I assume you are all familiar with the idea of a "cut": A non empty subset A of $\displaystyle \mathbb{Q}$ with the following properties:

(1) A is bounded above: $\displaystyle x \in A$ And $\displaystyle y \leq A \Rightarrow y \in A$

(2) A has no maximum (least upper bound)

(3) A is closed downwards

I'm not sure with the following things:

(a)Prove that every cut is "dense" (Jason: just comes out of the closed downwards property?)

(b)Prove that theunionof two cuts is always a cut.

(c)Let $\displaystyle \{ A_i : i \in I \}$ be any collection of cuts. Prove that their union

$\displaystyle U = \bigcup_{i \in I} A_i = \{x : x \in A_i for some i \in I\} $

is either a cut or else is equal to $\displaystyle \mathbb{Q}$.

(d)[Define what it means for a linearly ordered set to be "complete". (I know this,if every subset of X which is bounded above has the least upper bound, X is complete) ]

Recall that the set $\displaystyle \mathbb{R}$ of real numbers isdefined as the set of all cuts. Prove that $\displaystyle \mathbb{R}$ is complete.

(Jason: I know I have to use the previous part some how. Every subset of $\displaystyle \mathbb{R}$ is a set of cuts and these are all bounded above, so is the proof just that these cuts have the least upper bound in $\displaystyle \mathbb{R}$? And if so, how are they cuts?!)

(e)For a rational number r with the followingdefinition

$\displaystyle \overline{r} = \{x \in \mathbb{Q} : x < r\}$

I need to prove that $\displaystyle \overline{r}$ is acut.

(f)Let $\displaystyle \mathbb{Q}^+$ denotethe set of all positive rational numbers. Describe theintersection:

$\displaystyle \bigcap_{r \in \mathbb{Q}^+} \overline{r} $

is it a cut? (Jason: I thought this might be something similar to the empty set? In which case it's not a cut.)

(g)Prove thatevery set$\displaystyle A \subseteq \mathbb{R}$ thathas a lower bound, also has a greatest lower bound. (Jason: here I think because$\displaystyle \mathbb{R}$ iscompleteand then take the negative of allcuts?)

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Help on any of this ismuchappreciated.