$$\forall a,b\in \mathbb{Q}, [(a<b)\land( b\in L)\Longrightarrow a\in L]$$

b)
Is $P=\{x^4|x\in L\}$ a Dedekind cut?

P is proper:
$$(a\in L)\Longrightarrow (a^4\in P)\Longrightarrow P\neq \emptyset$$ $$(-1\in \mathbb{Q})\land (-1\notin P)\Longrightarrow P\neq \mathbb{Q}$$ So $P$ is a proper set.

P has no maximal element:
$$(\forall x \in L, x<d)\Longrightarrow( \forall x^4 \in P, x^4<d^4)$$ So $P$ has no maximal element.

P does not satisfy the 3rd property:
$$(-1\in \mathbb{Q})\land (a^4 \in P)\land(-1<a^4)\land (-1\notin P)$$ So $P$ is not a Dedekind cut.c)
Is $M=\{23x|x\in L\}$ a Dedekind cut?

M is proper:
$$(x\in L)\Longrightarrow (23x \in M)\Longrightarrow M\neq \emptyset$$ $$(L\neq \mathbb{Q})\Longrightarrow (d\notin L\quad \text{for some}\quad d\in\mathbb{Q})$$ $$\Longrightarrow( \mathbb{Q}\ni 23d\notin M)\Longrightarrow (M\neq \mathbb{Q})$$ So $M$ is a proper set.

M has no maximal element:
$$(\forall x\in L, x<d)\Longrightarrow (\forall 23x \in P, 23x<23d)$$ So $M$ has no maximal element.

M satisfies the 3rd property:
$L$ satisfies the 3rd property, so $$\forall a,b\in \mathbb{Q}, [(a<b)\land( b\in L)\Longrightarrow a\in L]$$ Multiplying the $a$s and $b$s by $23$ leads to $$\forall 23a, 23b\in \mathbb{Q}, [(23a<23b)\land (23b\in M) \Longrightarrow 23a\in M]$$
$M$ is a Dedekind cut.

 

[mathwonk]

You received no response since you asked no question, but if you want checking of your work, then in b, the argument for no maximal element contains a false claim, i.e. one can. have a<b but b^4< a^4, if a and b are negative.

In part c, the argument that the 3rd property holds is phrased incorrectly. namely one should assume that b is in L and a < 23b, then show a is in M.