I think most of my hatered to it comes from not studying to any math test in school, yet I didn't know geo from outside (uni) so I always got stuck at these questions..
@Studentmath what you're thinking about is probably plane geometry. differential geometry uses calculus and (I think) deals with things like smooth deformations of surfaces. algebraic geometry is a discipline of math invented to sound really fancy and sometimes study algebraic varieties.
Same, I have a professor talking to me if I need, but it utilizes self-study. You are not force-fed everything by a professor. @AlexanderGruber sounds fancy indeed, and yes, plane geometry is what I am talking about.
Also, @Mike , it's not that I think that self-study is better or anything, simply the only real option I had. Won't say that this format doesn't have its benefits (among others employers here sometimes rather people from there)
okay wait: if every homotopy of $h$ restricted to $K^{(m-1)}$ has an image which cannot be linearly extended to an $m$-face, then that means that $K$ must be retractable to $K^{(m-1)}$, which we can assume w.l.o.g. can't happen. i think. right?
76 users currently talking in 40 rooms. Soon it will be 1 person having a meaningful conversation about life, the universe and everything, with himself. Per room of course.
@Studentmath I probably misled you about the upper limit being $\omega^2$. We can certainly reach $\omega^2+n$ and maybe even much higher. I am currently taking another look at it.
@Studentmath I proved that if there is a set of order-type $\alpha$ in $(\mathbb Q,<)$ then there is one of order-type $\omega\cdot\alpha$. The main idea is based on the lemma: $[a,b)\cap\mathbb Q$ is order-isomorphic to $\{x\in\mathbb Q:x\ge 0\}$. Therefore, we can attain every ordinal smaller than $\omega^\omega$.
How is it we know that the rationals are not complete, but the reals are?
There is no least upper bound for the rationals, i.e. we can take the sqrt(2) as an upperbound, and there is no least upper boud.
bound*
But this only happens because we can imagine functions like the sqrt(2), that produce irrational numbers. How do we know there isn't something else we haven't thought of yet? How is that a proof?
Because that set I gave really does reside in $\mathbb Q$. So if I can prove that every subset of $\mathbb R$ has a LUB, then I'm good. There won't be any "holes" like $\sqrt 2$ is for $\mathbb Q$.
@Anthony Suppose there is a set $(X,<)$ such that $\mathbb R\subset X$ and the order agrees with the usual one on $\mathbb R$. Assume we have a bounded set $S=\{r\in\mathbb R:r<x\}$ such that $S$ has no LUB in $\mathbb R$, for some $x\in X$. Then, this would contradict the completeness of $\mathbb R$, since $S\subset\mathbb R$.
I'm going to keep parsing that, but two things- I thought completeness was what we were trying to show, and also, you said r in R, but for q, don't we choose a q outside of q, namely sqrt(2)?
@KarlKronenfeld Well, there are some things that amaze one when first reading about them, that seem mysterious or special, but after some time they become common, trivial.
@Anthony My infinitesimals example was used to show that completing is not the same as adding in elements willynilly. The resulting set $X$ after including the infinitesimals is not complete. In fact, any bounded subset of $\mathbb R$ does not have a LUB in $X$. Completeness is very much about the existence of holes (an internal property to your ordered set) rather than the existence of fillers (an external property).
So I guess I can see that, but I still don't see how this completeness makes sense. For the rationals we use the sqrt(2) as a counterexample, for Real numbers why can't we then use an infinitesimal as a counter example to completeness?
@Anthony Infinitesimals are not actually real numbers. We can say "there exists a number smaller than all positive real numbers", but if we ask "what is the greatest lower bound of the set of positive real numbers", then when we live in the reals, the answer is "0". The GLB exists, and it's 0.
The infinitesimals cause problems because all of a sudden, we've created things in between all positive real numbers and 0. So all of a sudden there is no greatest lower bound. We've destroyed our completeness.
Because $\mathbb Q$ actually has a hole where $\sqrt 2$ is. (I.e. there are sets of rational numbers like I provided above without a least rational upper bound.) On the other hand, $\mathbb R$ does not have a hole where some infinitesimal is by the definition of completeness.
Well, it has to do with the ability to take arbitrary unions of a bounded family of cuts and get a cut back. Then, you just have to establish the connection between $<$ and $\subset$.
That given cut that's bounding the collection from above is not necessarily the LUB.
It really does have to do with the fact that unions are lubs.
@Anthony You always work with bounded sets--it's in the definition of completeness. E.g., there is no new information when I say $\{x\in\mathbb Q:x>0,x^2<2\}$ is bounded above by $1000000$
I'm still confused. Stating there is an upper bounding cut, and that every cut must be less than it, and there those cuts form a cut... Why can't we make the same argument for rationals?
I'm just still confused by the fact that we chose a number outside of Q for a upper bound on Q, but we seem to ignore the possibility of a number outside of R for an upper bound on R.
@PaulEpstein Mainly the fact that if FLT is true, then the Frey curve cannot be rationally parametrized to the modular equation of j invariant. Wiles show that all almost all elliptic curves are modular, i.e., has a modular parametrization. A complete indirect mess.
I never understood the part with Hecke algebras. (i.e., the correction of Wiles' works by Taylor)
