1. ## Cuts

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

(1) A is bounded above
(2) A has no maximum (least upper bound)
(3) A is closed downwards
: $x \in A$ And $y \leq A \Rightarrow y \in A$

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 the union of two cuts is always a cut.

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

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

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

(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 $\mathbb{R}$ of real numbers is defined as the set of all cuts. Prove that $\mathbb{R}$ is complete.

(Jason: I know I have to use the previous part some how. Every subset of $\mathbb{R}$ 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 $\mathbb{R}$? And if so, how are they cuts?!)

(e) For a rational number r with the following definition

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

I need to prove that $\overline{r}$ is a cut.

(f) Let $\mathbb{Q}^+$ denote the set of all positive rational numbers. Describe the intersection:

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

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 that every set $A \subseteq \mathbb{R}$ that has a lower bound, also has a greatest lower bound. (Jason: here I think because $\mathbb{R}$ is complete and then take the negative of all cuts?)
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Help on any of this is much appreciated.

2. Originally Posted by Jason Bourne
I assume you are all familiar with the idea of a "cut": A non empty subset A of $\mathbb{Q}$ with the following properties:

(1) A is bounded above
(2) A has no maximum (least upper bound)
(3) A is closed downwards
: $x \in A$ And $y \leq A \Rightarrow y \in A$

I'm not sure with the following things:

(a) Prove that every cut is "dense" (Jason: just comes out of the closed downwards property?)
What do you mean by "dense" here? I am familar with the property of being dense in some set but what set do you intend here? No the set of real numbers or rational numbers- that's not true.

(b) Prove that the union of two cuts is always a cut.

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

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

is either a cut or else is equal to $\mathbb{Q}$.
It might help to prove first: If A and B are cuts then one and only one is true: A= B, A is a proper subset of B, or B is a proper subset of A.

[/quote](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 $\mathbb{R}$ of real numbers is defined as the set of all cuts. Prove that $\mathbb{R}$ is complete.

(Jason: I know I have to use the previous part some how. Every subset of $\mathbb{R}$ 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 $\mathbb{R}$? And if so, how are they cuts?!)p[./qu0te]
If A is such a set of cuts, then, as above, the union of all cuts in A is itself a cut. (You need the upper bound on A to show this union is not the set of all rational numbers. Show that that union is the least upper bound.

(e) For a rational number r with the following definition

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

I need to prove that $\overline{r}$ is a cut.
Well, that's straight forward. Just show that the three properties you listed originally are true for this set.

(f) Let $\mathbb{Q}^+$ denote the set of all positive rational numbers. Describe the intersection:

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

is it a cut? (Jason: I thought this might be something similar to the empty set? In which case it's not a cut.)
No, it's not the empty set. because r> 0, and $\overline{r}$ contains every rational number less than 0, the intersection contains every rational number less than 0.

(g) Prove that every set $A \subseteq \mathbb{R}$ that has a lower bound, also has a greatest lower bound. (Jason: here I think because $\mathbb{R}$ is complete and then take the negative of all cuts?)
Yes. If a is a lower bound for set A then -a is an upper bound for {-x| x in A}.
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Help on any of this is much appreciated.

3. Firstly, thank you very much for taking time to reply.

What do you mean by "dense" here? I am familar with the property of being dense in some set but what set do you intend here? No the set of real numbers or rational numbers- that's not true.
A totally ordered set X is called dense if, for all $x,y \in X$ with $x < y$ there exists $z \in X$ such that $x < z < y$. From this definition, the set $\mathbb{R}$ and $\mathbb{Q}$ would be dense. So does this mean that every cut is dense because every cut is closed downwards and every cut is composed of rationals(which are dense)?

It might help to prove first: If A and B are cuts then one and only one is true: A= B, A is a proper subset of B, or B is a proper subset of A.
Thanks, so I guess because every set is either a proper subset of one or the other, it follows immediately that the union of the two sets must satisfy the conditions of being a cut.

I'm not sure how $U = \bigcup_{i \in I} A_i = \{x : x \in A_i for some i \in I\}$ could be equal to $\mathbb{Q}$. If the set $\{ A_i : i \in I \}$ is NOT bounded above, then $U=\mathbb{Q}$? So is the union of a finite number of cuts is a cut, but the union of an infinite number of cuts is infact the set of rationals?

But here I would have to prove that the set
$U$ satisfies the three conditions to be a cut?

Show that that union is the least upper bound.
I think I understand what you are saying (this stuff does my head in ). The union must be the least upper bound for the set of cuts $\{ A_i : i \in I \}$, because otherwise you would have $\mathbb{Q}$. $\{ A_i : i \in I \} \subseteq \mathbb{R}$ and so $\mathbb{R}$ is complete.

How would I prove that $\overline{r}$ has no maximum? Because the set of rationals is dense?