@robjohn I'm not saying we can dispense with contradiction altogether. I dare anyone to give a direct proof that $\sqrt2$ is irrational. I typically do a proof by contradiction for the Lebesgue number lemma, but Munkres gives a direct proof in the second edition of his Topology.
I never remember. Which one is Heine-Borel?
Some of those proofs can proceed by successive bisection.
Random opinion: I think the proof* of the quotient manifold theorem is one of the coolest proofs about smooth manifolds out there (*at least the one in Lee's Smooth Manifolds book)
But I agree that the obvious proof is to suppose not and create a sequence.
If we have a countable open cover $\{U_i\}$ of $X$ and $X$ is contained in no $U_1\cup\dots\cup U_n$, then for each $n$ we pick $x_n\in X$ with $x_n\notin U_1\cup\dots\cup U_n$. The sequence has a convergent subsequence. Contradiction.
(So closed + bounded proves sequential compactness by a successive bisection argument. This is all in my book. :P)
Anyhow, yes, I'm just trying to get Koro to get rid of the redundant contradiction arguments, not solid ones.
I have a complex geometry question. Is there a simple proof that the degree of a polarization on an Abelian Variety is always a perfect square? I know a sheaf cohomology proof that works over general fields, but for the complex numbers there should be something simpler
@TedShifrin For me, the illumination of that fact came with ODE existence & uniqueness proofs. Having specific quantities over which the solution exists (given Lipschitz constant, etc) really show up how the various quantities impact the result. Sort of a loose 'sensitivity analysis'.
So if we have a complete DVR $R$ with finite residue field and uniformiser $\pi$, then by a simple exact sequence argument $R/\pi^n$ is finite for all $n$. So if we take any sequence $a_n$ in $R$, then there is a subsequence $a_{n,1}$ such that the reduction mod $\pi$ is constant. Then we find a subsequence $a_{n,2}$ of that such the reduction mod $\pi^2$ is constant etc. In the end, we take the diagonal subsequence
I don't get lunch for another hour, until our biweekly cleaning person leaves (for which I have to spend the morning cleaning the house, explain that to me)
going over my old solutions I think I didn't fully grasp the concept the first time around, so here is a solution to a simple question. Wanted to know if the reasoning is correct this time:
I have to determine if this is a smooth curve.
$F(x,y) = y^2 - x^3 - x^2 = 0$
$DF(x,y) = [-3x^2 - 2x, 2y]$
The two scenarios where $DF(x,y) = [0,0]$ are:
$x = 0, \frac{2}{3}$ and $y = 0$ or more cleanly $(0,0)$ and $(\frac{2}{3},0)$.
So now I inquire on whether these points are on my surface. $(0,0)$ is on the surface because it satisfies the equation, but $(\frac{2}{3},0)$ does not satisfy my equation so I can conclude it is not on the surface. But since $(0,0)$ is on the surface, this means that the equation $F(x,y)$ cannot be expressed as a $1$ dimensional manifold.
bought a shiny new laptop, but the pen does not work properly. i am loath to return it for something small, but after paying $$$ for it, i am loath to suffer the pen not working. 1st world problems.
dc3rd: the approach looks OK. the curve has a self-intersection at 0. locally in a parametrization it will look like a manifold, but the image will not. i forget this crap. is it an embedding, is it an immersion. i forget the extent to which this matters. i defer to ted.
@dc3rd The failure of the implicit function theorem hypotheses does not guarantee failure to be a manifold. It does indicate "trouble points" where you're not sure. But you need to use the other criterion (graph over one of the axes) we were discussing.
when you say use the criterion of "graph over one of the axes" that would mean I would have to "solve" for my equation explicitly where one variable would be the function of the others?.....
Summary: Given [a,b], a<b. Suppose I is an open covering of [a,b]. We can suppose I to be collection of open intervals. (For every open set in R is a collection of open intervals). Step 1: Finding a d>0 such that if S is any subinterval in [a,b] of length <d then S lies in some member of I. Step 2: Partitioning [a,b] in subintervals (each having length <d). Done. @robjohn
Got it......at this level of mathematics I really got to let go of the notion of "clear cut" answers. Not much is clean anymore....always approxiamtions.
Continuous function's property that: a continuous function on a closed interval attains minimum has been established by that point in the book. @robjohn
@robjohn: The argument is along these lines as I understood it: Suppose that $a'$ is mid-point of an interval (from the cover) that has a as its left end and also suppose that b' is the mid-point of an interval with b as its right end. Define $g:[a',b']\to \mathbb R$ as $g(x)=\sup\{y\in \mathbb R: (x-y,x+y) \text { is in some subinterval of the cover}\}$. Then g is shown to be continuous and positive on $[a',b']$. $g$ attains its minimum say r. Then d is chosen as $\min(2r, 2(a'-a),2(b-b'))$.
Unfortunately, many of my advisees didn't take me very seriously when I told them things like — you can't take three (or more) hard math classes at one time. "But I am behind and I need to graduate." Well, failing more of them won't help.
3
The typical adviser just signed the form and the students suffered for their "independence."
that was my worst advising experience during my postdoc. i told someone, this is too much. i'm not commenting on your abilities, this would be too much for anybody. i was ignored.
about a month later i was dealing with the consequences of someone wanting to drop out of everything.
i ended up negotiating something where someone dropped out of some additional obligations they had as part of some 'honors' program. they got through the regular stuff OK. they just didn't have the time.
completely set up for failure by whoever who had (not) advised them the year before.
I did honors advising of all math majors and also first- and second-year CS and physics majors. I inherited one advisee from a general college adviser one term, found out that she had been misadvised (shock!) and tried to explain it to her as nicely as I could, apologizing for the errors of others. Apparently, I upset her so much that her mother called me up a half hour later and yelled at me for 20 minutes (while I had an office full of students for office hours).
One of many times that helicopter parents tried to bully faculty for doing their job.
How could I ruin her poor little flower's life like that?
Sorry if the previous adviser told her to drop a course that was required for her major. Not my fault.
haha. my wife sometimes deals with that. she has the good fortune of being the chair of her department now. the numerous errors in advising eventually land on her doorstep.
For most faculty, knowing what is in the math major curriculum and what the requirements are is more than they can manage. Adding general education requirements , etc., makes it way too challenging.
This student was a CS major whom I inherited, not a math major. But at the time CS majors had to take through calc 3 and linear algebra. That changed subsequently to ... just one calculus! That infuriated me.
Well, part of good advising is gauging the student's strength(s) with who's teaching the course. If the teacher is a known disaster, it won't go well for a weak student.
Well, the discrete math for CS course is on top of that.
Calculus 1 is required for every science major.
But they took out two calculus and one linear algebra!
I mean, I can understand why business majors would do better with stat than with calculus. But CS? really?
I remember teaching a CS major years ago in my linear algebra class. The student struggled to get a D. But got an A simultaneously in the computer graphics class (um, guess what's used in computer graphics?).
Well, all the linear algebra was a canned black box in the CS course. And I was making students write basic proofs, along with all the stuff on projections, least squares, eigenvalues, diagonalization, etc.
depending on the CS class, maybe you can go a long way cut-pasting from stuff you don't fully understand. if you understand how to reduce problems to sub-problems and find the right code you might not have to do it yourself.
I spent many hours figuring out the linear/projective geometry of computer graphics. I wrote a section on it for our linear algebra book. Sadly, I never got to teach that material. Just no time.
But I was so thrilled when I figured out precisely how Mathematica draws figures from a given viewpoint. :)
a substantial amount of that stuff has been rolled into software libraries where maybe you don't need to know as much as you might have in 2000. and sometimes it's faster if you use out-of-the-box solutions than home brew. still helps to know it.
one of my first mathematical revelations was going nuts when my 3d rotation stuff seemed to distort a cube. i didn't realize something fundamental: the pixels weren't square, so x pixels by x pixels on the screen did not look like a square.
Humm. This does not look right. Fermat's last theorem at youtube, it says "What was the problem? For any three positive integers; a,b,c prove that a^N + b^N = c^N for N greater than 2". Here is the link. At 0:39 youtube.com/… it should be "no three positive integers a, b, and c satisfy the equation a^N + b^N = c^N for any integer value of n greater than 2"
@TedShifrin My first thought is that it is not well-defined. There are a sequence of singularities converging to $z_0$. How many are outside $\gamma_0$.
