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0

In order to show that the spaces mentionned in b), c) and d) are not compact, it is sufficient to exhib an unbounded sequence in these spaces. For example, consider respectively $$\left( \begin{matrix} n & 0 \\ 0 & n \end{matrix} \right), \ \left( \begin{matrix} n+1 & 1 \\ n & 1 \end{matrix} \right) \ \text{and} \ \left( \begin{matrix} 1 ...


0

Two norms, $||.||$ and $||.||^*$ on a vector space are said to be equivalent norms if there exists a constant $C > 0$ such that $\frac{1}{C}||x||\leq ||x||^* \leq C||x||$. It is a nice exercise to show that all norms on $\mathbb{R}^N$ are equivalent.


0

If $K$ and $L$ are compact metric spaces and $X$ is some metric space, then $C(K\times L,X)$ is actually isometric to $C(K,C(L,X))$. The natural definition would be the following: $$\phi:C(K\times L,X)\to C(K,C(L,X)),\qquad \phi(F)(k)(l)=F(k,l).$$ In principle, $\phi$ is actually just a function $C(K\times ...


0

Since clearly $\|f\|_1\le \|f\|_2$ for all $f\in X$, we need only find some $c$ such that $\|f\|_2\le c\|f\|_1$ for all $f\in X$. In fact $c=3$ works: Let $a=\|f\|_1$ and $x\in (0,1]$. Then by the IVT, we have $$f(x)=f(0)+(x-0)\cdot f'(y)$$ for some $y\in (0,x)$. From $f(0)\le a$, $f'(y)\le a$, and $0<x\le 1$ we conclude $f(x)\le 2a$, similarly $f(x)\ge ...


1

$\| f\|_1 \leq C \| f \|_2$ seems obvious because a pointwise value must be less than the sup. For the other direction, my hint is that $$\|f\|_{\infty} - |f(0)| = \sup_{x \in [0,1]} |f(x)| - |f(0)| \leq \sup_{x \in [0,1]} |f(x) - f(0)|$$ Now use the fundamental theorem of calculus and some obvious inequalities.


1

For boundedness, I suggest you set $\epsilon=1$, that is $|\xi_j| < 1 + k_{N+1}$ for some $k_{N+1}\in \mathbb{R}$,so that you get a fixed upper bound for $|\xi_j|$. For convergence, Since $(x_n)$, where $x_n \colon= (\xi_j^{(n)})$ is a Cauchy sequence in $\ell^\infty$. Then, given $\epsilon > 0$, there exists an integer $N$ such that $\forall m,n ...


0

Solved it with a lot of tinkering. To calculate the amount of objects i can fit around a certain path. Where distance between point A (center of circle) and point B (center of outer object) is known. I also knew the objects width i could do: objCount = (distance*2*pi)/objWidth This in turn had to be converted into the a actual angle, so objAngle = ...


0

Hint: Perhaps try the continuous real-valued functions on $X$ with a suitable norm.


0

$$d(Mv_1,Mv_2):=d\left((1,1)+\frac12(x_1,y_1)\,,\,\,(1,1)+\frac12(x_2,y_2)\right)=$$ $$=d\left(\left(1+\frac12x_1\,,\,1+\frac12y_1\right)\,,\,\,\left(1+\frac12x_2\,,\,1+\frac12y_2\right)\right)=$$ $$=\sqrt{\frac14(x_1-x_2)^2+\frac14(y_1-y_2)^2}=\frac12\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\le$$ $$\le\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=d\left(v_1\,,\,\,v_2\right)$$


1

Here's a sketch of the argument; see if you can fill in the details. For each $x\in B$ let $C_x=B\,\cap\succeq\!(x)$. Use the completeness of the preference relation to show that if $F$ is a finite subset of $B$, then $\bigcap_{x\in F}C_x\ne\varnothing$. Conclude that $\bigcap_{x\in B}C_x\ne\varnothing$, and show that if $b$ is any member of this ...


2

Abstract idea: connect two points $A,B$ by a countable family of arcs $\gamma_n$ so that the length of $\gamma_n$ is $1+1/n$. Concrete realization: in the Hilbert space $\ell^2$ with standard basis $\{e_0,e_1,\dots\}$, consider the points $A=e_0$, $B=-e_0$, and $C_n = (1+1/n)e_n$ for $n=1,2,\dots$. All these are at distance at least $1$ from one another. ...


1

Since your $f$ is continuous and $S$ is compact so is $f(S)$ but in $\mathbb{R}$. Hence $f(S)$ contains its infimum.


3

(By request of the OP.) If $x$ is a point different from $P$, and $\epsilon\le\|x-P\|$, the $\epsilon$-ball centred at $x$ is an open interval of length $2\epsilon$ centred at $x$ on the line $\overline{Px}$: If $\epsilon>\|x-P\|$, the $\epsilon$-ball at $x$ consists of the segment $\overline{Px}$, the open ray of length $\epsilon$ extending from $x$ ...


0

Hint: (I'm not sure about your notation, but I'll write $B_r(x,y)$ for the ball centred at $(x,y)$ with radius $r$.) You don't need to find the boundary of the set: you just have to find a way to exactly cover the set with countably many open balls. You're right in that the boundary is really where the problem is at though. For everything not "close" to ...


0

You have that $$d({\bf x}, {\bf y}) = \frac{1}{2} \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}$$ This way, if ${\bf x} \in B(0,1)$, you have: $$\frac{1}{2} \sqrt{x_1^2 + x_2^2} < 1 \implies x_1^2 + x_2^2 < 4,$$which is the equation of a circle centered in the origin with radius $2$.


-1

First of all you need to prove that this suprimum is attained for some $y \in A $ since $A$ is a compact subspace of a Hausdorff space. Proof: Let $f(x)=\gamma$. Now consider the sets $V_1=X-B(x,\gamma-{1 \over 2})$ and $U_1=V_1\cap A$. [Here $B(x,r)$ is an open ball of radius $r$.] Since $\gamma$ is the supremum $U_1$ is non-empty. (Verify) Similarly ...


1

$\overline{(0,1) \cap (1,2)}=\overline{\emptyset}=\emptyset$ While we have $\overline{(0,1)}\cap \overline{(1,2)}=[0,1]\cap[1,2]=\{1\}$ Also: $Int([0,1]\cup[1,2])=(0,2)$ and $Int([0,1])\cup Int([1,2])=(0,1)\cup(1,2)$


1

You're right, $d^{-1}(V)$ is a subset of $X^2$, and unless you define a metric on $X^2$ (which would be a function from $X^4 \to [0,\infty)$) it does not make sense to talk about $d^{-1}(V)$ being a ball. What we can say is that $d^{-1}(V)$ is open in the product topology on $X^2$.


2

The inequality which you want to show is often called the Minkowski inequality. An unusual proof can be found here Link. Below I will show a standard one which involves Hölder's inequality. We want to show the triangle inequality, that is, that for any $f$, $g \in C([a, b])$ $$\begin{align*}\| f+g\|_p = \left(\int_{a}^b |f(x)+g(x)|^{p} \ ...


-1

Firstly, I don't know if this approachment may be helpful, but we can show that $f(A)$ is a compact subset of $\mathbb R$. Then we have that $f:A\to f(A)$ is continuous, since both $A$ and $f(A)$ are compact. We have that $f(A)=\sup\{d(x,y): x\in A, y\in A\}$. Also, $A$ is compact in $X \implies A\times A$ is compact in $X\times X$. We consider ...


1

Let $O \subseteq \mathbb R$ be open and $x_0 \in f^{-1}O$. The goal is to show that $f^{-1}O$ is open so we want to find an open ball around $x_0$ that is contained in $f^{-1}O$. Since $O$ is open there exists $\varepsilon > 0$ such that $(f(x_0)-\varepsilon, f(x_0)+\varepsilon)\subseteq O$. Now let $\delta = {\varepsilon \over 2}$. Then (same argument ...


4

I think I got it. There is no such counterexample. I welcome and appreciate any comments on the proof below, which hinges on ideas presented by Klee (1952). $\textbf{Claim:}\quad$If $(X,\|\cdot\|)$ is a completely metrizable normed vector space, then it is a Banach space. Proof:$\quad$Suppose that $(X,\|\cdot\|)$ is a normed vector space and $d$ is such a ...


1

Observe that the inequality follows first from Cauchy-Schwartz inequality and then by AM_GM inequality.


1

Hint: For $x,y\in\mathbb{R}$ and $a>0$ we have \begin{align*} 0&\leq\left(\dfrac{x}{\sqrt{a}}\pm\dfrac{\sqrt{a}y}{2}\right)^2\\ \text{i.e.}\qquad0&\leq\dfrac{1}{a}x^2\pm xy+\dfrac{a}{4}y^2 \\ |xy| &\leq\dfrac{1}{a}x^2+\dfrac{a}{4}y^2 \end{align*}


2

You could start with the other end, use AM-GM: $$\dfrac{1}{a} \sum_{i=1}^n {x_i}^2 + \dfrac{a}{4}\sum_{i=1}^n {y_i}^2 \ge \left( \sum_{i=1}^n {x_i}^2 \right)^{1/2} \left( \sum_{i=1}^n {y_i}^2 \right)^{1/2}$$ and finish off as you did with Cauchy-Schwarz.


2

$$\left|\sum_{i=1}^n x_i y_i \right| \le \sum_{i=1}^n\left| x_i y_i \right| $$ And for each $i$ $$|x_iy_i| \le \frac{1}{a}x_i^2 + \frac{a}{4}y_i^2$$ because $x + y \geq 2\sqrt{xy}$ when $x, y \geq 0$


1

No it is not a norm. If $\|x\|\ne 0$, then $$ \varphi(2x)=\frac{\|2x\|}{1+\|2x\|}=\frac{2\|x\|}{1+2\|x\|}\ne 2\frac{\|x\|}{1+\|x\|}=2\varphi(x). $$


1

Two things: a. $f(x)=0$ iff $x=0$. b. $f(x+y)\le f(x)+f(y)$. They are necessary and sufficient conditions.


8

The answer is yes, any such space must be a Banach space. This result was proved by Victor Klee in 1952 and answered a question first asked by Banach in 1932. V. L. Klee, Invariant metrics in groups (Solution of a problem of Banach), Proc. Amer. Math. Soc., 3 (1952), 484–487.


1

Hint: $l = \inf C \iff \forall \epsilon > 0 \, \exists c \in C : c < l + \epsilon$, now consider $\epsilon = \frac{1}{n}$ for $n \in \mathbb{N}$ Edit: I just realized you might not have been given this as a definition for $\inf$, if this is the case let me know what your definition is and I'll connect it to the one I used.


1

You can consider the following open cover for the plane: $A_{1} = D_{1}$ (the open disk of radius 1 centered at (0,0)) $A_{2} = D_{2} - \overline{D}_{1 - \frac{1}{2}}$ (the open annulus that overlaps with $A_{1}$) $A_{3} = D_{3} - \overline{D}_{2 - \frac{1}{3}}$ ... $A_{n} = D_{n} - \overline{D}_{(n-1) - \frac{1}{n}}$ You see, the area of overlap ...


0

This is true not only for compact metric spaces but also for separable ones - a metric space $(X,\varrho)$ is called separable if it has a countable dense subset So suppose that $D=\{q_n\}_{n\in\mathbb N}$ be a dense subset of $X$. Then define the sequence: $$ q_1,q_1,q_2,q_1,q_2,q_3,q_1,q_2,q_3,q_4,\cdots, q_1,q_2,\cdots,q_n,q_1,\cdots, $$ Then, all the ...


1

I will assume that a "limit point" of a sequence means what I call a partial limit: namely the limit of some convergent subsequence of the sequence. Theorem: Let $X$ be a metric space. a) The following are equivalent: (i) $X$ is compact. (ii) Every sequence in $X$ has at least one partial limit. b) For a nonempty subset $Y \subset X$, the ...


0

Yes, there always exists such a sequence. First, let $\{B_1(x) : x \in X\}$ be a cover of $X$ by open balls of radius $1$. By compactness, this has a finite sub-cover $\{B_1(x_1),...,B_1(x_n)\}$. Let $(x_1,...x_n)$ be the first $n$ terms of the sequence. Next, let $\{B_{1/2}(y) : y \in X\}$ be a cover of $X$ by open balls of radius $\frac12$. This has a ...


1

consider the following open covering of the plane: $$\{ (-\infty, 3) \times \mathbb{R}\} \cup\{ (p_n, p_{n+2}) \times\mathbb{R}|n = 1, 2, ... \}$$ where $p_n = 1 + \frac{1}{2}+ \cdots + \frac{1}{n}$. There is no "Lebesgue number" for this covering, so the Lebesgue Lemma fails.


0

Hints: Let $X=[0,1]\times[0,1]$ with usual topology is a compact metric space. So the sequence that $\{(\frac1n,0)\}$ is the requried sequence. Since each point that $(\frac1n,0)$ is a limit point. It is the limit of this sequence $\{(\frac1n, \frac1m)\}_{m\ge 1}$


0

This follows from $\sup X\le \sup Y$ for $X\subseteq Y$. Which again follows from $\forall x\in Y\colon x\le s\implies \forall x\in X\colon x\le s$


0

If $x_{k(n)}$ had a subsequence converging to $x^*$, then the inequality $d(x_{k(n)},x^*)<\epsilon_0$ would hold for all sufficiently large $n$. (Why? Review the definition of limit). But this is not the case.


2

You're proof seems mostly all right, though I find it hard to follow towards the end. If I were to write a proof along the same lines, I would write it as follows: Let $\{ y_{n} \}_{n=1}^{\infty}$ be a sequence in $X$. $X$ is totally bounded, so we can find finitely many $x_{i} \in X$ ($i = 1, \dots, n $) such that $X = B(x_{1}, \frac{1}{2}) \cup \dots ...


1

The basic idea is entirely correct, but it’s a bit sloppy in spots. I would do something like this. Since $X$ is totally bounded, there are an $x_0\in X$ and an infinite $N_0\subseteq\Bbb N$ such that $y_n\in B(x_0,1)$ for all $n\in N_0$. Given an infinite $N_k\subseteq\Bbb N$ and an $x_k\in X$ such that $y_n\in B(x_k,2^{-k})$ for all $n\in N_k$, total ...


2

When you are given just a vector space $X$, then you have as many non-equivalent norms on $X$ as many non-isomorphic normed spaces you can find with the same linear dimension. This is the standard `structure transport' argument. For instance, suppose that you are given a vector space of dimension continuum. Then for each infinite-dimensional Banach space $Y$ ...


0

For the first one, suppose not, then $A\cap C\neq \emptyset$ and $A\setminus C\neq \emptyset$, which are open set in $A$ separates $A$. Then it contradicts $A$ is connected. For the second one, you can prove if $A$ is connected, $x$ is a limit point of $A$, then $A\cup \{x\}$ is also connected, since if it's not, there are two open sets $G_1,G_2$ separate ...


1

Where $\phi$ is latitude and $\theta$ is longitude, and your chosen coordinate setup: $$(\cos\phi\times\sin\theta, \cos\phi\times\cos\theta, \sin\phi)$$ To see how this works, consider the circle that is the parallel of latitude. What $z$ coordinate does it have? What is its radius? Major Warning: math libraries tend to work in radians, not degrees, for ...


2

Let $h:A\to[0,1]$ be a homeomorphism such that $h(v)=1$. By the Tietze extension theorem there is a continuous $H:X\to[0,1]$ such that $H\upharpoonright A=h$. $X$ is metrizable, so there is a continuous $g:X\to[0,1]$ such that $A=g^{-1}[\{1\}]$; let $$f:X\to[0,1]:x\mapsto H(x)g(x)\;;$$ clearly $f$ is continuous, $f\upharpoonright A=h$, and $f(x)=1$ iff ...


4

Let $K$ be compact, then $$ \{B(x,\varepsilon) : x\in K\} $$ is an open cover of $K$, and hence it has a finite subcover: $$ K\subset B(x_1,\varepsilon)\cup\cdots\cup B(x_n,\varepsilon). $$ Note however, that every non-compact subset of this compact set $K$ has the same property, as it can also be covered by finitely many such balls. In particular, every ...


0

For 1,2 no. Consider $f(x)=\frac{1}{x}$ on $(0,1]$ For 3, no. Consider identity map from $[0,1]$ to $[0,1]$ , where the first one with discrete metric and the second one with standard metric.


3

Yes. An ordered field will work. http://en.wikipedia.org/wiki/Generalised_metric


0

A closed set contains its limit points. To prove that we need to show that the set, $\{y ∈ X | r ≤ d(x,y) ≤ s\}$, contains its limit points $r$ and $s$. This is obvious since any open ball of radius $\epsilon$ centered at $r$ or $s$ will have a nonempty intersection with both the set and the complement of the set $\forall$ $\epsilon>0$. Thus, it can be ...


3

There are plenty of non-equivalent norms. Let $X$ be an infinite-dimensional normed space with norm $\|\cdot\|_X$. Let $Y$ be another normed space with norm $\|\cdot\|_Y$. Let $T\in \mathcal L(X,Y)$ be compact and injective. Then $$ \|x\|_T:=\|Tx\|_Y $$ is a norm on $X$. Moreover, $\|x\|_T \le \|T\|_{\mathcal L(X,Y)} \|x\|_X$. However, both norms cannot ...



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