# Contruction of the Reals?

1. Sep 26, 2005

### Gale

yes... thats possibly the subject of my homework... anyway, i'm gonna try this...

this is the actual assignment:
Homework page

this are the notes from class:
Notes

i don't have a book to work from all i have are these notes, which i also have written in my notebook with occassional supplements, but not much.

So, i looked at problem 1, i don't know what it means to have B's complement... or to be an interval actually... so i really didn't know where to begin. i know a dedekin cut is like, a point on the number line, and then everything to the left of it? i think... so... i'm not sure what that has to do with anything, except with helping to order the number line...

number two, i think i can do, so i'm gonna work on that while someone maybe helps with 1? but thing is, i've never had a notation class like this before... i vaguely know what all the symbols mean, but the proofs and stuff... i'm really struggling, so any pointers also welcome.

2. Sep 26, 2005

### Gale

actually, i don't really understand how to go about proving any of this stuff at all. i'm looking at two, and i'm not really sure what N=N(epsilon) means... i see it in the notes, and i realize i should just be plugging this function in somewhere and finding the limit... or ya, the limit right? eh... i know it converges to zero... but what do i do with that?

so, i know whatever i do for the first part... i do something similar to show its a cauchy sequence... cept instead of using a limit i use two points in the function or something right? help!

3. Sep 26, 2005

### Tom Mattson

Staff Emeritus
Consider two sets $\mathbb{Q}$ and $B$, such that $B\subset\mathbb{Q}$. Then the complement of $B$ in $\mathbb{Q}$, $B^c$, is the set difference $\mathbb{Q}-B$. That is, $B^c$ is the set of all elements that are in $\mathbb{Q}$ and not in $B$.

The definition of an interval on the rationals is given in the problem statement. What about that definition is giving you heartburn?

4. Sep 26, 2005

### Gale

Oooooook... i've just never done proofs like this before i guess... its weird to me. So, i show that if p is an element of B and q is an element in B as well, and r comes between them... B is an interval... and i know p and q are in B because... if i chose a p in B, then i chose a q less than p, then i know q is also in B and since they're both in b, and ordered, there exists an r between them,... thus its an interval?

and B complement is as well because... i'm not quite sure... actually, i'm a little confused about what it means to be a dedekin cut, and what B actually is.

5. Sep 26, 2005

### Tom Mattson

Staff Emeritus
No. If you have a subset $B$ of the rationals, and you know that $p,q\in\mathbb{Q}$, $p<q$, then to show that $B$ is an interval in $\mathbb{Q}$ you would have to show that all of the rationals between $p$ and $q$ are elements of $B$.

If on the other hand you are told that $B$ is an interval in $\mathbb{Q}$, and you are told the endpoints $p$ and $q$, then you can conclude that any $r$ between $p$ and $q$ is in the interval.

Regarding complements, what about the definition I stated is giving you trouble?

Regarding Dedekind cuts, what definition are you using? (I haven't looked at the class notes you posted).

Regarding B, it's just a subset of the rationals. Do you know the definition of subset?

Last edited: Sep 26, 2005
6. Sep 26, 2005

### Gale

i've rectified my issue with compliments, i thought they had to have a least lower bound, which, they don't. regarding dedekin cuts, my definition was mostly intuitive from discussion in class and i couldn't make sense of the symbols attatched, but i think i get it now. and the definition of a subset is um, a set whose elements are all in some other set... eh?

anyways, regarding question two. i understand the idea, but i don't know how to express it. which i think is my more general problem, is i can't take whats written and compose it into concepts, or take concepts i understand, and write it down the way i'm intended to...

i'm supposed to show that the function converges using the definition of a limit basically? i can't figure out how to do that. do i plug infinity into the function? what do i do with the epsilon?

7. Sep 26, 2005

### Gale

actually, hey, know what? i can't do this stuff. my knowledge of sequences and series and convergence and everything about that is soooo weak. plus i've never formally learned all this notation yet, and i'm real uncomfortable with it. and i've got all the new stuff from class we learned.... there's just no way i can do this.

8. Sep 26, 2005

### Hurkyl

Staff Emeritus
You can do this... it will just take some time. :tongue2:

(But you shouldn't feel pressured to stick to it -- if you are totally lost, don't be afraid to withdraw, but do it soon enough to get a full refund! You can always come back after you have more experience under your belt!)

I find that an "algebraic" definition of things are quite useful for actually proving things. For example, when I see $A \subseteq B$, I think:

If $x \in A$ then $x \in B$.

Which is, of course, the definition... but I think of it more like an "algebraic" rule: If I know that $A \subseteq B$, and I manage to deduce that $x \in A$, then I can write $x \in B$ as my next step.

I've found this way of thinking to be very useful in doing "computation" with the elementary set operations. (e.g. union, intersection, complement, subset, ...)

Actually, I find this useful for just about anything confusing.

For (2), remember that the definition of the "limit of a sequence {an} is L":

For all ε > 0
There exists an N > 0 such that
For all n > N
We have |an - L| < ε

One interpretation I've seen people use of this sort of thing is to think of a game:

On the first move, your opponent picks a positive value for ε.
On the second move, you pick an integer, N.
On the third move, your opponent picks an integer n bigger than N.
You win if |an - L| < ε.

The goal is to find a strategy so that you will always win, no matter what your opponent does. (Or at least prove such a strategy exists)

I don't know if this will help, but I figured I'd throw it out there. This "game" interpretation works for any statement involving phrases like "for all" and "there exists".