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## Hot answers tagged metric-spaces

4

We have $$||x_n||_\infty\le \sup_{t\in[0,1]}|t^{2n}|+\sup_{t\in[0,1]}|t^{3n}|=2$$ hence $(x_n(t))$ is bounded on $[0,1]$.

3

Let us take a closer look at $\lVert\,\cdot\,\rVert_0$. Since for $x\in X$ we have $\hat{x} = \{ x-y : y \in Y\}$, we can write $$\lVert \hat{x}\rVert_0 = \inf_{z\in\hat{x}} \lVert z\rVert = \inf \{ \lVert x-y\rVert : y \in Y\}.\tag{1}$$ So we have $\lVert \hat{x}\rVert_0 = 0$ if and only if for every $\varepsilon > 0$ there exists a $y_\varepsilon \in ... 3 Hint: can be the complementary (rationals of interval) countable intersection of open sets? Also: Baire Category Theorem. 2 The usual definition of completion of a normed space is obtained by considering equivalence classes$[\{ x_{n} \}]$of Cauchy sequences, grouped according to the equivalence relation that $$\{ x_{n} \} \sim \{ y_{n} \} \iff \lim_{n}(x_{n}-y_{n})=0.$$ This$\sim$is an equivalence relation because it is (a) reflexive (b) symmetric and (c) transitive: (a) ... 2 Your property$(\ast)$is precisely uniform continuity. The proof that uniform continuity implies$(\ast)$is easy: according to the definition of uniform continuity, for every$\varepsilon > 0$there is a$\delta > 0$with$d(x,y) < \delta \implies d(f(x),f(y)) < \varepsilon, and from that follows that $$f(A\oplus\delta) \subset f(A) \oplus ... 2 You can use the same idea that works for +:$$\|(\alpha,x)\|_{K\times X}=|\alpha|+\|x\|.$$In any case, you can endow K\times X with the product topology and no explicit norm is required. EDIT:$$\eqalign{\|\alpha x - \alpha_0 x_0\|_X & = \|\alpha x - \alpha_0 x + \alpha_0 x - \alpha_0 x_0\|_X\cr &\le\|\alpha x - \alpha_0 x\|_X + \|\alpha_0 ... 2 What does it mean for a sequence to be Cauchy? It means for each\epsilon > 0$, there is some$N \in \mathbb{N}$such that$n, m \geq N$implies$d(x_{n}, x_{m}) < \epsilon$. Since this is true for every$\epsilon > 0$, it is true if$\epsilon = \frac{1}{2}$, for example. That is, there is some$N_{1} \in \mathbb{N}$such that$n, m \geq N_{1}$... 2 For metric spaces: Take$X = (0..1)$and take the inclusion$ι\colon X → ℝ$. Now take$A = X$. Or even take$f\colon X → ℝ,~x↦ \frac{x}{1-x}$and$A = X$if you want$f(A)$to be neither bounded nor closed in$ℝ$. 2 How to prove this depends a great deal on what tools you already have. The first answer assumes that you know that a metric metric space is compact if and only if it’s complete and totally bounded, but I shouldn’t be at all surprised if this exercise were intended to prepare for that result. Suppose that whenever$A$is an infinite subset of$X$and ... 2 Hint: Suppose A and B have a point in common, and B and C have a point in common, and A and C have no points in common. Spoiler: 2 Everything except the triangle inequality is obvious, I'll leave them for you. For the triangle inequality, it is a matter of applying$\sup_nin the right order. We have: \left|\sum_{k = 1}^n (x_k - y_k) \right|\leq \left|\sum_{k = 1}^n (x_k - z_k) \right|+\left|\sum_{k = 1}^n (y_k - z_k) \right| \leq \sup_n\left|\sum_{k = 1}^n (x_k - z_k) ... 2 If V and W are normed vector spaces of the same finite dimension n, then V and W are isomorphic as topological vector spaces, i.e., there is a a vector space isomorphism f : V \rightarrow W such that both f and f^{-1} are continuous. However, if n \ge 2, V and W may not be isometrically equivalent, i.e., it may be impossible to choose ... 1 Let F,U,x,y be as in your attempt. For every \theta the projection map \operatorname{proj}_\theta is continuous; hence, \operatorname{proj}_\theta (U) is connected. A connected subset of a line with two distinct points has positive length. For all \theta\in [0,\pi) except one (the one parallel to the line segment xy), ... 1 Try to create some extreme situations where the claim fails. Take \mathbb R with the Euclidean metic, and consider A and C to be some really far away subsets, say even just a singleton each, but far away. Clearly, d(A,C)>0. Now for B you can choose all of \mathbb R. What happens then? Remark: the minimum in the definition of distance between ... 1 The only nontrivial thing to prove here is the triangle inequality. Hint: \begin{align*} d\left(x,z\right) & =\sup_{n}\left|\sum_{k=1}^{n}x_{k}-z_{k}\right|\\ & =\sup_{n}\left|\sum_{k=1}^{n}x_{k}-y_{k}+y_{k}-z_{k}\right|\\ & =\sup_{n}\left|\sum_{k=1}^{n}x_{k}-y_{k}+\sum_{k=1}^{n}y_{k}-z_{k}\right| \end{align*} 1 The nub of it is that total boundedness is an intrinsic property of a metric [more generally, uniform] space. It does not depend on whether the space is a subspace of some larger space, and if so, what that larger space is, all that matters is the metric. We have two metric spaces, X = A^\ast(S^\ast), with the metric induced by the norm on E^\ast, and ... 1 For the first questions, notice that A is the intersection of \mathrm{ev}_0(\{0\}) and \mathrm{ev}_1(\{1\}), where \mathrm{ev}_x\colon X\to [0,1] is evluation at x. Can you show that \mathrm{ev}_x is continuous? For the second part, one can easily show that d(\tilde f,\tilde g)=d(f,g)/2. The last claim then follows from the Banach fixed-point ... 1 To prove that f is well-defined, you want to show that if you pick (x_n) \in l^1, then f\big((x_n) \big) \in l^2, that is, the map actually does what it says it will do in terms of mapping an element from its domain to it's codomain. Theorem: Suppose 1\le p_1<p_2\le\infty. Prove that \ell^{p_1}\subseteq \ell^{p_2} by proving that ... 1 Your original idea is right on the mark. Let z=re^{it}. Then the mapz\mapsto \frac{1}{z}$$or equivalently$$z\mapsto \frac{1}{r}e^{-it}$$will do the trick. If we want to "keep it real," the map is$$(r\cos(t),r\sin(t))\mapsto ( \frac{1}{r} \cos(t) , -\frac{1}{r} \sin(t))$$The minus sign is for convenience in the easier definition but is not ... 1 A compact metric space is totally bounded. Inasmuch as total boundedness is hereditary, a metric space which is isometrically embeddable in a compact space must be totally bounded. Conversely, if a metric space is totally bounded, then its completion is totally bounded and (of course) complete; and a totally bounded complete metric space is compact. (This ... 1 In many sources, simple functions are those measurable functions that have a finite set of values. But here a countably infinite set of values is allowed. This allows one easily approximate any measurable function f uniformly by simple functions: for example, let$$ f_n(x) = \frac{\lfloor n f(x)\rfloor }{n} $$and observe that f-\frac{1}{n}\le f_n\le f ... 1 Daniel is right, Banach's fixed point theorem is the way to go. Defining A as he did, we see that A: C([0,\pi/2]) \rightarrow C([0,\pi/2]). In addition, ||Af-Ag||_{C([0,\pi/2])} = \sup_{t \in [0,\pi/2]} \left|\int_0^{\pi/2} \arctan\left(\frac{f(s)}{2} +t\right) - \arctan\left(\frac{g(s)}{2} + t\right) ds\right| Since arctan is Lipschitz continuous ... 1 The easier way is to note that x_n(t) is bounded between -1 and 1. Therefore (x_n) is bounded in the sup norm. If x_n converges uniformly to some function f then it is easy to see that f must be zero, by taking the limit pointwise. Thus it suffices to look at the maximum of |x_n| on [0,1]. You can find that the maximum is taken in ... 1 The fields \mathbb R and \mathbb C are equipped with a standard topology (derived from the standard metric and the standard absolute value). 1 About compact: take the open cover (and prove it is such)$$\left\{\;\left(\frac1n\;,\;\;\sqrt2-\frac1n\right)\cap\Bbb Q\;\right\}_{n\in\Bbb N\setminus\{1\}}$Resuming (see the comments): it is closed, bounded and not compact. 1 The closed unit ball is not compact in any infinite dimensional Banach space. Thus the closure of unit ball is not compact. If the unit sphere in$\ell_2 $were compact then the sequence$\{e_n\}$of vectors of orthogonal basis of$\ell_2$should has a convergent subsequence but this is impossible since$||e_i -e_j| =\sqrt{2}.$1 Uniform continuity implies$(*)$. In particular$(*)$is equivalent with continuity if$X$is compact. Proof: Suppose$B,B'\subset Y$are such that$B\oplus r\subset B'$for some$r>0$. To show that$f^{-1}(B)\oplus s\subset f^{-1}(B')$(for a suitable$s>0$), we need to show that for any$x\in f^{-1}(B)$we have$B_s(x)\subset f^{-1}(B')$. By uniform ... 1 For compactness, recall that$\Bbb R = \pi(P)$where$P$is the graph of parabola, and$\pi$is the projection on the first coordinate. If$P$would be compact, so would be$\pi(P)$since$\pi$is a continuous map. 1 Define the metric$d(x,y) = \left|\frac{1}{x} - \frac{1}{y}\right|$on$X = [-1,0) \cup (0,1]$. It's easy to check that this is a metric. The function$f(x) = \frac{1}{x}$is (by construction) an isometry between$(X,d)$and$Y = (-\infty,1] \cup [1, +\infty)$where the latter space has the usual Euclidean metric. And$Y$is complete (being closed in ... 1 A less trivial example may be$X=\mathbb R^+_0$,$d(x,y):=|\sqrt{x}-\sqrt{y}|$and$x_n:=n\$.

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