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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 $\sup_n$ in 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 map $$z\mapsto \frac{1}{z}$$ 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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