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Hot answers tagged metric-spaces

6

So, I must prove that if I have an open cover in $M\times N$, with a finite open subcover, I must have an open cover in $M$ and $N$ with finite open subcovers. However, I don't have any idea on how to prove that. No, that is not what you must prove. What you must prove is that assuming that $M\times N$ is compact, if you have an open cover of $M$, then it ...

5

No, it's possible to construct an uncountable closed set containing only irrationals. Let $\{p_n\}$ be an enumeration of $\mathbb{Q}$. Put $I_n = (p_n - 2^{-n-1},p_n+2^{-n-1})$. Then each $I_n$ has measure $2^{-n}$ (if you haven't learnt Lebesgue measure before, it's just the length of $I_n$). Their union $U$ is an open set containing all rationals, but has ...

4

No. Let $X=[1,\infty)$. Define $$d_1(x,y)=|x-y|,\quad d_2(x,y)=\left|\frac 1x-\frac 1y\right|.$$Then $d_1$ is certainly complete, but $(x_n)$ is a Cauchy sequence with respect to $d_2$ if $x_n=n$.

4

No. Consider as your model set $M$ a tripod. (A tripod is a graph with one vertex of degree three and three vertices of degree one attached to it.) For your different length metrics, just assign the edges of the tripods various different lengths, and equip the space with the length metric induced by those edge lengths (making each edge into an isometric ...

3

Take $X$ to be an uncountable set with the cocountable topology (a set is open if and only if its complement is countable). You can verify that the compact subsets of $X$ are finite sets with the discrete topology. Hence, take any $A\subseteq X$ that is not open. Then, for any compact $K\subseteq X$, $A\cap K$ is open in $K$.

3

Your notation suggested another alternative: Let $g^2 = (g^1)^2$ -- a local extremum of $g^1$ is also a local extremum of its square. In fact, let $\phi$ be strictly increasing on $[ 0, \infty )$, then $g^2 = \phi(g^1)$ replicates the extrema of $g^1$. So geodesics (paths whose length is first order stationary under perturbation, hence correspond to local ...

3

It is clear that $d$ is defined on $\mathbb R^2$, because no matter what $u_1,u_2, v_1, v_2$ are, you can calculate $d(u,v)$. You show that $d$ is a metric by showing that it satisfies all the properties that metrices must satisfy: Show that $d(u, v)\geq 0$ for all $u,v$. Show that $d(u,v)=0 \implies u=v$. Show the triangle inequality.

3

$T$ cannot be injective. If $T$ is constant this is obvious, so assume $T$ is not constant. Then, as you explained already, $T$ attains its minimum an maximum in $x_0$ and $x_1$, say, respectively. Now choose two different paths from $x_0$ to $x_1$ which are disjoint with the exception of the common endpoints. The image of each covers the interval $[T(x_0), ... 2 The inverse image of a closed set under a continuous map is closed and since$\{y \in \mathbf{R}: y \leq \epsilon\}$is closed, it follows that$\{x \in M: \phi(x) \leq \epsilon\}$is closed. 2 Assuming that you’ve reproduced it accurately, that proof makes life difficult by using$d$both for the metric on$M\times K$and for the metric on$M$. In fact the metrics just get in the way: the result is true for an arbitrary space$M$and an arbitrary compact space$K$, and I think that the proof is actually a bit clearer in that setting, so let me ... 2 It’s basically correct. There’s a typo at the very beginning, where you meant to write ‘For each$n$there exists’ (instead of$j$). And you need to pass to a convergent subsequence only once: the tail sequence$\langle x_n:n\ge k(j)\rangle$already converges to$\hat x$, since it’s a subsequence of a sequence converging to$\hat x$. For a proof in case$X$... 1 A space$X$is anticompact if and only if the only compact subsets of$X$are the finite subsets. Let$X$be$T_1$, anticompact, and not discrete. Since$X$is not discrete, it has a non-empty subset$A$that is not open.$X$is$T_1$, so the relative topology on each finite subset of$X$is discrete, and therefore$A\cap K$is open in$K$for each compact$...

1

The question seems a bit unclear to me, but I guess you are asking: How to see that the Baire space has a countable base. The standard base for the product topology on $\newcommand{\N}{\mathbb N}\N^{\N}$ is the base consisting of sets $$\mathscr N_{a_0,a_1,\dots,a_n}=\{x\in\N^{\N}; x_0=a_0, x_1=a_1,\dots,x_n=a_n\}.$$ I.e. the basic sets consist of sequences ...

1

You know that given $\epsilon>0$ there is $N$ such that $$|f_n(x)-f_{n+p}(x)|\le\epsilon\quad\forall x\in E,\quad\forall p\in\mathbb{N}.\tag{1}$$ You also know $\lim_{n\to\infty}f_n(x)=f(x)$. Now let $p\to\infty$ in (1).

1

All correct (I think I wrote that on wikipedia) $$d(O, Q) = \frac{1}{2} \ln \frac{1+r}{1-r} = \operatorname{artanh} r$$ see also the formulas for tanh and artanh at en.wikipedia.org/wiki/Hyperbolic_function I have added it to the wikipedia page ps related: For the Poincare disk model he hyperbolic distance between O the center of unit disk and a point ...

1

Let us prove the statement for a fixed $n$. Denote$$f_n(x_1,x_2,\ldots,x_n,x) = \frac{d(x_1, x) + d(x_2, x) + \ldots + d(x_n, x)}{n},$$$$M(x_1,x_2,\ldots,x_n)=\max_{x\in X}f_n(x_1,x_2,\ldots,x_n, x),$$$$m(x_1,x_2,\ldots,x_n)=\min_{x\in X}f_n(x_1,x_2,\ldots,x_n,x),$$and let $I(x_1, x_2, \ldots, x_n)$ be the closed interval $[m(x_1, x_2, \ldots, x_n), M(x_1, ... 1 Your reasoning for$d(A,B)=1$is flawed.$B$does not need to be a subset of the$2\delta$-interval centered at$0$. Instead,$B$needs to be a subset of$A_\delta$which is equal to the union of$2\delta$intervals centered on all the points of$A$, that is$$B \subset [0-\delta,0+\delta] \cup \bigcup_{n \in \mathbb{N}} \,\, \biggl[\frac{1}{n}-\delta,\... 1 Let$(X,d_X)$and$(Y,d_Y)$be two metric spaces, with$X$compact, and let$f:X\to Y$be a continuous bijection. Definition: The function$f$is continuous if for every set$O \subset Y$which is open in$Y$,$f^{-1}(O)$is open in$X$. Lemma: The function$f$is continuous if and only if for every set$C \subset Y$which is closed in$Y$,$f^{-1}(C)$is ... 1 I won't insert all the details into a hint, but consider this oft-used technique:$|x(t_2)-x(t_1)|<|x(t_2)-x_m(t_2)|+|x_m(t_2)-x_m(t_1)|+|x_m(t_1)-x(t_1)|$, the RHS of which is arbitrarily small. The existence of the function$x(t)$, the 'limit' of the Cauchy sequence, or more formally that function for which$\sup_{0\leq t\leq T}|x(t)-x_m(t)|$is ... 1 Writing$(-\infty,-1]\cup\{-1\}$is redundant since$-1\in(-\infty,-1]$. When you write cl in LaTeX, write \operatorname{cl} instead; it will give it the right styling. Lastly,$\operatorname{cl}\mathbb{Q}=\mathbb{R}$, since each sequence in$\mathbb{Q}\$ converges to a real number and any real number can be written as a convergent sequence of rationals.

1

I would like to say a few things here...are you studying analysis or topology? If you are studying analysis then you can talk about closure which directly means closure in standard topology but if you are studying topology you need to ask first is it standard topology or discrete topology or lower limit topology or K topology and stuff like that. Assuming ...

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