7

The set $[0,1]$ is an open subset of $[0,1]$ with respect to the Euclidean metric. It is true that $[0,1]$ is not an open subset of $\Bbb R$, but that's not relevant here.


5

The family $$\mathcal H:=\left\{\left(\frac12,1\right); \left(\frac13,\frac12\right); \left(\frac13,\frac14\right);\cdots\right\}=\left\{\left(\frac1{n+2},\frac1{n+1}\right)\,:\, n\in\Bbb N\right\}$$ seems quite countable to me and $S=[0,1]\setminus\bigcup\mathcal H$, where $$\bigcup\mathcal H:=\{x\in\Bbb R\,:\, \exists I\in \mathcal H,\, x\in I\}=\left\{x\...


3

Yes $f: \Bbb R \to \Bbb R$ defined by: $$f(x) = \begin{cases} 1 & x \in \Bbb Q\\ 0 & x \notin \Bbb Q\\ \end{cases}$$ is everywhere discontinuous but constant (so continuous) on both dense sets $\Bbb Q$ and $\Bbb R\setminus \Bbb Q$.


3

$\Rightarrow)$ Suppose $A'$ is dense in $X$. Fix $x\in X$ and $\varepsilon>0$. By assumption, there is some $b\in A'$ such that $0<d(x,b)<\frac{\varepsilon}{2}$. Since $b\in A'$, there is some $a\in A$ with $0<d(a,b)<\frac{\varepsilon}{2}$. Note $d(x,a)<\varepsilon$ by the triangle inequality. This shows that $A$ is dense. The other ...


3

As pointed out in the comments, Heine Borel does not apply. Let $x := \lim x_n$ and $X$ be the set in question. Let $\cal U$ be an open cover of $X$. There exists some $U_0 \in \cal U$ containing $x$. Since $U_0$ is open, there exists some $N \in \Bbb N$ such that $x_n \in U_0$ for all $n > N.$ Thus, $U_0 \setminus X$ has at most $N$ elements. For each ...


2

The problem with this idea is that $d_Z$ is not a metric unless $Y$ consists of a single point: if $y_0$ and $y_1$ are distinct points of $Y$, and $x\in X$, then $$d_Z(\langle x,y_0\rangle,\langle x,y_1\rangle)=0$$ even though $\langle x,y_0\rangle\ne\langle x,y_1\rangle$. However, it is true that $S_1\times Y$ and $S_2\times Y$ are open in $Z$. This is ...


2

Here is a proof that uses continuous functions: since $X$ is disconnected, there is a continuous surjection $f:X\rightarrow \{0,1\}$ where the latter is equipped with the discrete topology. Then $g:X\times Y \rightarrow \{0,1\}$ given by $g(x,y)=f(x)$ is a continuous surjection as well (it is equal to $f$ composed with the projection map onto the first ...


2

The continuous image of a connected space is connected, so from this the "positive" version of your statement follows: If $X \times Y$ is connected then $X$ and $Y$ are connected, which is logically equivalent to $X$ disconnected or $Y$ disconnected implies $X \times Y$ disconnected. (let $p="X \text{ connected}"$, $q = "Y \text{ connected}"$, $...


2

I begin by commenting something that really got me confused at first. In the statements, you really need $\subset$ without equality being allowed, otherwise the statement as you put it is not true: take $X = A= [0, 1]$ with $d$ the regular metric. This $X$ has certainly interior points, but the distance between any two points is bounded by $1$, which means ...


2

Not true: Take $n=1, M=(0,1)$ and $g_2(x)=|\frac 1 x-\frac 1 y|$. Then $g_2$ has the same open sets as $g_1$ and the inclusion you have written says that $\frac 1 x$ is uniformly continuous on $(0,1)$ which is false.


2

Strictly speaking, we should always refer to "metric tensor field" because it's actually a tensor field, although for convenience we simply refer to "metric tensor". You're right that the metric tensor field (usually denoted as $g$) assigns to each point $p$ of the manifold an inner product (which is of course a $(0,2)$ tensor) $g_p$ on ...


2

A point on the line has coordinates $P(t,3-t)$ The squared distance from the center of the circle $O(0,0)$ is $$OP^2=f(t)=t^2+(3-t)^2=2t^2-6t+9$$ This is minimum when $f'(t)=0$ and $f''(t)>0$ $f'(t)=4t-6=0\to t=\frac32$ $f''(t)=4$ so $P^*\left(\frac{3}{2},\frac{3}{2}\right)$ is the point of the line closest to the circle. $OP^*=\frac{3}{\sqrt{2}}$ So the ...


2

These things are often called "annuli" (in the singular, "annulus"), from the Latin for something that looks like a ring. It makes intuitive sense that these things are arcwise connected because there is nothing stopping you from drawing a curve from any point in one of these things to any other without leaving the annulus. If you want ...


2

The Heine-Borel theorem does not hold in metric spaces in general. However, the result is easy to prove directly from the definition of compactness: start with an open cover $\mathscr{U}$ of the set in question, and show that it has a finite subcover. HINT: Some member of $U$ must contain the limit of the sequence.


2

There are metric spaces with finite open sets (e.g. the discrete metric on any set). Also, the complement of a closed set is always open. (2) and (3) follow directly from the definitions: $x\in X\setminus A^\circ$ means that every open neighborhood $U$ of $x$ is not entirely in $A$, that is, intersects $X\setminus A$. And the closure $\overline B$ of any ...


1

First of all, I highly recommend Burago, D.; Burago, Yu.; Ivanov, S., A course in metric geometry, Graduate Studies in Mathematics. 33. Providence, RI: American Mathematical Society (AMS). xiv, 415 p. (2001). ZBL0981.51016. specifically, Chapter 7, for a treatment of the GH-distance and convergence. If you want an example of a sequence of metric spaces which ...


1

Your proof is almost correct, except for the justification of openness of $(S_1\times Y)$-- the space $X\times Y$ comes equipped with a very specific topology (the product topology) and your task is to show $(S_1\times Y)$ is open in that particular topology. (Otherwise, you could just say "let $d_Z$ be the discrete metric, then every set is open hence ...


1

HINT: Try $X=[0,2]$, $Y=[0,1)\cup[2,3]$, and $$f:X\to Y:x\mapsto\begin{cases} x,&\text{if }0\le x<1\\ x+1,&\text{if }1\le x\le 2\,. \end{cases}$$ Here $X$ and $Y$ have their usual metrics as subspaces of $\Bbb R$ with its usual metric.


1

Edit: Apologies to @Raffaele I took too long writing this and by the time I'd finished you'd already posted the answer. Leaving mine up in case it's useful, but please don't award me points that should go elsewhere! Points on the line are given by $(x, f(x))$ where $f(x) = 3-x$. If we have some other point $(a, b)$, we can find the distance from this point ...


1

You can take $d\bigl((x,y),(z,t)\bigr)=|x-z|+(y-t)^2$. The restriction to $\{(x,0)\mid x\in\Bbb R\}$ is the usual distance there. But\begin{align}d\bigl((0,0),(0,1)\bigr)+d\bigl((0,1),(0,2)\bigr)&=2\\&<4\\&=d\bigl((0,0),(0,2)\bigr).\end{align}


1

Given $r>0$, pick $s > 0$ so that $s^2 = \frac{1}{2}r^2$, explicitly we define $s=\sqrt{\frac{r^2}{2}}$. If then $\textbf{y} \in B^{d_1}_s(\textbf{x})$ we know that $$d_S(x_1, y_1) \le d_1(\textbf{x},\textbf{y}) < s$$ and also $$d_T(x_2, y_2) \le d_1(\textbf{x},\textbf{y}) < s$$ So that $$d_S(x_1, y_1)^2 + d_T(x_2, y_2)^2 < 2s^2 = 2\cdot \frac{...


1

Let $f: \Bbb N \to \Bbb Q$ be a bijection. Define $d(n,m)=| f(n)-f(m)|$.


1

You are almost done: Observe that $$d(x_k,x_l)\leq \mathrm{diam}(E)$$ holds (by using the definitions of $\mathrm{diam}$). Thus, we have $$ d(x_k,z) \leq \mathrm{diam}(E)+ d(x_l,z). $$ By taking the infimum over $x_l\in E$, we get $$ d(x_k,z)= \inf_{x_l\in E} d(x_k,z) \leq \mathrm{diam}(E)+ \inf_{x_l\in E} d(x_l,z) = \mathrm{diam}(E)+ \mathrm{dist}(z,E) = r. ...


1

Suppose that $\textrm{diam}(E) = C < +\infty$, so $E$ is bounded in the diameter sense, let $z=x_1$ and $r= C+1 >0$. Then I claim that all points of $(x_n)_{n \in \Bbb N}$ are in $B_r(z)$: Let $x_m$ be arbitrary, then $d(x_1, x_m) \le \textrm{diam}(E)$ as the diameter is an upper bound for all the distances between members of $E$. In particular $d(z,...


1

Actually, the sequence $(1/n)_{n \geq 1}$ is Cauchy in $\mathbb{R}$, and $\exp$ is uniformly continuous over $[0,1]$, so the sequence $(\exp(1/n))_{n \geq 1}$ is also Cauchy in $\mathbb{R}$, which means that $(1/n)_{n \geq 1}$ is Cauchy for your metric $d$.


1

Your approach won't work since $e^{1/n} > e^0 = 1$ for any $n \in \mathbb{N}$, and we can find no bound $e^{1/n_\epsilon} < \epsilon <1$. Instead, for $l > k > n$, we have $$\left|e^{\frac{1}{l}} - e^{\frac{1}{k}} \right|= e^{\frac{1}{k}}(1 - e^{\frac{1}{l}- \frac{1}{k}}) \leqslant e \cdot \left(1 - e^{-\left(\frac{1}{k}- \frac{1}{l}\right)} \...


1

I assume you have proved that there is a finite set $S$ such that every point of $X$ is within distance $\varepsilon$ of some point of $S$ (cover every point with $\varepsilon$-balls, extract a finite subcover). For the second part, try induction on the size of the covering. A valid inductive hypothesis might be, "suppose that for any compact metric ...


1

Let $f_n(t): \frac{1}{n} \sin (nt).$ Then $||f_n||_{\infty} \le \frac{1}{n}$ for all $n$. Hence $$ ||f_n -0||_{\infty} \to 0.$$ But we have $(h(f_n))(t)= \cos (nt)$ and therefore $(h(f_n)))$ does not converge to $(h(0))$ in the norm $||\cdot||_\infty$


1

Continuity of $h$ means given $f$ and $\epsilon >0$ there exist $\delta >0$ such that $\|g-f\|_{\infty} <\delta$ implies $\|h(f)-h(g)\|_{\infty} <\epsilon$. In your example $\|h-f\|_{\infty} <\delta$ is not satisfied for any $\delta \in (0,1)$ so the example says nothing about continuity of $h$. [BTW, you should avoid using $h$ with two ...


1

You can just define: \begin{equation} f(x) = \begin{cases} 1 & \text{if $\ x \in D $}\\ 0 & \text{if $\ x \in \mathbb{R} \setminus D$} \end{cases} \end{equation} It is a discontinuous function at each point of $D$ and it is also continuous at each point of $\mathbb{R} \setminus D$ assuming $D$ is a finite set or an infinite set without any ...


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