Cauchy sequence... I need to prove 03 and 04.

Question

usukidoll · Answer

Definition of a Cauchy sequence.\

$[a_n]$ is a cauchy sequence if $ \forall 3 >0 \exists N 3 i: \mid a_n-a_m \mid < E \forall n,m \geq N$\

Equivalence classes def and how to work with them\

-Axioms for ordered field \

(01) $x > y $ or $ y >x$ or $x=y$ for all $x,y \in F$\
(02) $ x >y$ and $y > z \rightarrow x > z $ transitivity \
(03) $ x > y \rightarrow x+z > y+z$\
(04) $ x > y$ and $ z >0 \rightarrow xz>yz$ \

usukidoll · Answer

@KingGeorge

kinggeorge · Answer

I'm a bit confused about what you need to prove, and what you're assuming. Do you need to show that the equivalence classes of limits of Cauchy sequences satisfy those 4 axioms?

usukidoll · Answer

so here's the first one... 

01) $ x>y$ or $ y>x $ or $x=y$ for all $ x,y \in F$\

Let $[a_n]$ and $[b_n]$ be Cauchy sequences. Then $[(a_n)-(b_n)]$ could converge to zero in which case $[(a_n) = [b_n]$ or $[a_n-b_n]$ could be bounded > E > 0 for some N on. \

Then we said $[a_n-b_n] > 0$ and defined this as $[a_n] > [b_n]$ or $[a_n-b_n]$ could be less than E, E > 0 for some N on \

usukidoll · Answer

yes the equivalence classes ^^

usukidoll · Answer

I'm still typing my notes..

usukidoll · Answer

01 and 02 were given and I attempted 03 and 04 and then I got lost a bit

usukidoll · Answer

this is the 02 proof

(02) $ x>y$ and $y > z \rightarrow x > z $ transitivity \

Suppose we let $ x =[a_n], y = [b_n], and z =[c_n]$\

$ x >y $ means $ a_n-b_n >E>0$ for some N on.\

$y > z $ means $b_n-c_n >E>0$ from some $\bar{N}$  on.\

$ x > z $ means $ a_n-c_n>E>0$ from some N on\

Suppose we have $\bar{E} =$ min $[E, \bar{E}]$ and max$ [N, \bar{N}$\

Then $a_n - b_n \bar{E}$ and $b_n-c_n > \bar{E}$ for all $ n > \bar{N}$\

$a_n-c_n > \bar{E}$ for all $n > \bar{N}$ \

So, $[a_n]>[c_n]$ which makes $ x > z$\

usukidoll · Answer

So the third one is $$x > y \rightarrow x + z > y +z$$

but all I can get from this so far is that if we let $$x = [a_n]$$, $$y = [b_n]$$ and $$z = [c_n]$$

then I need to prove that $$[a_n]>[b_n] \rightarrow [a_n]+[c_n ] > [b_n]+[c_n]$$

usukidoll · Answer

so for the $$x > y$$ that would mean that there exists a $$a_n-b_n >E>0$$ from some N on .. by the way that E is epsilon

kinggeorge · Answer

Right. So $[a_n]>[b_n]$ implies that for $n>N$ for $N$ large enough, $a_n-b_n>\epsilon$. We want to prove that for $M$ large enough, then for all $m>M$ we have$$(a_m+c_m)-(b_m+c_m)>\epsilon$$

usukidoll · Answer

O_o what the heck where did you get the +'s?!

usukidoll · Answer

oh so we need an M like ummm $$b_n-c_n?$$

usukidoll · Answer

oh hey long time no see @Alchemista

kinggeorge · Answer

The pluses come from $x+z=[a_n]+[c_n]=[a_n+c_n]$.

And in this case, I think you can just take $M=N$. To see that, all you need to do, is add and subtract $c_m$. We have$$a_n-b_n>\epsilon$$so for any $c_m$, this means that$$a_n-b_n=a_n-b_n+c_n-c_n>\epsilon$$Rearranging the terms gives$$(a_n+c_n)-(b_n+c_n)>\epsilon$$for $n$ large enough

usukidoll · Answer

for the fourth one since x> y there exists a $$a_n-b_n >E>0 $$for some N on . and since z> 0 that would mean ... well assuming that it's a Cauchy sequences such that z = 0 and it converges to zero. 

but that's all I"m getting for 

$$x > y $$ and $$z>0 \rightarrow xz>yz$$

usukidoll · Answer

so... there's some multiplication.. if we let $$x=a_n,y=b_n,z=c_n$$

usukidoll · Answer

hmmm let's see what if I have 
$$[a_n-b_n] c_n $$ 
$$[a_n c_n -b_n c_n ]$$ (by distributive law)
$$[xy-yz]$$...........

usukidoll · Answer

is this even right? how do I get rid of the subtraction sign?!

anonymous · Answer

That is the right idea.. I think.
$a_n = x$
$b_n = y$
$c_n = z$

$a_nc_n - b_nc_n$
$c_n(a_n - b_n)$

If $c_n$ coverges to something greater than zero and so does $a_n - b_n$ then 
$a_nc_n - b_nc_n$ does as well.

kinggeorge · Answer

That looks basically right to me.

anonymous · Answer

Just out of curiosity. Is this a real analysis course and you are discussing the completion of metric spaces?

usukidoll · Answer

O____________________O! wow I can't believe my attempt was right XD
@Alchemista it's an Intro to Advanced Mathematics course... supposedly it's a class designed on how to master proof writing

kinggeorge · Answer

This looks like one of the constructions of the real numbers to me. Probably one of the simplest.

usukidoll · Answer

remember last year with my elementary linear algebra proof and I had no idea what was going on? I didn't take the proof writing course.. until now.. I mean I think it sort of works but it depends on the professor who sucks at the moment... you know something is up when everyone failed the two midterms.

usukidoll · Answer

The only thing right now that is helping me out of that course is the easy proofs. If there's like 1 definition involved I can handle it, but if there is a problem with 3 or 4 definitions... then I'm screwed.

anonymous · Answer

You can use the technique to complete arbitrary metric spaces.
I think a one more specific to the real numbers is the dedekind cut method

anonymous · Answer

http://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_by_Dedekind_cuts

usukidoll · Answer

I so need more time to adjust to this...one semester is obviously not enough...

usukidoll · Answer

but I'm able to do the easy proofs now... just got to practice and understand more on the medium and hard proofs

anonymous · Answer

I think the best approach is to try and understand whatever mathematical object(s) you are working with as best as you can. When you understand something well enough coming up with proofs about the properties of the objects you are working with come naturally.

usukidoll · Answer

true... unfortunately, I only know 5 out of the 8 chapters... I am forever stuck on relations, cardinality, and functions (the second half of the chapter x.x)

usukidoll · Answer

now I just got to figure out how to write this. x.x because a symbols only proof is a bad idea.

usukidoll · Answer

(04) $ x >y$ and $ z >0 \rightarrow xz>yz$\

We are going to use the Cauchy Sequence theorem which states that the terms of the sequence are arbitrarily close to one another. Given $E >0$, there exists $N$ such that if $m,n > N$, then $ \mid a_m-a_n \mid < E$.\

Suppose we have $x>y$. We let $ x =[a_n]$ and $y =[b_n]$ be a Cauchy sequence, so there exists a $ a_n - b_n >E>0$ from some $N$ on. For $ z > 0$, we let $z=[c_n]$ and be a Cauchy sequence. Then $z=0$ which converges to zero.\

If $c_n$ converges to something greater than zero, then $a_n-b_n$ and $a_nc_n-b_nc_n$ converges as well\

$a_n > b_n$ and $c_n>0$\
$[\mid a_n - b_n \mid]c_n$\
$[\mid a_nc_n - b_nc_n \mid]$\
$[\mid xz - yz \mid]$\
We have proven that $ x >y$ and $z >0 \rightarrow xz>yz$\

usukidoll · Answer

does that sound right @KingGeorge

kinggeorge · Answer

That looks mostly alright to me. Those last three sequences should all be greater than 0 correct?

usukidoll · Answer

you mean after the c_n > 0 line?

usukidoll · Answer

$a_n > b_n$ and $c_n>0$\
$[\mid a_n - b_n \mid]c_n>0$\
$[\mid a_nc_n - b_nc_n \mid>0]$\
$[\mid xz - yz \mid]>0$\

usukidoll · Answer

???????

kinggeorge · Answer

Yes. That seems to make more sense to me.

usukidoll · Answer

I'm just going to attempt to type the proof for 03... so far I got

(03) $ x>y \rightarrow x+z > y+z$\

We are going to use the Cauchy Sequence theorem which states that the terms of the sequence are arbitrarily close to one another. Given $E >0$, there exists $N$ such that if $m,n > N$, then $ \mid a_m-a_n \mid < E$.\

Let $ x =[a_n], y =[b_n]$ and $ z = [c_n]$ be Cauchy sequences. Then, for $ x >y$ there exists a $a_n-b_n >E>0$ from some $N$ on. \

kinggeorge · Answer

Looks good so far.

usukidoll · Answer

(03) $ x>y \rightarrow x+z > y+z$\

We are going to use the Cauchy Sequence theorem which states that the terms of the sequence are arbitrarily close to one another. Given $E >0$, there exists $N$ such that if $m,n > N$, then $ \mid a_m-a_n \mid < E$.\

Let $ x =[a_n], y =[b_n]$ and $ z = [c_n]$ be Cauchy sequences. Then, for $ x >y$ there exists a $a_n-b_n >E>0$ from some $N$ on. We need to prove that there is some $M$ large enough so for all $m > M$ we have $(a_m+c_m) -(b_m+c_m) > E.$. Now suppose $M=N$. Then,
$a_n-b_n > E$, and for any $c_n$, we have 

$a_n-b_n=a_n-b_n+c_n-c_n >E$\

Rearranging the terms gives us \

$(a_n+c_n)-(b_n+c_n) > E$\
for n large enough. 

For m large enough, we have$a_m-b_m > E$\. Also, for any $c_m$, we have 
\

$a_m-b_m=a_m-b_m+c_m-c_m >E$\

and rearranging the terms gives us $(a_m+c_m)-(b_m+c_m) > E$\

We have proven that 
$ x>y \rightarrow x+z > y+z$\

kinggeorge · Answer

That looks decent enough.

usukidoll · Answer

yay ... would you take over my class? my professor is a total dxxxxe... like seriously no one in my class got anything right during today's lecture

kinggeorge · Answer

To be totally honest, I probably wouldn't be the best. I'm not super great at writing proofs. I do know a few people that are though.

usukidoll · Answer

oh x.x but it seems that I'm learning more about the material with you than some over 60 year old tenure abuser bloke.

kinggeorge · Answer

A good place to learn proof writing, is simply from reading other people's proofs. This may be from a book, a paper, or even just solutions online.

usukidoll · Answer

ah.. k .. I have one more... I was wondering if you could make heads or tails on this one http://math.stackexchange.com/questions/751410/prove-f-infty-a-infty-rightarrow-b-infty-is-a-bijection

kinggeorge · Answer

Some of those things look scary...

What are the definitions of $A_\infty$, $B_\infty$, and $f_\infty$?

usukidoll · Answer

errrrrrrrr I lost my notes so I have no idea... but I need to get some revision out there x.x

usukidoll · Answer

I do remember this though|dw:1398918262698:dw|