# Tag Info

0

I try to show it directly. Not sure it is right... Let $y$ be an arbitrary point in $M$. And $i_{n_k}(p) \to i$ uniformly. Isometry is a homeomorphism, it follows that there must be unique $p$ and $q$ such that $i_{n_k}(p)=y$, and $i(q)=y$. Now $$d(i_{n_k}^{-1}(y),i^{-1}(y))=d(p,q)$$ If we can show $d(p,q)<\epsilon$, for every $\epsilon > 0$, for ...

2

Here is a hint. If $f(p)\neq s$ for any $p\in M$, then $$f^{-1}((s,\infty)) \cup f^{-1}((-\infty,s)) = M$$

0

See page 6 of S.Lang, "Differential and Riemannian manifolds". He defines differentiability (in the sense you are asking, which is stronger than Frechet) for general topological vector spaces. Definition. Let $E, F$ be topological vector spaces and $\phi: U\subset E\to F$ is a map of a neighborhood $U$ of $0\in E$. Then $\phi$ is said to be tangent at $0$ ...

0

A particular element $\partial D^m_\alpha \times D^n_\beta$ of $\partial(D^m_\alpha \times D^n_\beta)$, is mapped to $\varphi_\alpha(\partial D^m_\alpha) \times D^n_\beta$ which is part of the $(m + n − 1)$ skeleton.

4

Let us define the functionals $\lambda_N\in C(K)^*=M(K)$ by $$\lambda_N=\sum_{n=1}^N a_n\delta_{x_n}.$$ Your first assumption implies, through the uniform boundedness principle that $$\sup_N\|\lambda_N\|_{M(K)}<\infty.$$ Now recall that $$\sup\{|\lambda_N(f)|: \|f\|_{C(K)}\le 1\}= \|\lambda_N\|_{M(K)}\le \sum_{n=1}^N|a_n|.$$ Consider the function $f$ ...

4

Consider the pushout $$\begin{matrix} [0,1) &\to & [0,1] \\ \downarrow & & \downarrow \\ [0,1] &\to & X \end{matrix}$$ where all maps are inclusions. Here $X$ (an interval "with two $1$'s") is not Hausdorff.

3

The notes in this PDF have a decent discussion. In its terminology your first definition is of a Kelley subnet and your second of a Willard subnet. There is a third definition, of what in these notes is called an AA-subnet, that is actually better in many ways than either of these: Let $\langle I,\preceq_I\rangle$ and $\langle J,\preceq_J\rangle$ be ...

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I'm having a bit of trouble thinking of a good way to write this down. The idea behind all of this is not complicated, however complicated it may seem because of my notational difficulties. Let $B_1 = \bigcap_{i\in I} B_{p_i}(x_1, \varepsilon_i), \quad B_2 = \bigcap_{j\in J} B_{p_j}(x_2, \varepsilon_j)$, where $I$ and $J$ are of course finite. Let $x \in ... 0$W_4$is also$K_4$of course. However the answer is straightforward; the tetrahedron provides an example that$m=3$is feasible, and arranging the vertices in two back-to-back equilateral triangles demonstrates that$m=2$is not feasible as the last edge cannot be added.$m=3$is also feasible (and probably required) for$W_5$and$W_6$, for example ... 1$Y_1$is indeed open, in both$X$and$\mathbb{R}$, but not for the reason you say. It's open, because every point is an interior point, or you could look at the complement, which in$\mathbb{R}$is$(-\infty,1] \cup [2, \infty)$, which is indeed closed (it contains all its limit points). But openness is more direct, as$Y_1$is an open interval which is ... 0 Because$X$is closed in$\mathbb R$all closed subsets of$X$are also closed in$\mathbb R$. Proof: Take$A \subseteq X$close. Thus$X\setminus A$is open. Also$\mathbb R\setminus X$is open because$X$is close in$\mathbb R$. So$\mathbb R \setminus A = (\mathbb R\setminus X) \cup (X\setminus A)$is open which proves that$A$is close in$\mathbb R$. ... 2$T^3$is a Lie group,so your space$T^3\times T^3-\Delta$is a trivial bundle over$T^3$with fiber$T^3-\{\ast\}$(google"configuration space" for more details).So $$\pi_1(T^3\times T^3-\Delta)\cong \pi_1(T^3\times (T^3-\{\ast\}))\cong Z^3\times Z^3\cong Z^6$$ 1 What i understood is that your are trying to find a homotopy of$id_{S^1}$to$a:S^1\to S^1$such that$a(x)=-x$. This is the antipodal map. Such a homotopy is$F:S^1\times I\to S^1$with$F(x,t)=e^{i\pi t}x$. It's is clearly continuous,$F(x,0)=x$and$F(x,1)=-x=a(x)$. Also the projective plane$\mathbb RP^2$is the quotient space$S^1/_{x\equiv a(x)}$0 The definition of Hausdorff metric I will use, starting from a metric space$(X,d)$, is$d_H(K_1, K_2) = \max(\sup_{x \in K_1} d(x, K_2), \sup_{x \in K_2} d(x, K_1))$. So suppose$K = \cap_n K_n$, where$K_{n+1} \subseteq K_n$for all$n$, and all$K_n$are non-empty compact (so$K$is too). For$x \in K$,$x \in K_n$as well, so$d(x, K_n) = 0$, so ... 1 The intuitive idea that topology captures is the idea of separation (in some sense) of points and subsets of a set. The intuition is most obvious in the most basic (least interesting and most extreme) topologies; the discrete topology (the collection of all subsets of a given set) where every two points and every two subsets can be separated and the ... 1 Here are some hints: (Whenever I speak of a map betweens spaces, I'll mean a continuous function.) Note that the compact subsets of$X$are exactly the images$t(C)$of maps$t:C→X$where$C$is a compact space (call such a map a test map). Then$X_c$has the final topology for all test maps to$X$, which are the same as the test maps to$X_c$, so$X_c$is ... 0 For proper functions$f: D\to D$you can define the degree as the cardinality of$f^{-1}(z)$for$z$which is not a critical value of$f$. (Here$D$is the open unit disk.) To see that this is independent of$z$, prove that the restriction$f: D'\to D''$is a covering map, where$D'\subset D$is the set of regular values and$D''=f^{-1}(D')$. To prove the ... 4 In this case$F$is injective : if$F(x) = F(y)$then$0 = d(F(x),F(y) \geq L d(x,y)$implies that$d(x,y) = 0$, that is, that$x=y$. As$F$is surjective, it is bijective, and its inverse$F^{-1}$is Lipschitz with coefficient$1/L < 1$, so as$X$is complete, this inverse$F^{-1}$has a fixed point$x_0$, which is also a fixed point for$F$. Finally, ... 1 This property has been called hypocompactness; it is strictly weaker than compactness and strictly stronger than paracompactness. The Sorgenfrey line is hypocompact but not compact: every open cover has a disjoint clopen refinement. Any hedgehog space$X$of uncountable spininess is an example of a paracompact space that is not hypocompact.$X$is ... 1 Let$\tau$be any topology on$X$with the stated property; what you’re being asked to prove is that$\tau\supseteq\mathcal{T}$, which amounts to proving that the sets$\{x\in X:x<a\}$and$\{x\in X:a<x\}$belong to$\tau$for each$a\in X$. For convenience let$(\leftarrow,a)=\{x\in X:x<a\}$and$(a,\to)=\{x\in X:a<x\}$. HINT: Fix$a\in X$. For ... 0 At each stage of the construction of the middle-thirds Cantor set we remove only finitely many intervals, so there are countably infinitely many removed intervals. Your set$A$contains one point from each of them, so$A$is countable. Every non-empty open subset of$\Bbb R$is uncountable — in fact of cardinality$2^\omega=\mathfrak{c}$— so the interior of ... 1 HINT: Let$\tau$be the order topology on$Y$, and let$\tau_Y$be the subspace topology on$Y$that it inherits from$X$. You’re asked to show that$\tau\subseteq\tau_Y$, and that there are examples in which$\tau\subsetneqq\tau_Y$. Showing that$\tau\subseteq\tau_Y$is straightforward: you need only show that for each$y\in Y$,$\{z\in Y:z<y\}$and ... 5 Define$F(x,t)=(1-t)x$. Note that it is not true that a space is necessarily contractible if it has trivial fundamental group. For example, any sphere of dimension greater than 1 has trivial fundamental group and is not contractible. 0 Let$X$and$Y$be CW complexes. Let$X\times Y$be the usual product (in$\mathbf{Top}$, the category of all spaces), and let$X\times_k Y$be the$k$-ification of$X\times Y$, so$X\times_k Y$is the product in$\mathscr U$, the category of$k$-spaces. By$X\times_c Y$we will denote the CW structure on the product. So we want to show that the identity ... 0 Theorem. Every locally-compact second-countable Hausdorff space is a Polish space. I thought I'd quote a sketch of proof of above fact from somewhere else. The following sketch is from https://golem.ph.utexas.edu/category/2008/08/polish_spaces.html if X is second countable locally compact Hausdorff, then the one-point compactification X+ is ... 0 The converse of this theorem does not hold. As an example consider the set of all real numbers$ \mathbb{R} $with the discrete topology$ \tau_{d} $. Clearly$ (\mathbb{R},\tau_{d}) $is metrizable and the discrete metric$ \rho_{d} $is the metrization of$ \tau_{d} $. Also$ \mathbb{R} $is regular with respect to$ \tau_{d} $since every closed subset ... 1$\|(A+B)-(C+D)\| = \|(A-C)+(B-D)\| \le \|A-C\| + \|B-D\|$. Hence if$\|A-C\|, \|B-D\| < {1 \over 2} \epsilon$, then$\|(A+B)-(C+D)\| < \epsilon$.$\|AB-CD\| = \|AB-CB+CB-CD\| \le \|AB-CB\|+\|CB-CD\|$. Now suppose$\|A\|,\|B\| < M$and$\|A-C\|, \|B-D\| < \min(1,{1 \over 2(M+1)} \epsilon)$, then$\|AB-CD\| \le \|B\| \|A-C\| + \|C\| \|B-D\| < ...

1

For $0<p<1$ the $p$-norm is only a quasinorm. Nevertheless, it is not hard to check that all of them give the same topology as the $\infty$-norm $$||(x_1, \ldots, x_n)||_{\infty} = \max|x_i|$$ Let $x$ with $||x||_{\infty} < \epsilon$. Then $\sum_{i=1}^n |x_i|^p \le n \epsilon^p$ and so $||x||_{p} < n^{\frac{1}{p}}\cdot \epsilon$. Therefore ...

2

One way of proving that norms give the same topology is to find inequalities $$c||x||_1\leq ||x||_2\leq C||x||_1\ \ \ \ \text{ with }c>0.$$ $$\text{If }\ \ q>p\geq1\ \ \ \ \ \ \ \ \ \text{ then }\ \ \ \ \ \ \ \ \ \frac{n^{1/q}}{n^{1/p}}||x||_p\leq||x||_q\leq||x||_p.$$ The first inequality is the generalized mean inequality and the second is because ...

0

Here is a possible proof (credit to user pappus here): Recall that a manifold is orientable iff there exists a volume form on it. So in order to show that $K^2$ is non-orientable, it is enough to show that any $2$-form $\omega$ on $K^2$ must vanish at some point. Let $K^2$ be defined here as the quotient of $\mathbb{R}^2$ by the subgroup of diffeomorphisms ...

1

if one topology is finer then for $x,y\in X$ you can take the same open sets that are disjoint neighbourhoods of $x$ and $y$ in both topologies.

0

You need to fill in some important details, but the basic idea is fine. You might start like this: Let $\langle x_n:n\in\Bbb N\rangle$ be a Cauchy sequence in $\langle X,e\rangle$. For $n\in\Bbb N$ let $t_n=\frac1{x_n}$. Then by definition for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that $|t_m-t_n|<\epsilon$ whenever $n\ge ... 0 I think what you're trying to say is this: if two metric spaces are isomorphic (that is, there exists a set bijection between them that plays well with the two metrics), then if one is complete then so is the other one. This is a true fact (it's a good exercise to prove it). And you're also saying that the map from$X$in this problem, to$[1,\infty)$with ... 3 The idea is OK, but should be formalised a bit more. Let$Y = [1,\infty)$with the Euclidean metric$d$. Clearly,$Y$is complete as it is a closed subset of the complete$\mathbb{R}$in the Euclidean metric. Then$f(x) = \frac{1}{x}$from$(X,e)$to$(Y,d)$. Then$d(f(x), f(y)) = |f(x) - f(y)| = |\frac{1}{x} - \frac{1}{y}| = e(x,y)$for all$,y \in X$. ... 0 For part b , consider the ball centered at (x,y) with radius x/2. For the second part of part c, take another point$p$in the$\delta$-ball and consider the ball with radius half the minimum of the distances$|p-(x,y)|,\delta-|p-(x,y)|$. 3 The complement of any neighbourhood of$(0,0)$in$Y_1$is a finite collection of line segments, while there are neighbourhoods of$(0,0)\in Y_2$where that isn't the case. One other property that differentiated them is that$Y_2$isn't locally connected. For instance any small enough neighbourhood of$(1/2,0)$consists of infinitely many line segments. 0 Let us look at$(n,n+1)$it is open in$\mathbb R$for any$n\in\mathbb Z$, hence$U=\cup _{n\in\mathbb Z}(n,n+1)$is open in$\mathbb R$(any union of open sets is open) thus$\mathbb R\setminus U$is close in$\mathbb R$(a compliment of an open set is a close set) but$\mathbb R\setminus U=\mathbb Z$thus$\mathbb Z$is a close set in$\mathbb R$1$X$is not a complete metric space. Observe the sequence $$p_n(x) \;\; =\;\; \sum_{k=0}^n \frac{x^k}{k!}.$$ Each term is a finite sum of polynomials but in their limit$p_n(x) \to e^x$. This remains true independent of the norm put on$X$. Yes,$T$is continuous. Let$\epsilon > 0$and let$p(x) = a_0 + a_1x + \ldots + a_nx^n$and$q(x) = b_0 + ...

2

Suppose not, we have a subsequence of $f(x_n)$ s.t. $d'(f(x_{n_k}),f(x))\ge \epsilon$.WOLG, we assume the subsequence is $f(x_n)$ and $x_n\not =x\forall n$ (otherwise restrict our discussion to suitable subsequence). Then set $B=\{f(x_n)\}$, hence $\{(x_n)\}\subset f^{-1} ( B),\{(x_n)\}\cup \{x\}\subset \overline{f^{-1} ( B)}$ since $x_n\to x$ By ...

0

Calm down. For any $x\in (0,1]$ , let $\delta < |x|$ , then $e(x,y) = |\frac{1}{x} - \frac{1}{y}|=|\frac {x-y}{xy}|\le \frac {\delta}{(x-\delta)^2}, \ \forall y\in (x-\delta,x+\delta)$ Let $\delta \rightarrow 0 \Rightarrow \frac {\delta}{(x-\delta)^2}\rightarrow 0$ Thus , $\forall r>0 , \ \exists \delta>0 \ s.t \ \frac {\delta}{(x-\delta)^2}\le ... 0 To show$B_d (x,r) \subset B_e(x,t)$, we need to show given$x\in (0,1],t>0,\exists r>0$, s.t.$d(x,y)=|x-y|<r\implies e(x,y)=\frac{|x-y|}{|xy|}<t$As you can see, the key is to obtain a lower bound for the denominator, and since$x$is fixed, we need to bound$|y|$away from$0$using suitable$r$. Then can we? Yes, set$r=\frac{|x|}{2}$, then ... 0 You need to show that for every$r>0$, there is an$s$such that$|x-y|<s\to |1/x-1/y|<r$. Let$s=\min(1,xr)\frac{x}2$. Since$s\le\frac x2$and$|x-y|<s$, you have$y\ge\frac x2$as well. Then $$\left|\frac1x-\frac1y\right|=\left|\frac{x-y}{xy}\right|\le\frac{|x-y|}{x(x/2)}=\frac2{x^2}|x-y|<\frac2{x^2}\cdot xr\cdot\frac x2=r.$$ (Note that ... 0 Check out the portal Applied Topology, which arose from Gunnar Carlsson's research group in Stanford. There are many applications areas mentioned with relations to statistics, data-mining, biology etc. 0 Topology is at least partially built into human intuition because it talks about invariants - general properties and classification independent of fine details - exactly what humans are best at! There are many real-life examples, for instance, the hairy ball theorem is encountered every time you wonder how come we can't have a map of the earth without ... 0 Topology also has applications within computer science. Directed algebraic topology is a branch of algebraic topology that has applications in concurrency theory when trying to avoid and resolve deadlocks and starvation. See for example here. Topological data analysis is an alternative to standard data mining, which allows one to infer global and ... 0 Kuratowski's Theorem characterizing planar graphs (http://en.wikipedia.org/wiki/Kuratowski%27s_theorem) has application to circuit boards where certain nodes must be connected but without any edge crossings. 1 Your approach works if you combine your two ideas: Form an open set around each point in$K$, such that it contains only finitely many points of$S$. These open sets cover$K$, and since$K$is compact, extract a finite subcover, each element of which only contains finitely many points of$S$. 0 An infinite subset of a compact set has a limit point (Rudin). 0 Let$X$be the compact set and$S$be a set with no accumulation points. Let$T = X \cap S$. If$T$is infinite: Let$x = ( x_k : k \in \mathbb{N} )$be an infinite sequence of distinct points in$T$. Then$x$has a subsequence$( y_k : k \in \mathbb{N} )$that converges in$X$. Can you see that we get a contradiction? 0 Alternately, stereographic projection maps circles to circles, and so for topological reasons$f$maps open disks in the planes into open discs on$S^n - \{p\}$(regions with circular boundary). But the set of open disks in the plane is a basis for the standard topology on the plane, and hence (once you know it exists)$f^{-1}\$ is continuous. Similarly, so ...

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