I'd like to prove that the hyperbolic plane is complete in a short, nice way. Here's my proof, but I'm not convinced of its correctness yet. Let $\mathbb{H}=\left\{(x,y)\in\mathbb{R}^2\big\vert y&gt;0\right\}$. Since this is a connected, Riemannian manifold with infinitesimal metric $dh^2=ds^2/y^2)$ (where $ds$ is the usual Euclidean metric) we can use the Hopf-Rinow theorem and say that completeness is equivalent to having geodesics defined from the whole real line. Hence, it suffices to show that every point $p\in\mathbb{H}$ is at infinite distance from the boundary. Take $q\in\partial\mathbb{H}$. By moving the points via isometries, we can assume that $p$ lies in the positive, pure imaginary, open semi-line and have coordinates $p=(0,\bar{y})$ and that $q$ is the point $\infty$. Then we may finish the proof just computing the integral $$\int_{\bar{y}}^\infty \sqrt{\frac{dy^2}{y^2}}=log(|y|)\Big\vert_{\bar{y}}^\infty=\infty.$$ Is there any wrong step? Do you know any other concise proof of the completeness of the hyperbolic plane?
122231
Consider the upper half plane model of the hyperbolic space ($\mathbb{H}$ with the riemannian metric $g=\frac{dx^2+dy^2}{y^2}$). It is known that $(\mathbb{H},g)$ is geodesically complete, which means that no geodesic can reach the border $\partial \mathbb{H}$ in a finite time.   Why is that?  Of course if I consider particular geodesics such as $t\mapsto (0,e^{-t})$ this is true, but I can't figure out a general proof for this fact which doesn't rely on the particular form of the geodesic considered. My intuition is that it must depend on the fact that, as a geodesic approaches the border, the denominator of $g$ goes to $0$. Can you point me out an explanatory proof which shows that the length of a hyperbolic geodesic which goes to the border can't be finite?    EDIT: Thank you so much for the answers, I realized I didn't make myself clear. What I mean is: is there a way to prove directly for any hyperbolic geodesic (using the fact that $g$ has $y^2$ at the denominator) that its length is infinite?
55910
I need to show that the Hopf map $H : S^3 \to S^2$ is a submersion. There are many ways to define it, but for now I've gone with the restriction of the function $F : \mathbb{R}^4 \to \mathbb{R}^3$ mapping $$(x,y,u,v)\mapsto (2(xu+yv), 2(xv-yu), x^2+y^2-u^2-v^2).$$ It wasn't too bad to show that $F$ is a submersion, but is there a way to go from here to show that $H$ is a submersion? I can write down $H$ as the composition $$S^3 \overset{\iota}{\hookrightarrow} \mathbb{R}^4 \overset{F}{\to} \mathbb{R}^3 \overset{\pi}{\to} S^2,$$ and define $\pi(p) = p/\lVert p\rVert$, but I don't really know where to go after computing the Jacobian of $\pi$, and I don't know if this will even work for my purposes (I don't see how to argue that the composition of the corresponding differentials here isn't rank $1$ — I know $d\iota$ has trivial kernel but things seem to break down once I get to $d\pi$). I'm really just trying to avoid having to compute the Jacobian of $H$ with respect to the stereographic projections and show that that beast of a thing is full rank. Am I on the right track, or is there a better way?
428
Let $p &lt; q &lt; r &lt; s$ be real numbers. Let $l$ be the geodesic with endpoints at $p$ and $q$ and let $m$ be the geodesic with endpoints at $r$ and $s$. (a) Prove that there is a unique geodesic segment from $l$ to $m$ that is perpendicular to both. (We sometimes say $l$ and $m$ are ultraparallel since they are not only parallel but they do not share endpoints also.)  (b) Now suppose $p &lt; q &lt; r$ are real numbers and $l$ is a geodesic whose endpoints are $p$ and $q$ and $m$ is a geodesic whose endpoints are $q$ and $r$. Then $l$ and $m$ are parallel, but prove that there is not geodesic segment from $m$ to $l$ that is perpendicular to both. That is, unlike the ultraparallel case, there are no points on $l$ and $m$ that realize the shortest distance from $l$ to $m$.
48616
Can anyone please tell me whether the following function $d:\mathbb{R}^2 \times \mathbb{R}^2 \to  \mathbb{R}$ is really a metric or not on $\mathbb{R}^2$.  $$ d((x_1,x_2),(y_1,y_2))= \begin{cases} |x_1-y_1|, &amp;\text{ if } x_2=y_2\\ |x_1|+|y_1|+|x_2-y_2|, &amp;\text{ if } x_2 \neq y_2. \end{cases} $$  If it really is indeed a metric, please give the proof of triangle inequality.(I'm OK with the other two conditions).
123187
I am trying to work out the details of following example from page 101 of Tu's An Introduction to Manifolds:     Example 9.3. Let $\Gamma$ be the graph of the function $f(x) =  \sin(1/x)$ on the interval $]0, 1[$, and let $S$ be the union of   $\Gamma$ and the open interval $I=\{(0,y)\in \mathbb{R^2} |−1&lt;y&lt;1\}$.      The subset $S$ of $\mathbb{R^2}$ is not a regular submanifold for the   following reason: if $p$ is in the interval $I$, then there is no   adapted chart containing $p$, since any sufficiently small   neighborhood $U$ of $p$ in $\mathbb{R^2}$ intersects $S$ in infinitely   many components.   Using Tu's definitions, I need to show that given a neighborhood $U$ of $p$, there exists no homeomorphism $\phi$ from $U$ onto an open set of $V \subset\mathbb{R^2}$ with the property that $U \cap S$ is the pre-image with respect to $\phi$ of the $x$ or $y$ axes intersected with $V$.   I am not sure where the fact that $U \cap S$ has infinitely many components comes into play. I tried to derive a contradiction using this fact; but even if restrict my attention to connected neighborhoods $U$ of $p$, the intersection of the connected set $\phi(U)$ with the $x$ or $y$ axes might have infinitely many components (I think), so there's no contradiction with the fact that homeomorphism preserves components.  I would appreciate any help!
61233
I am trying to show that the sphere $S^2$ and $\mathbb{R}^2$ are not homeomorphic.I understand that you can't 'compress' a 3D shape into a 2D plane but I don't know how I would express this formally.  $S^2 = \{(x, y, z) ∈ \mathbb{R}^3: x^2 + y^2 + z^2 = 1\}$  As always, any help is appreciated!
26068
Let $M$ be a smooth, finite-dimensional manifold. Suppose $M$ is also a metric space, with a given distance function $d: M \times M \rightarrow \mathbb{R}_{+}$, which is compatible with the original (manifold) topology on $M$.  Question: is there a Riemannian metric $g$ on $M$ such that the distance $$d_g(p, q) = \inf_{\gamma \in \Omega(p, q)} L_g(\gamma)$$ coincides with $d$?  I believe that this setting is very standard, but for the sake of completeness: $L_g$ denotes the Riemannian length of the curve $\gamma$, and $\Omega(p, q)$ the set of all piecewise smooth curves $\gamma : [a, b] \rightarrow M$ s.t. $\gamma(a) = p$ and $\gamma(b) = q$. Thanks in advance.
65740
Let $M$ be a smooth manifold and let $N$ be a manifold with boundary, both with the same dimension $n$. If $dF_{p}$ is an isomorphism, then $F\left(p\right)\in\mbox{int}N  $.  I am trying to prove this theorem to prove a result about smooth embeddings.  Here is how I am thinking the problem could be solved. Assume thet $F(p)$ is a boundary point of $N$. Then there exists a chart $(V,\psi)$ at $F(p)$ such that $\psi(V) $ is an open subset of the upper half space $\mathbb{H}^{n}$. I guess we have to use some fact about $M$ having no boundary and the the fact that $dF_p$ is an isomorphism to show that there is a contradiction, but I cannot figure that out.
111002
Define the following function on $\mathbb{R}$ by $d(x,y)=\dfrac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$. Prove that this is metric.  I proved the first two properties of metric. But how to prove that $d(x,y)+d(y,z)\geqslant d(x,z)$ for any $x,y,z\in \mathbb{R}$
82848
I always heard the eigenfunctions of a  form a complete basis. Where can I find a proof in infinite dimension space? Presumably readable for physicists.
58439
Someone posted a question on the notice board at my University's library. I've been thinking about it for a while, but fail to see how it is possible. Could someone verify that this is a valid question and point me in the right direction?  'Given an infinite set of points in a plane, if the distance between any two points is an integer, prove that all these points lie on a straight line.'
103809
How am I to "[u]se induction to show that $n$ straight lines in the plane divide the plane into $\frac{n^2+n+2}{2}$ regions"? It is assumed here that no two lines are parallel and that no three lines have a common point. Further, this is not a non-Euclidean problem, but I wouldn't mind a discussion on the non-Euclidean nature of the problem.  I was thinking it might be easier to show using the unit sphere.    A GOOD REFERENCE I FOUND THAT WILL BE USEFUL TO ANYONE COMING ACROSS THIS PROBLEM IS Concrete Mathematics, Graham, Knuth, Patashnik pp. 4–8.
13415
Given the metric space $(X,d)$ with $X := (0,\infty)$ and $d:=|\ln(x)-\ln(y)|$, how can I show that $(X,d)$ is complete?  I need to prove that any Cauchy sequence converges, so:  If $(x_n)$ is a Cauchy sequence in X, then it follows for all $\epsilon \gt 0$ that there exists an $n_0 \in \mathbb{R} : \forall n,m \leq n_0: d(x_n,x_m) \lt \epsilon$.  I couldn't find a direct way to prove this, I guess an indirect approach might go as follows:  Let $(x_n)$ be a Cauchy sequence in X and assume that it does not converge, then it follows that there exists an $\epsilon &gt; 0$ such that for an arbitrary $x \in X : \forall n \in \mathbb{N}: d(x_n - x) \geq \epsilon$ and this would contradict the fact that $(x_n)$ is a cauchy sequence. I'm assuming that this is incorrect since I didn't even use the given metric and this proof would mean that no cauchy sequence converges.  So how can I prove that this metric space is complete? And in general is there a way how to approach these completeness proofs?
38299
I want to prove that the geometric Hahn Banach theorem implies the analytic one.   Edit:  To avoid confusion I will state the vesion of H.B theorems im familiar with:  Analytic H.B:   Let $X$ be a linear space(over $\Bbb R$) and $Y\subset X$ a subspace. Let $p:X \to \Bbb R$ be a sub-additive function. Suppose $f:Y\to \Bbb R$ is a linear map s.t $f(y) \le p(y)$ for all $y\in Y$ then there exists an extention $g:X\to \Bbb R$ of $f$ s.t. $g(x)\le p(x) $ for all $x\in X$.  Geometric H.B:  Let $X$ be a linear space. $K \subset X$ convex s.t. each point in $K$ is an internal point. Let $D$ be a plane disjoint from $K$ then there exists hyperplane that contains $D$ and disjoint from $K$.  In my problem $X$ is NOT a normed space so there are no open sets.   Given $X$ a linear space and $Y\subset X$ a subspace $p:X\to \Bbb R$ sub-additive, and $f:Y\to \Bbb R$ linear s.t $f(y)\le p(y)$ for $y\in Y$ we need to extend $f$ to $g:X\to \Bbb R $ and $g(x)\le p(x)$ fo r all $x\in X$  So, we look at $X \times \Bbb R $and define $K = \{(x,t) : t&gt;p(x)\}$.  The fact that $K$ is convex is easy. How can I show directly that every point in $K$ is internal? (can't say that $K$ is open).  Now after showing that, we can look at $Graph(f)$ and observe that $Graph(f)\cap K = \emptyset$.  So by the geometric H.B theorem we have a hyperplane (which is a maximal subspace in this case because $(0,0)\in Graph(f)$ ) M containing $Graph(f) $ and disjoint from $K$.  Now my problem is to show that each $x\in X$ has a unique $t\in \Bbb R$ s.t. $(x,t) \in M$.  (I need it in order to extend $f$).  I know that if we take $v_0\notin M$ then each $v \in X \times \Bbb R$ has a unique representation as $v = \alpha v_0 + m$ for $m \in M$ , this is because $M$ is a maximal subspace. not sure if that helps.  Thanks for helping!
106549
I'm trying to find the total curvature (or equivalently, rotation index, winding number etc.) of a plane curve (closed plane curves) given by $$\gamma(t)=(\cos(t),\sin(nt)), 0\le t\le 2\pi$$for each positive integer $n$.  Looking at the image of thess curves makes me believe that the answer is $0$ when $n$ is even and $2\pi$ when $n$ is odd, but how do I prove this? Calculating the integral of the curvature is extremely complicated!  I learned in my class that total curvature is invariant under regular homotopy, but how do I construct such a homotopy?
69748
There is something I don't understand about the proof that . The same proof is present in Rudin's Principles of Mathematical Analysis.  Do we assume that our construction of $U_n$ must contain all points of $S$? What if we are only collecting evenly-indexed points of $S$ ($x_{2n}$)? We would still get an infinitely countable subset of $S$, and the rest of $S$ can be used to provide points for $V$. What am I missing?
85442
There's a question in my complex analysis book:     Let $G$ be a region and let $\gamma_0$ and $\gamma_1$ be two closed smooth curves in $G$. Suppose $\gamma_0\sim\gamma_1$ and $\Gamma$ is a homotopy between them. Also suppose that $\gamma_t(s)=\Gamma(s,t)$ is smooth for each $t$. If $w\in \mathbb{C}-G$, define $h(t)=n(\gamma_t;w)$ and show that $h:[0,1]\rightarrow\mathbb{Z}$ is continuous.    This is an exercise right after the section that proves the following version of Cauchy's theorem:     If $\gamma_0$ and $\gamma_1$ are two closed rectifiable curves in $G$ and $\gamma_0 \sim \gamma_1$ then $$\int_{\gamma_{0}}f=\int_{\gamma_1} f$$ for every function $f$ analytic on $G.$   I must be missing something. It seems like you can just say the following:      'Proof:' Since $\frac{1}{z-w}$ is analytic in $G$ ($w$ isn't in $G$), and $\gamma_0\sim\gamma_t\ \forall t\in[0,1]$ (consider a rescaled version of $\Gamma$), then by Cauchy's theorem, $n(\gamma_t;w)=n(\gamma_0;w)=n(\gamma_1;w)$. Thus, the winding number is constant along $t$ and thus certainly continuous.   My problem with that is that this doesn't require the assumption that $\Gamma$ is a smooth homotopy. I think the author had meant for the reader to work harder for this result. What is the proof 'supposed' to look like?