Indeed, suppose three are collinear (wlog the first three). That is, suppose PQA, PQB, and PQC are each collinear triples. Then ABC are collinear, contradicting Ax 5
Yeah
Wait I noticed a snag
I want to guarantee the existence of a plane containing P and Q which is not Sigma
I want to avoid PQA collinear, PQB collinear, PQC det. Sigma, PQD det. Sigma
(det = 'determine', Ax 2)
but I feel like I would have to assume the conclusion to show that that configuration is impossible
@Fargle Ah, I was overcomplicating things, the accepted answer is pretty simple lol
Yeah, the trick of letting T be more arbitrary does a lot of work here.
I think to be careful as possible, one would be best suited to consider the cases A in PQ, A not in PQ with extreme caution, but I believe it all works out.
In fact, if we can assume that then we can do it much quicker: Suppose $\overline{PQ}$ is not contained in $\Sigma$. Then there is a point $A$ in $\overline{PQ}$ not in $\Sigma$. By mystery assumption, $P,Q,A$ are contained in a plane $T$; since $T$ contains $A$ and $\Sigma$ doesn't, $T\ne\Sigma$, so by your proof in the OP $T\cap\Sigma$ is a line containing $PQ$ and so must be $\overline{PQ}$.
@leslietownes as you said "it's in definition 1.7. from this you can deduce that a lower bound of B, a subset of an ordered set S, is among other things an element of S." I wonder how do you know it's subset of same set S? I know it has order but how about lub property?
imagine rudin wrote a line at the beginning of those two or three pages of discussion along the lines of: in the following, the ordered set S will be fixed throughout
if you're worrying about order theoretic properties of subsets depending on which ordered sets you consider them to be part of, that's a valid concern, but it's definitely not one that rudin wants you to consider in that discussion, where everything is a subset of the same ordered set S
because i've read the rest of chapter 1, and several books on analysis and adjacent topics in order theory, and can see what he's getting at. it's also the interpretation you need for the notion to make sense.
if your brain feels like it's throwing the equivalent of a 'divide by zero' exception when understanding a definition, vs. there's some simple interpretation that makes it go away, it's safe to ignore the exception.
different authors pay varying degrees of attention to how carefully they use english and symbols in ways that are completely unambiguous, vs. only unambiguous if viewed in context. rudin is actually pretty good at this. what he's also good at is the formally unambiguous but completely inscrutable exposition of mathematical ideas.
ahlfors' complex analysis book is a book that is 'too far over the line' into informality for my taste. there are a lot of natural assumptions that are unstated and a number of concepts and definitions are summarized in paragraphs instead of the formalism i would expect. but it's still a better book than some more recent and formal books.
TND stands for "tertium non datur", Latin for "there's no third option" (literally "a third is not given"). It outputs $P\lor(P\rightarrow\bot)$ for any $P$
Is there any good source to practice smooth manifold theory? (Problem set for example) Many of the exercise problems in Lee's book are about calculations.
Proving the theorems in the book is a good exercise.
Corrected question: Is there any $r$ such that there is a FLT that maps the annulus ${1\over 2}<|z|<1$ to the region bounded by $|z-1/4| = 1/4$ and $|z|$?
@Peter I am not promoting drug use, however, I don't think we understand the operation of the brain, especially creativity, well enough to make such a statement.
I know that taking a shower or a long walk boost my creativity.
a similar reasoning applies, if you have objects with parameters $l_1,....,l_n$ and $k_1,l_2,...,l_n$ the you expect the volume of the sum (disjoint union) to be $l_1+k_1,l_2,...,l_n$.
@Shin In my early lectures, I showed explicitly how to get the alternating multilinear property for signed area of parallelograms. The key thing from my perspective is that $T(v,w+cv)=T(v,w)$ is Cavalieri's principle.
@XanderHenderson It depends on the field, I think. I did hear a story that one works on Fourier–Mukai transforms in algebraic geometry without knowing what Fourier transform is on the real line.
The only place I'm not using orientation is $T(v,w+cv)=T(v,w)$.
I think that to get $T(v+v',w)=T(v,w)+T(v',w)$ you already need the signed area. There are nice pictures for this, but you might run into orientations right there.
That is only part of it, Shin. $T(cv,w)=cT(v,w)$ is "obvious" when $c>0$ and needs the sign for $c<0$. So $T(v,w+cv)=T(v,w)$ follows from $T(v,w+w') = T(v,w)+T(v,w')$ once we show $T(v,v)=0$ (that's where alternating comes in). Can we go backwards here?
So I need to know a little bit of linear algebra to go backwards here. I need to know that when $v,w$ are linearly independent, any $w'$ can be written as a linear combination of $v$ and $w$. Then $T(v,w+w') = T(v,w+(cv+dw)) = T(v,(1+d)w) = T(v,w) + T(v,dw) = T(v,w)+T(v,w')$.
Yeah I know, basically the det remains unchanged. It’s slightly different from multilinearty cause we only require that the det is equal under replacement of columns
the geodesic compactification of the symmetric space $\Bbb R^3$ is denoted as $\Bbb R^3 \cup \Bbb R^3(\infty).$ It's constructed by adding limit points to certain geodesics in $\Bbb R^3.$ How do you determine these "certain geodesics?"
Our definition: A sequence $\{f_k\}$ converges to zero in $\mathcal S(\mathbb R^n)$ if for all polynomial $P(x)$ and all differential operator $L$ with constant coefficients, the sequence $\{P(\cdot)L(f_k(\cdot))\}$ converges uniformly to $0$ in $\mathbb R^n$.
I want to see how $\mathcal D(\mathbb R^n)\hookrightarrow \mathcal S(\mathbb R^n)$
@geocalc You need to specify how it's a symmetric space. You're talking about geodesics. With respect to what metric? I assume this is not the usual flat metric.
Well, to prove continuity you want to fix $f_k\in \mathcal{D}_K$ for some compact $K$ such that $f_k\to 0$ uniformly together with all of its derivatives
And then prove that it converges to 0 in the Schwartz space
Basically, an operator $T:\mathcal{D}(\mathbb{R}^n)\to X$ is continuous iff $T\restriction\mathcal{D}_K$ is continuous for every compact $K$
@PNDas the space of smooth functions with support contained in $K$, equipped with a family of seminorms which correspond to convergence of all of the derivatives
The inclusion can never be an isomorphism onto its image, though, since $\mathcal{D}(\mathbb{R}^n)$ is not even metrizable
That's why they don't introduce the topology on it, probably. On all things linear it acts nicely, but take anything non-linear and you'll probably have a bad time
I've got what's probably a simple question which I can't seem to see my way out of. A module $M$ is called finitely related if there is a free module $F$ and surjective homomorphism $f:F\to M$ such that the kernel of $f$ is finitely generated. I suspect that the prufer p-group is not finitely related as a $\mathbb Z$ module but I feel at sea trying to prove it.
Yesterday, I found a paper (may be some lecture note I don't remember) online which talks about different spaces etc and related order of a distribution with their colimits or something. I just skimmed through it. But I was reading the paper in incognito mode and by mistake, I deleted the whole incognito window. I never found that paper again.
@TedShifrin I was thinking of selecting some straight line geodesics using the usual flat metric and adding 2 limit points but I didn't know how to choose them
I use incognito mode almost every time I need to search something. I don't even know why I do this. I always clear history and cache data after I search something in normal mode.
It may be because, I used to use a phone with only 1GB storage. I always get a storage full warning. So I started clearing cached data of every app after I use them.
isn't every finitely related module just a direct sum of a finitely presented and a free one?
cause if $F\rightarrow M$ is our epimorphism and $K$ the kernel, pick a basis for $F$. since $K$ is f.g., finitely many basis elements will suffice to generate all elements of $K$, so $K\subseteq F^{\prime}$ with $F^{\prime}$. since $F^{\prime}$ was chosen generated by a subset of a basis, it has a free complement $F^{\prime\prime}$, and we obtain $M=F^{\prime}/K\oplus F^{\prime\prime}$, the former is f.p. and the latter free
@Thorgott I bought your argument, so I started searching on the web to see if it's a route thing and hit this so I think you're right! It's something I didn't know...
@Thorgott but the prufer p group is indecomposable, not free and not finitely presented. So I guess that does it!
We’ll, with a little more work…
You just say one of the halves has to be zero, and the remaining one is indecomposable. That leaves an indecomposable fg module or Z, neither one a valid option.
@Thorgott Actually, I think we can say simply any indecomposable finitely related module is finitely presented. Starting from where we left off, it is either the finitely presented piece, or else it is a summand of $R_R$, but each summand of $R_R$ is clearly finitely presented ($(1-e)R\to R\to eR$ given an appropriate idempotent.) Do you agree?
I don't think we need indecomposability. The Prüfer $p$-group is torsion, so the free summand can be only $0$. I.e. if it were finitely related, it would be finitely presented, hence finitely generated. But a finitely generated torsion group is finite (only for abelian groups, of course), and the Prüfer $p$-group is infinite.
Let $M$ be $[0,1] \times S^1$ accumulating to 2 points. What metric can I put on $M?$ I was thinking product metric but not sure if that will work given the 2 limit points in play...
twice pointed sphere though because @XanderHenderson the two limit points are singular. I see what you're saying though... when I was trying to explain better I changed what i was saying a little
@geocalc33 I don't understand. You have given a construction for a topological space (a cylinder, with each boundary circle identified to a point). This space is a sphere (or, at least, homeomorphic to a sphere). You then asked what metrics you could place on that space. "Pointedness" assumes additional structure which you have not explained.
Ugh... COVID forced us to move away from placement exams (which I am not entirely opposed to), and into a model of "directed self-placement". However, one of our high schools has taken that to mean "We can place our students into any class we like!"
So we have about a dozen students who have taken AP Calculus, and are asking to be placed into "College Mathematics".
Which isn't even a transferable course. It is, essentially, remediation with a fancy name. The universities will take it as a math elective, but it doesn't apply to any major.
Also, I am salty that this high school is teaching AP Calc, when those students could take calculus from either myself or one of my colleagues, and actually earn real, honest-to-goodness college credit.
@rschwieb it can't be a summand of $R$, right? the module has to be either the f.p. piece or the free piece, but if it's the latter, indecomposability would force that $R$ is indecomposable over itself, i.e. that it has no idempotents, right?
But physics is required for high school graduation.
@leslietownes I have one in my back yard. It is about five feet tall, feeble, and rarely produces fruit.
The lack of qualified physics instructors in the area has made me seriously consider applying to graduate programs in physics, where, if my application is successful, I can take the requisite 18 hours, then drop out.
@robjohn @Xander @leslie Would anyone more of an analyst than I care to reassure this OP that iterated integrals (in the context of Fubini) can be over something other than a fixed interval?
@Xander I don't know any state where physics (as opposed to some physical science) is required for high school graduation.
