@TedShifrin Pardon for the bad usage of the term "two variable". I mean its image is in R^2 and the domain is [0,1]. I look the curve as a function of t and take its derivative with regard to any t in the domain. Let me emphasise that this is not necessarily the same as the slope of tangent line. Like the circle (sin(t^2),cos(t^2)) that has tangent line everywhere but (in my definition of) derivative has a (0,0) derivative at t=0.
I mean taking simply the derivative of the components at each point. This is how Pugh defines derivative; from the domain to the Linear transformations, and each linear transformation is represented by a matrix (Here it is a 1x2 matrix)
@TedShifrin So it seems your definition of derivative is different from what is in my mind, and perhaps that's the root of confusion. What you mean by derivative of the curve is having tangent line at each point with slope dy/dx regardless of parametrization?
@shintuku I didn't talk with you at the first place. I thought Ted was ignoring and I sent him a message. You jumped in the middle from nowhere talking about freedom and the ways of the world.
And I never insulted you in the first place. Getting some downvotes (which nevertheless I will remove right now, for you to feel happy) is nothing compared to calling me a monkey or other things.
@shintuku Bro, read my message. Why did you at the first place intervene in a thing which was not related to you? It was an issue between me and Ted, and finished peacefully. Don't continue.
What does "differentiability of a curve" mean? Typically it means someone has given you a parametrization of the curve, and that parametric map is differentiable. I know of no other definition.
The correct notion is that of a "differentiable submanifold".
hm and then again a single line through a random point somewhere on the plane could have a line intersecting it and we wouldn't want to call that tangent
It is bad to be wrong. It is worse to be wrong while agreeing to help someone. But the worst case scenario is to be wrong while agreeing to help someone, and then insulting them when they're actually saying the right thing, or even if they're just confused!
@BalarkaSen OK I read this. Thanks Balarka for clarification. In the second part, when you say "such that U∩C is a graph of a smooth function defined on ℓ∩U" what you mean is considering ℓ as diffeomorphic to an interval on the y-axis or x-axis
@FarhadRouhbakhsh You may choose an identification like so, but you may work abstractly as well. $\ell \cap U$ is a Euclidean line segment on $U$. Choose an orthogonal line $\nu$ to $\ell$ through $p$. $\nu \cap U$ is another Euclidean line segment on $U$. These pair of segments give a coordinate system on $U$, with $\ell \cap U$ as the $x$-axis and $\nu \cap U$ as the $y$-axis. There should be a smooth function $f : \ell \cap U \to \nu \cap U$ such that $C$ is the graph of $f$ over $\ell \cap U$.
(By the way, I'm very happy you said the word "diffeomorphism", because once you confirm that the above is OK I will give you a different, much shorter definition of a smooth submanifold, in terms of that)
@BalarkaSen Perfect. I understood. So, if these two statements are equivalent, then I think in the problem we have assumed the \textbf{first part of} the first statement ,i.e, that the parametrized curve is infinitely differentiable and the higher order derivatives match up at 0 and 1. And then, adding the \textbf{second part}, that is, the regularity condition, we can use statement 2, that says at each point there exists a tangent line. And then of course the solution of the professor works.
One needs to demonstrate an equivalence. Ted is avoiding going into details because this is usually first lecture material on an intro to smooth manifolds course.
@FarhadRouhbakhsh: I re-read Ted's answer; the second half of the answer where he justifies carefully essentially gives a proof. I will leave it to you to verify how.
@BalarkaSen So far so good. Now the point where I get deeply confused is this: suppose we are given a curve $γ(t)$, but this time with a non-regular parametrization. That is, the curve is smooth with this parametrization, and derivatives of every order match up at 0 and 1, but there exists a point like p in [0,1] such that γ'(p) = (0,0). Then what happens? I want to know your comments up to here and after that I ask my other questions in this regard.
@FarhadRouhbakhsh Then you cannot guarantee situation (2). It may or may not happen. You gave the example (t^2, t^3) where it does not happen, there is no "tangent line" over which the little arc of the curve is a graph. But for your other example (cos(t^2), sin(t^2)) it does happen, there is a "tangent line" at (1, 0) over which the little arc of the curve is a graph.
It depends on the geometry of the image of the curve.
@BalarkaSen Ok, you are talking about the existence of the tangent line. What I ask is about the result of the problem. That is, is there an example in which $γ(t)$ is smooth, derivatives match up, but γ'(p) = (0,0) AND that we CAN NOT partition it to finite arcs, each of which are smooth functions of one variable?
@Balarka Obviously, if the tangent line exists at every point, like the case of (cos(t^2), sin(t^2)), the answer is that WE CAN ALWAYS DO THE PARTITION. How about if it doesn't exist in one point?
@FarhadRouhbakhsh: OK, I understand your question now. Yes, there are examples (and Ted may have talked about it before). The immediate thing that comes to mind is a a cuspidal curve, with infinitely many cusps accumulating to one point.
@BalarkaSen I want your comment on the example I had in mind. (Ted said it doesn't work), but I don't understand why.
" For each interval in the x-axis like [1/n, 1/(n+1)], I define my curve as a semi-circle (drawn above the x-axis) with the diameter equal to the length of the interval. The semi-circles keep shrinking up to zero. At the endpoints (where x = 1/n) the curve has cusps, but I think it is also smooth there (not sure about that) because it is where the two tangent semicircles meet.
@BalarkaSen I want to have a smooth parametrization of this curve (the image I sent) such that for all natural number n, the derivative at a point t in [0,1] such that γ(t) =(1/n,0) is equal to (0,0). Is it doable?
Yes. As I said, model each cusp where two adjacent semicircles intersect as $(t^3, t^2)$.
The only caveat is that the norm of the derivatives on each of the semicircles approaching $(0, 0)$ should become smaller and smaller (ie if you imagine a parametrization as a particle travelling on the curve going from $(1, 0)$ to $(0, 0)$ along your semicircular curve, the particle should keep slowing down as you approach $(0, 0)$)
This is to ensure $\lim_{t \to 0} \gamma'(t) = \gamma'(0)$.
@BalarkaSen Pardon for asking, but what do you mean by "modelling each cusp"? Do you mean writing an anylytical definition for it in each interval [1/n, 1/(n+1)], as for example we write "analytically" $(t^3, t^2)$ for for example t in [-1,1] and in the image of the curve we get a cusp at t = 0?
Of course, I am giving you a rough argument telling you that it's possible if you do everything right. It's annoying to actually do the writing; personally, I find the rough argument convincing enough.
yes, thanks for giving me the direction. The point is, I don't know how on the world to parametrize a curve like this. I don't know even how to start. Yeah, analytically it's hard.
But it seems I got my answer, nevertheless. The "character" of the examples are like this, having infinite cusps, regardless of how you write their smooth parametrization. Perhaps it is "easier" to write for the zig zag. But I don't know how to parametrize such that the zig zag stays smooth when changing direction. The reason I chose semi-circles instead of zigzag was that I thought with myself that circles "feel" more smooth.
@BalarkaSen @TedShifrin Thanks for your guidance guys. I learned a lot from you. I wish all of my professors or teacher assistants or even my friends here in Iran were like you too, helping and open to silly questions XD.
breaking the problem into smaller chunks: is there a way to sum or multiply elements of $(1/2)\mathbb Z$ to obtain a rational number with a denominator that is not a power of $2$?
Given a separated morphism $f:X\to Y$ of schemes, is it then true that $X$ is automatically separated?
the book I'm reading (Liu) seems to use this, but without explanation
I guess I could try to see if we have a map $X\times_Y X\to X\times X$ that is a closed immersion
Since by definition of separatedness we have that the diagonal map $X\to X\times_Y X$ is a closed immersion
hm I guess if we look locally, given a morphism of rings $A\to B$, the question is whether the canonical map $B\otimes_\mathbb Z B\to B\otimes_A B$ is surjective
I think it's an isomorphism, so that seems to work?
actually, I don't even need to check this
all I need to check is that the image of the diagonal map $X\to X\times X$ is closed. We know it's closed in $X\times_Y X$ by separatedness of $f$. So I guess I should argue that $X\times_Y X\to X\times X$ is a closed map topologically
Actually its not clear to me why it's closed either, because fibered product of schemes is different from the fibered product of the underlying sets IIRC
I think it may not be true without "Y is separated" as hypothesis. What about the identity map X -> X for a non-separated scheme? If there is any truth to the world, the identity map is separated.
But again, $\beta X = \mathrm{mSpec} C(X)$ seems like a fancier reformulation of $\beta X$ as the closure of $X$ embedded in a product of intervals by all the continuous functions on $X$
The extra maximal ideals come from "the points at infinity"
The proof might be involved but it seems very believable once you take the latter as the definition
how I learned you produce compactifications for (separable) metric spaces is so called Wallman compactifications, which is a twist on constructions with ultrafilters
you just specify a basis for closed sets that satisfies extra conditions and consider filter associated to it
in fact, the definition of Stone-Cech compactification I'm using uses filters too
I learnt a bit of nets and filters because my undergrad topology course skipped the proof of Tychonoff, and I didn't really want to read the long proof in Munkres. I learnt it, proved the Tychonoff theorem using nets/filters, was shocked at how easy it is, and then never encountered more about nets/filters ever again.
The next I heard about them was in grad school; apparently one of many components in the proof of a long-standing open problem uses ultrafilters :P
Let $I=[0,1]$ be the closed unit interval. Suppose $f$ is a continuous mapping of $I$ into $I$. Prove that $f(x)=x$ for at least one $x\in I$. Proof. Consider the line $y=x$ in $I \times I$. If the claim was false, then $f(0) > 0$ and $f(1) < 1$. By continuity, the set of points $f(x)>x$ would then be above $y=x$ and the set of points $f(x)<x$ would then be below $y=x$. The only possible point to connect these sets lies in $y=x$, so they cannot be connected. This contradicts connectedness of $f(I)$.
Bottom line, I think point set topology is great. I would like to see more applications of the weird point set topological nonsense being used to study nicer spaces I care about.
Ah, now I remember. I realized that the ultralimit of bounded sequence $x_n$ wrt ultrafilter $\mathcal{U}$ corresponds to the value of continuous extension $\beta\mathbb{N}\to \mathbb{R}$ at the point corresponding to $\mathcal{U}$.
is there a way to sum and multiply elements of $(1/2)\mathbb Z$ to obtain a rational number with a denominator that is not a power of $2$?
answer: no
proof attempt: let $r$ be the result of sums and multiplications between elements of $(1/2)\mathbb Z$. then $r$ is the sum of simple terms $s_i = z_i/2$ with nonsimple terms $s'_i$, which are products of sums of simple terms.
Notice: any sums $\sum s_i$ of simple terms $s_i$ will be equivalent to an irreducible fraction with denominator 1 or 2. Therefore, any products $s'_i$ of $\sum s_i$ will have as a denominator a power of $2$.
Let $T$ be the nonsimple term of $r$ with the greatest denominator, and set all other terms to the same denominator. Then, $r$ is a fraction that has a denomina…
now this is clearly enormous, there has to be a better way of proving this
Is there a way to characterize $\beta X$ in terms of all the continuous maps $\beta X \to Y$ out of it? Are these exactly the same as all set-theoretic maps from $X$ to $Y$?
That can't be right.
$X$ is a subspace of $\beta X$, so that conjecture makes no sense. There must be something else
OK, If $Y$ is compact, then every continuous map $X \to Y$ compactifies to a continuous map $\beta X \to Y$. Conversely, one can restrict.
Bit bizarre lol. $\mathrm{Hom}(\beta X, Y) = \mathrm{Hom}(X, Y)$??
$\beta X$ is the unique compact Hausdorff space in which $X$ is densely embedded, and such that any continuous $X\to Y$ into compact Hausdorff space $Y$ has (unique) extension $\beta X\to Y$
A map from $\beta X$ to a non-compact space can be made into a map from $\beta X$ to compact space by considering image of $\beta X$, is that what you mean? I'm not sure
is there any circumstance where analytically computing the derivative of a function is easier to do by hand than by computer?
i am trying to understand what result automatic differentiation gets you; sources online make it sound like automatic differentiation yields the exact derivative, but I guess I am skeptical of this
We have points such that $f(\alpha_x) = \min f(x)$ and $f(\beta_x) = \max f(x)$. Consider ${[\alpha_x, \beta_x] \times [\alpha, \beta]}$. The intersection of this set with the line $y=x$ is not empty. By IVT, we know that $f([\alpha_x, \beta_x])$ is in $[\alpha, \beta]$ and is connected. From the intersection, there is a point $(x,f(x))$ lying in $y=x$.
Is it better? I suspect, no. Still seems muddled.
What was the thing with Mobius strip comment? Was it related to my proposed orientations of my sets? I guess, you meant that "what if we twist that unit square"?
