# Tag Info

5

This cover has no finite subcover : $$\Bbb Q \cap [0,1] = \bigcup_{n} \left( \Bbb Q \cap \left( \left[0,\ln(2)-\frac{1}{n}\right[\;\cup\; \left]\ln(2)+\frac{1}{n},1\right] \right)\right)$$ Edit : forgot to add the $\frac{1}{n}$

3

If we are allowed to take steps in any of the eight directions you describe, the number of steps it takes to get from $(x_1,y_1)$ to $(x_2,y_2)$ is the maximum of $|x_1-x_2|$ and $|y_1-y_2|$. This is often called the $L^\infty$ metric.

3

Actually the problem has nothing to do with square roots except for the property of $x\mapsto\sqrt x$ to be a bijection. If $f:X\to(Y,d)$ is any injective function, where $X$ is a set and $Y$ is a space with metric $d$, and if we equip $X$ with the metric $d_f(x,x')=d(f(x),f(x'))$, then $f$ becomes an isometry $(X,d_f)\to(Y,d)$, i.e. a map which preserves ...

2

HINT: One direction is immediate. For the other direction, suppose that for each $\epsilon>0$ there is a countable cover of $X$ by open $\epsilon$-balls. By taking $\epsilon=\frac1n$ for each $n\in\Bbb Z^+$, show that $X$ is separable. Then show that a separable metric space is second countable and therefore Lindelöf. Note, though, that your definition ...

2

Let $x,y,u,v \ge 0$, we first want to prove that $$\sqrt{(x + y)^2 + (u + v)^2} \le \sqrt{x^2 + u^2} + \sqrt{y^2 + v^2}. \tag 1$$ Squaring both sides and simplifying, the previous inequality is equivalent to $$xy + uv \le \sqrt{(x^2 + u^2)(y^2 + v^2)}.$$ Squaring again we obtain \begin{align} x^2y^2 + u^2v^2 + 2xyuv \le &\ x^2y^2 + x^2v^2 + ...

2

Hint: it's all in the last step. What's an equivalent condition to $|e^x-e^y|=0$? What is special about the exponential function that allows you to conclude from this that $x=y$? Try to answer without mentioning logarithms!

2

Judging from your proof, you've completely misunderstood the problem. You're supposed to show that every Cauchy sequence $(f_n)_{n = 1}^\infty$ in $C[0,1]$ converges to an element $f$ of $C[0,1]$. Given a Cauchy sequence $(f_n)$ in $C[0,1]$, we have for each $x\in [0,1]$, $(f_n(x))_{n = 1}^\infty$ is Cauchy in $\Bbb R$. Since $\Bbb R$ is complete, there ...

2

If $f$ has no infimum $> 0$, then $f^{-1}((\epsilon, \infty)) \ne M$ for any $\epsilon > 0$.

2

$(0,0)$ is a limit point but $(0,0)\notin M$ (Put $x_n=\frac{1}{2n\pi} \Rightarrow x_n\to 0 , \sin x_n=\sin \frac{1}{\frac{1}{2n\pi}}=\sin 2n\pi=0$ Theorem Subspace of Complete Metric Space is Closed if and only if Complete.

2

You could use $$\max(a,b) = \frac{a + b + |a - b|}{2},$$ where $a = |x_1 - y_1|$ and $b = |x_2 - y_2|$.

2

In short, yes if something in math has a property that has a definition, then the definition can be used to prove that property. It would be a bit odd for a set to be closed, yet you would be unable to prove it is closed using the definition. That being said, there are multiple equivalent definitions of what it means to be closed and you mentioned two of ...

2

It is proven by Swenson in E. Swenson, A cut point theorem for CAT(0) groups, J. Differential Geom. 53 (1999), no. 2, 327–358. that each infinite CAT(0) group contains an infinite order element.

1

Yes you are right the set of cluster points of your given set is $\{(x,y) \in \mathbb{R}^2 , x \geq 0, y \geq 0 \}$ (i.e. all the points of 1 st quadrant of \mathbb{R}^2 with points on the axes). To prove this formaly you may use the definition used here Definition of cluster point .

1

Consider $A^C$. Let, $p$ be a limit point of $A^C$ and let $\{x^{(n)}\}$ be a sequence of points in $A^C$ converging to $p$. Then, $\lim_{n\to \infty}x^{(n)}_k=p_k,\ k=1,2,\cdots$. Since $x^{(n)}\in A^C$, $$x^{(n)}_1\ge \sum_{k\ge 1}x^{(n)}_k2^{-k}\implies \lim_{n\to \infty}x^{(n)}_1\ge \sum_{k\ge 1}\lim_{n\to \infty}x^{(n)}_k 2^{-k}\implies p_1\ge ... 1 Hint for convergence:$$\|T(x) - T(y)\|^p = \sum_j |x_j - y_j|^p |x_j + y_j|^p \le \|x - y\|^p \sup_j |x_j + y_j|^p$$Hint for T(A): a sequence in T(A) has a subsequence that converges elementwise. Does it converge in norm? 1 It's perfectly fine to show that a finite set is closed because it's equal to its own closure. But to compute the closure, you have to take the set of all boundary points of A union A, not the set of interior points! So you need to show that a finite set has no boundary points, in other words, that no sequence contained in a finite set converges outside ... 1 The taxi-cab metric is isometric to the max metric only for dimensions 1 and 2, and for higher dimensions they are not isometric because their unit balls have different number of extreme points (isometries of normed spaces carry extreme points to extreme points of the corresponding unit balls). The Euclidean metric is not isometric to none of the above ... 1 This is wrong. The statement is clearly false. For example, Let M = N = \Bbb R, and let$$f(x) =\begin{cases} 1 & x \in [0, 1] \\ 0 & x \notin [0,1]\end{cases}.$$Then f is continuous on [0,1], but is not continuous on \Bbb R. More generally note that there is absolutely nothing given that restricts the function outside of K. You can give ... 1 Using your condition that X is compact if and only if X is complete and totally bounded. Assume that X is not compact, then X is either not complete or not totally bounded. If X is not complete, there is a Cauchy sequence \{x_n \} so that it has no limit. Let C_n = \{ x_k: k\ge n\}. Then C_n are all closed (if C_n has a limit point x, ... 1 First of all, note that$$\begin{align}d(x,y) \\&=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2} \\&\le \sqrt{2\max\{|x_1-y_1|^2,|x_2-y_2|^2\}}\\&=\sqrt2 \max\{|x_1-y_1|,|x_2-y_2|\}\\&=\sqrt2 d^*(x,y)\end{align}.$$On the other hand,$$\begin{align}d^*(x,y)\\&=\max\{|x_1-y_1|,|x_2-y_2|\}\\&=\sqrt{\max\{|x_1-y_1|^2,|x_2-y_2|^2\}}\\&\le ...

1

$U$ could be any open set containing $(1/2, 1]$. So $U$ could be, say, $(1/2, 2)$ which is open in $\mathbb R$. So $B = U \cap A$ so $B$ is open in $A$. The point is an open set is that for every point $x$ (even the 1) of there is an $\epsilon$ where all $y$ in $A$ that $|y - x| < \epsilon$. This is true for the point $1$ as there are no points larger ...

1

First note that it is always true, that $D(x,z) \leq d(x,z)$, $D(x,z)\leq 1$ and that $d(x,z) \leq 1 \Rightarrow d(x,z)=D(x,y)$. You need to look at two cases: Case 1: If $D(x,y)+D(y,z) <1$ then you know that both $D(x,y) <1$ and $D(y,z) <1$. Because $d$ satisfies the triangle inequality, we find that: $D(x,z)\leq d(x,z) \leq d(x,y)+d(y,z) = ... 1 As Daniel wrote, you need a locally compact space for that. Local compactness is certainly necessary, as for$K=\{x\}$and$x\in U$, your property implies that$x$has a compact neighborhood$D$contained in$U$. But it is also sufficient. If$K\subseteq U$, then each point$x\in K$possesses an open neighborhood$U_x$contained in a compact set ... 1 The space$A=[0,\infty)$with the norm$d(x,y)=|\sqrt{x}-\sqrt{y}|$is complete. Here's a more elementary proof. Let$(a_n)$be a Cauchy sequence in$(A,d)$, and$\varepsilon$a positive real number. Then, there is a positive integer$N_\varepsilon$such that $$d(a_n,a_m)\le \varepsilon \quad \forall m,n\ge N_\varepsilon.$$ For$\varepsilon=1$, we get$$... 1 HINT: Show that$Z$is dense in$B\$.

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