The three dimensional torus  (the periodic domain in Navier-Stokes equations $\mathbb{T}^3=(\mathbb{R}/\mathbb{Z})^3$) is compact.  Can anyone here give us a detailed proof.
123089
I am having difficuties in showing a torus is compact. Initially I wanted to use Heine-Borel theorem, but after that I realise we are not working in $\mathbb{R}^n$ space. So a simple way to show torus is compact is by definition. But after defining an open cover for torus, I don't know how to proceed. Can anyone guide me?
117928
Let $A \in M_n(\bf C)$ and let   $S=\{ A \in M_n(\bf C) : A=A ^*$ and  $\rho (A) \leq 1 \}$, $\rho(A)$ denotes spectral radius of $A$.     Show that $S$ is compact subset of $M_n(\bf C) \cong \bf C^{n^2} $ with usual topology.    $S$ is subset of $\bf O_n(C)$ which is compact so I think its enough to show that $S$ is closed.Please help.
124247
This isn't homework. I have no idea what theorems I should be looking at to solve this. Guidance, partial and total solutions are all welcomed.     Let $f$ be a locally lipschitz function in an open set $G\subseteq \Bbb R^n$. Consider the autonomous system $y'=f(y)$.      Let $y(x,\xi)$ be the value at point $x$ of the maximal solution that satisfies the initial condition $y(0)=\xi$.         Prove that the domain of $y(\cdot, y(s, \xi))$ is $I-s$, where $I$ is the domain of $y(\cdot ,\xi)$.   Prove that for all $s, t$ such that $y(s, \xi)$ and $y(t+s, \xi)$ exist, then $y(t,y(s,\xi))$ also exists and $y(t,y(s,\xi))=y(t+s, \xi)$.   If $y$ is a maximal solution and there exists $T&gt;0$ such that $y(0)=y(T)$ and $f(y(0))\neq 0$, then $y$ is a periodic solution and not constant.   If $y$ is a solution whose domain is $(a,+\infty)$, if $\eta:=\lim _{x\to +\infty}y(x)$ and $\eta \in G$, then $f(\eta)=0$.      EDIT  I found an alternative solution for 3. Please check my proof and give feedback in comments:  let $u$ be the restriction of $y$ to $[0,T]$, now let $\overline u$ be the periodic extension of $u$ to $\mathbb R$. It is easy to see that $\overline u$ is a solution to the given differential equation. But since $f$ is locally lipschitz, $\overline u$ must coincide with $y$ wherever they are both defined. Since $y$ is a maximal solution, it must be $\overline u$, so $y$ is defined on $\mathbb R$.
42854
Please let me know if you find my proof satisfactory.  Let $A\subset B\subset \mathbb{R}^n$, where $\bar{A}$ is the closure of $A$.  If $\vec{x}\in \bar{A}$ then $\vec{x}\in A^\circ$ or $\vec{x}\in ∂A$ (in the boundary of $A$). If $\vec{x}\in A^\circ$ then $\vec{x}\in A \implies \vec{x}\in B\subset \bar{B}$. If $\vec{x} \in ∂A$ then $B_\varepsilon(\vec{x})\cap A\ne \emptyset \ne B_\varepsilon(\vec{x})\cap B$ (for any $\varepsilon &gt; 0$). Thus, since $A\subset B$, $\vec{x}\in B \implies \bar{A}\subset \bar{B}$.
65149
Let $S=\{\frac{1}{n}:n\in\mathbb{Z}\}\cup\{0\}$ be a subset of $\mathbb{R}$. I have to prove using the open cover definition that this is compact. Could you help me, please?
31985
Compute the fundamental group of the space obtained from two tori $S^1✕S^1$ by identifying a circle $S^1✕${$x_0$} in one torus with the corresponding circle $S^1✕${$x_0$} in the other.  Using van Kampen's theorem, I can fairly quickly show that if $T_1$ is one torus and $T_2$ is the other, then the group has to be isomorphic to $((π_1(T_1) ∗ π_1(T_2)))/N ≅ (\mathbb{Z} ✕ \mathbb{Z})∗ (\mathbb{Z} ✕ \mathbb{Z})/N$, where N is generated by elements of the form $i_{12}(w)i_{21}(w)^{-1}$, w is in $\pi_1(S^1✕${$x_0$}$) ≅ \mathbb{Z}$ and $i_{12}$, $i_{21}$ are the maps from $\pi_1(S^1✕${$x_0$}$)$ to $π_1(T_1)$, $π_1(T_2)$ respectively determined by inclusion. My problem is, as it seems to always be with van Kampen-related problems, figuring out what the elements of N look like. My attempt to do this in a pure-algebra way, as far as I can tell, failed me: I tried examining the case where w represents a loop that goes n (some integer) times around the circle, but then it seems like $i_{12}(w)$ &quot;$=$&quot; $(n, 0)$ in $T_1$ and $i_{21}(w)$ &quot;$=$&quot; $(n, 0)$ in $T_2$, which says to me that $i_{12}(w)i_{21}(w)^{-1} $&quot;$=$&quot;$ (n, 0) + (-n, 0) = 0$. The notation here isn't right, since $(n, 0)$ in $T_1$ is not the same as $(n, 0)$ in $T_2$ and their addition (composition?) shouldn't be written quite that way, hence my quotes around the equals signs, but certainly no loop in $S^1✕${$x_0$} is going to magically include anything but the 0 loop in either torus's second component, right? But this doesn't seem right to me, since if a, b are the generators of the circles in $T_1$ and d, e are the generators of the circles in $T_2$, and c is the constant path then the identification of the circle should make, say, $a = d$, so for example $(a, b)*(d, e)$ $ = (a^2,b)*(c, e)$ $ = (c, b)*(d^2,e)$. This doesn't match with what I just worked out N might be, but I also can't figure out what N should look like to make this equivalence make sense. My attempt to understand this geometrically has gone even worse. What I know for sure is that the resulting figure isn't the &quot;2-torus&quot;, 2 tori glued together at a disc. What I don't know for sure is whether this identification should actually create the hypertorus $S^1✕S^1✕S^1$ or something completely different.
96000
Let $I$ be the unit interval, and define the torus using the usual identifications on $I \times I$. I've shown that $I \times I - \{x\}$ (where $x$ is a point not on the boundary of $I \times I$) is homotopy equivalent to its boundary (and that it even retracts onto its boundary). Let $p: I \times I \rightarrow I \times I /\sim$ be the canonical quotient map. Using the retraction above, call it $r$, is there a way I can explicitly show that $p(I \times I - \{x\})$ is homotopy equivalent to $p(\text{boundary of } I \times I)$, i.e., that torus minus a point to a figure-eight?  More generally, if $A \subset X$, and $X$ is homotopy equivalent to $A$ (or $X$ retracts to $A$), for which maps $p$ can I say that $p[X]$ is homotopy equivalent to $p[A]$?   Hints appreciated!
90859
Let $\mathrm{M}_{d\times d}\left(\mathbb{C}\right)$  denote the set of all $d\times d$-matrices with complex entries. My goal is to show that the set $\mathcal{M}:= \left\{ \rho\in \mathrm{M}_{d\times d}\left(\mathbb{C}\right) \left|\right. \ \rho\geq0 \ , \  {\rm tr}(\rho) = 1 \right\} $ can be endowed with the structure of a smooth manifold (most likely with boundary).   I have proven that $\mathcal{M}':= \left\{ \rho\in \mathrm{M}_{d\times d}\left(\mathbb{C}\right) \left|\right. \ \rho&gt;0 \ , \  {\rm tr}(\rho) = 1 \right\}$ is indeed a smooth manifold. My argument roughly goes like this:   First we notice that $\mathrm{M}_{d\times d}\left(\mathbb{C}\right)$ is a $2d^{2}$-dimensional real vector space and the set $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$ of all hermitian $d\times d$ matrices with complex entries is a $d^{2}$ dimensional real vector subspace. In particular $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$ trivially forms a $d^{2}$-dimensional manifold since it's globally diffeomorphic to $\mathbb{R}^{d^{2}}$.  Next we want to show that the set $\mathrm{P}_{d\times d}\left(\mathbb{C}\right)$ of all positive $d\times d$-matrices (which is a subset of $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$) is open in the relative topology of $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$. Therefor it constitutes a $d^{2}$-dimensional manifold. To see this we can consider the map  \begin{align} f: \mathrm{H}_{d\times d}\left(\mathbb{C}\right) &amp;\rightarrow \mathbb{R}^{d^{2}} \nonumber\\ A &amp;\mapsto \left(\lambda_{1},...,\lambda_{d^{2}}\right) \end{align} which maps a hermitian matrix to its eigenvalues. This map is continuous. Since $\mathbb{R}_{&gt;0}^{d^{2}}$ is an open set in $\mathbb{R}^{d^{2}}$ it follows that the preimage  $f^{-1}\left(\mathbb{R}_{&gt;0}^{d^{2}}\right)=\mathrm{P}_{d\times d}\left(\mathbb{C}\right)$  is also open in $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$. Now consider the map  \begin{align} g: \mathrm{H}_{d\times d}\left(\mathbb{C}\right) &amp;\rightarrow \mathbb{R} \nonumber\\ \rho &amp;\mapsto {\rm tr}(\rho). \end{align} We need to show that 1 is a regular value of g. To do so we pick a $\rho\in g^{-1}(1)$.  Next we choose a curve $\gamma: (-\epsilon,\epsilon) \rightarrow \mathcal{M}$ with $\gamma(0) = \rho$ and $\partial_{t}\gamma(0) = A$ for some $A\in T_{\rho}\mathrm{P}_{d\times d}\left(\mathbb{C}\right)$.  We calculate that  \begin{equation} dg_{\rho}A := \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} {\rm tr}(\gamma(t)) ={\rm tr}(A). \end{equation} Certainly $dg_{\rho}$ is surjective. This proofs the result.  I'm wondering whether this proof can be extended to positive-semidefinite matrices since a positive-semidefinite matrix can be approximated by a sequence of positive definite matrices. I'm however not sure whether the "boundary" that we will obtain is still regular in the sense that it contains no "sharp edges".   I would be more than happy for some suggestions and ideas.
77685
Write $S^3=\{(z_1,z_2)\in\mathbb{C}^2:|z_1|^2+|z_2|^2=1\}$. Let $S^1$ act on $S^3$ by $z\cdot(z_1,z_2)=(zz_1,zz_2)$.   (a) Show that $f:S^3\rightarrow \mathbb{C}\times\mathbb{R}$ given by $f(z_1,z_2)=(z_1\bar{z_2},|z_1|^2)$ induces a homeomorphism $\bar f:S^3/S^1\rightarrow K$, where $K$ a compact subset of $\mathbb{C}\times\mathbb{R}$.  (b) Show that $\bar f(S^3/S^1)\cap(\mathbb{C}\times\{t\})$ is homeomorphic to a circle for all $t\in(0,1)$.  (c) Show that $S^3/S^1$ is homeomorphic to $S^2$.  For (a), the equivalence classes of $S^3/S^1$ are $[x]=\{y:\exists t\in[0,1)\;\; \mathrm{with}\;\; e^{2\pi it}y=x\}$. But I'm not sure how to show it is a homeomorphism. $K$ should be closed and bounded so that it is compact by Heine-Borel.  For the rest, I'm pretty stumped.
110270
Problem: Let $f(z)$ be a meromorphic function on the complex torus $\mathbb C/\Lambda$ that as a function on $\mathbb C$ has no zeros and no poles on $\partial P$, the boundary of the fundamental parallelogram $P$. Show that  \begin{equation}                 \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz \in \Lambda \, . \end{equation}  Thoughts: By the Residue Theorem  \begin{equation} \frac{1}{2 \pi i}\int_{\partial P}z\frac {f'(z)} {f(z)} dz = \sum_{z_0 \in \text{ Int }P} v_{z_0}(f)z_0. \end{equation}  I don't know why the latter is on the lattice.  Thanks!
91666
Consider the Laplacian on the $n$ dimensional torus $T$, given by $-\Delta : L^2 \rightarrow L^2$. Let the domain of $-\Delta$ be all $C^\infty$ functions initially. Now consider the Friedrichs extension of $-\Delta$, still called $-\Delta$. I want to find explicitly what the domain of $(-\Delta)^{1/2}$ is. Intuitively it seems to me that it should be all $L^2$ functions $u$ such that $\frac{\partial}{\partial x_k} u \in L^2, k = 1, 2,...,n$, but cannot prove it. Any help would be appreciated. Thanks!  Edit after Neal's comment: Instead of $-\Delta$, let's say, we consider the following operator $-A = \sum^n_{i = 1} X_i^2$, where $X_i$ are vector fields generating the tangent space at each point, and $A \geq 0$, symmetric and densely defined on $L^2$, so that we can use the Friedrich's extension on $A$. Can we still say that the domain of $A^{1/2}$ consists exactly of all $L^2$ functions $u$ such that $X_i u \in L^2$ for all $i$? Thanks for your time.
35179
I've read many times that 'compactness' is such an extremely important and useful concept, though it's still not very apparent why. The only theorems I've seen concerning it are the Heine-Borel theorem, and a proof continuous functions on R from closed subintervals of R are bounded. It seems like such a strange thing to define; why would the fact every open cover admits a finite refinement be so useful? Especially as stating "for every" open cover makes compactness a concept that must be very difficult thing to prove in general - what makes it worth the effort?  If it helps answering, I am about to enter my third year of my undergraduate degree, and came to wonder this upon preliminary reading of introductory topology, where I first found the definition of compactness.
63709
This is a reworking of a previous question  which was marked as a duplicate. Some nice folks have referred me to solutions to similar problems. I still have a couple of questions, since one of the solutions used the Poisson kernel, which is not the desired solution in the text, because the Poisson kernel has not been introduced yet. The other solutions were for problems that were similar, but whose solutions were only partially applicable, as I understand them.  I'm working on a problem from p. 166 of Lars Ahlfors' Complex Analysis:     If $u$ is harmonic and bounded in $0 &lt; |z| &lt; \rho$, show that the origin is a removable singularity in the sense that $u$ becomes harmonic in $|z|&lt;\rho$ when $u(0)$ is properly defined.   My Solution So Far:   Here's what I've got for a non-Poisson kernel solution: We want to find a harmonic conjugate for $u$ in the punctured disk, and then we can apply big Picard to show that $0$ is a removable singularity. The problem is that the punctured disc is not simply connected.   If we take $\sigma = (u \circ \exp)(z): L \to \mathbb{C}$, where $L$ is the left half-plane, now we have $\sigma$ harmonic, and defined in a simply connected set $L$, so it has a harmonic conjugate $\tau$. Now we wish to take $v = (\tau \circ \log):\{0 &lt; |z| &lt; \rho\}\to \mathbb{C}$, but in order to do so we need $\tau$ to be $2\pi i$-periodic. This is where my proof breaks down.  One of the similar problems which readers referred me to used the following strategy: note that since $\sigma$ is $2\pi i$-periodic, and $\sigma, \tau$ satisfy the Cauchy-Riemann equations, we have that $\nabla\tau(z) = \nabla\tau(z + 2 \pi i)$ for all $z\in L$, or in other words, $\tau(z) - \tau(z+2\pi i)$ is a constant function, say, equal to $\alpha \in \mathbb{R}$. Then, the solution suggests, we can set $$g(z) = u + iv - \frac{c}{2\pi}z,$$  in order to make $g$ $2\pi i$-periodic, and then we can use $(g\circ \log)$ on $\{0 &lt; |z| &lt; \rho\}$. But the reason why this won't work for my purposes is: now we have changed $u$! On the other hand, we can't simply subtract $\frac{c}{2\pi}\Im{(z)}$, and try to leave $u$ untouched, because now $g$ is not analytic. How can we get around this?  Another strategy entirely might be to try to use the mean-value property. Ahlfors discusses that a lot in the preceding section, and his main example is the punctured disc. Since $u$ is harmonic, we know that the average over circles centered at zero is a constant...
114947
Let $\mathbb{T}^2$ be the torus and have $\mathbb{Z}_2$ action on the torus by permuting the coordinates. I am trying to prove the orbit space is congruent to the mobius strip $\mathbb{T}^2/\mathbb{Z}^2 \cong M$.  I am writing $M\cong I^2/\alpha$, where $I=[0,1]$, $\alpha(x)=1-x$ and we identify the points $(x,0)\sim (\alpha(x),1)$. I know it is useful for this problem to identify the torus as the quotient $I^2/\sim_0$ with left glued to right and top glued to bottom (referring to sides) and then the action on the torus corresponds to switching coordinates on the square, which is reflecting over the diagonal $(x,x)$. What I don't understand is how to combine these relations of gluing and permuting to show this space is homeomorphic to the quotient above with $\alpha$.   I think that with both relations we will identify the edges together since $(x,0)\sim_0(x,1)\sim(1,x) \sim_0(0,x)$. Then points not on the edges will either be identified with themselves if they are on the diagonal or just their permutation. But I can't figure out how to map this to the Mobius strip...  How do you proceed formally with proving these spaces are homeomorphic? Obviously I'm not looking for all the details but I'm stuck both trying to visualize it (i get to folding the square over the diagonal $(x,x)$ but then I cant picture how to get the mobius strip from there on). Any help is appreciative.   Also a different generalization than the natural one: instead of permuting the coordinates in $\mathbb{T}^k$ and taking the action by $S_n$ consider the action by reversing the coordinates, which can still be done by using $\mathbb{Z}_2$. What is this orbit space homeomorphic to? Would it be $M^k$ when $k$ is even and something else for odd $k$?
80703
Suppose that $T$ is continuous map from an interval $I$ to itself. Moreover, suppose that there exists $x_1 &lt; x_2 &lt; x_3 &lt; x_4 $ such that   $$T(x_1) = x_2, T(x_2) = x_3, T(x_3) = x_4\ \  \text{ and }\ \ T(x_4) \le x_1.$$   Prove that $T$ has an orbit of period 3.   I don't really know how to start the problem. I am wondering how do I apply Sharkovskii's Theorem to the problem? Helps are much appreciated!  As we are given $x_1 &lt; x_2 &lt; x_3 &lt; x_4 $ such that $T(x_1) = x_2, T(x_2) = x_3, T(x_3) = x_4 $ and $T(x_4) \le x_1$, we get  $ x_1 &lt; x_2 = T(x_1) &lt; x_3 = T(x_2) &lt; x_4 = T(x_3)$. Also note that $T(x_4) \le x_1$, then $T(x_4) \le x_1 &lt; T(x_3)$.
109940
In the book of PDE, evans, the proof of the characterization of $W^{1,\infty}$ uses the following argument:  Suppose $u$ has compact support and Lipchitz continuous. Then  $$ \Vert D_i^{-h}u\Vert_{L^\infty(\mathbb{R}^n)}\leq Lip(u), $$ where $D_i^{-h}u$ the difference quotient defined by  $$ D_i^hu=\frac{u(x+he_i)-u(x)}{h}. $$ Then there exists a function $v_i\in L^\infty(\mathbb{R}^n)$ and a subsequence $h_k\rightarrow0$ such that  $$ D_i^{-h_k}u\rightharpoonup v_i \quad\text{weakly in }L^2_{loc}(\mathbb{R}^n). $$  Question: I don't know why such a subsequence exists. Since $u$ has some compact support $\Omega$, then $u\in L^2(\Omega)$. Then there exists a function $v_i\in L^2(\Omega)$ such that  $$ D_i^{-h_k}u\rightharpoonup v_i \quad\text{weakly in }L^2(\Omega). $$ But I don't know why this $v_i\in L^\infty$.
84916
This is one of the exercises of 'Do Carmo' (Section 3.5, 12)     How do you prove that there are no compact (i.e., bounded and closed in $\mathbb{R}^3$) minimal surfaces?   Thanks!