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0

Suppose you have two circles, $C_1$ and $C_2,$ with respective radii $r_1$ and $r_2.$. Without loss of generality, suppose $r_1\le r_2.$ Let $C_1'$ and $C_2'$ be the circles' respective translations to circles centered at the origin. Then radial projection will allow you to map $C_1'$ to $C_2'.$ Can you put the pieces together to construct a formula that ...


5

Consult your definitions—the interior of a set is its maximal open subset. If your topology is e.g. discrete, then $\{ x \}$ is open, so it is its own interior. But in the standard topology on the reals, it's not open. In this case, the empty set is its only open subset, so that is its interior. Both of the above cases are metrisable, so the restriction to ...


0

The $\epsilon_i$ at the very end of the displayed inequality $$\sum_{i=0}^mk^{m-i}\epsilon_{i}\leq k^{m-N}\sum_{i=0}^Nk^{N-i}\epsilon_{i}+\epsilon\sum_{i=N+1}^mk^{m-i}\epsilon_{i}$$ should not be there: it’s $$\sum_{i=0}^mk^{m-i}\epsilon_{i}\leq k^{m-N}\sum_{i=0}^Nk^{N-i}\epsilon_{i}+\epsilon\sum_{i=N+1}^mk^{m-i}\;.$$ Also, ...


2

The metric $d_2$ needs to be defined on the whole space. Are you taking $\arctan(\infty) = \frac \pi 2$? I'll assume that's the case. In order for $\{\infty\}$ to be open, there must be $\epsilon > 0$ satisfying $B(\infty,\epsilon) \subset \{\infty\}$ However, for every $\epsilon > 0$ you have $$B(\infty,\epsilon) = \{b \in X : d_x(\infty,b) < ...


0

A priori, $\arctan \infty$ is not defined, hence you failed to define $d_2$ where one of $a,b$ is infty. However, we may use the limit to make the pretty definitoin $\arctan\infty:=\frac\pi 2$. With this definition, $\{\infty\}$ is not open. The $\epsilon$-ball around $\infty$ is precisely the set of $a$ with $\arctan a>\frac\pi2-\epsilon$, i.e. the set ...


1

Consider the functions $f_n(x)$ where they take on value $0$ for $x< \frac{1}{2} - \frac{1}{n}$, $1$ for $x > \frac{1}{2} + \frac{1}{n}$ and grow linearly from $0$ to $1$ on $\frac{1}{2} - \frac{1}{n} < x < \frac{1}{2} + \frac{1}{n}$. Then, $d(f_n, f_m) \leq 4 \frac{1}{n}$ for $n <m$ (since each $f_n$ is bounded between $0$ and $1$ and the ...


2

Going from $f\big|_{A_{i+1}}$ continuous to $f$ continuous at $x$ requires some argument. Since $f:A_{i+1} \rightarrow Y$ is continuous, $f:$ Int$(A_{i+1}) \rightarrow Y$ is continuous. Therefore, for any neighborhood $V$ of $f(x)$, there exists a neighborhood $U$ of $x$ open in Int$(A_{i+1})$ such that $f(U)\subset V$. Since Int$(A_{i+1})$ is open in $X$, ...


2

Since you call the metric space $(X,k)$, I guess $k$ is the distance function? All right, then define $$M=\{x\in X:k(x,Y)\lt k(x,Z)\},$$ $$N=\{x\in X:k(x,Z)\lt k(x,Y)\}.$$


1

all $y \in Y$ there exist $\delta_y$ such that $$ B(y,\delta_y) \cap Z =\emptyset, $$ since $y$ is not in the closure of $Z.$ Take $M=\cup B(y,\delta_y/2)$. Define $N$ analogously. If $p \in M \cap N,$ then we have $y \in Y,z \in Z$ such that $$ p \in B(y,\delta_y/2) \cap B(z,\delta_z/2). $$ If $\delta_y \leq \delta_z$ then $y \in B(z,\delta_z)$ which ...


0

Notice first that $$ \left(\frac{1}{n+1},\frac1n\right)\subset [0,1] \quad \forall n\ge 1, $$ we have $$ E=\bigcup_{n\ge 1}\left(\frac{1}{n+1},\frac1n\right)\subset [0,1], $$ and therefore $$ \overline{E}\subset \overline{[0,1]}=[0,1]. $$ Since $$ \left(\frac{1}{n+1},\frac1n\right)\subset E \quad \forall n\ge 1, $$ we have $$ ...


2

To get a broader context for the well above relation you can consult any introductory text on domain theory. However, the context of continuity spaces is a bit different, and I prefer to have the intuition for Flagg's value quantales come directly from their intended role. So, the way I think about the well above relation is that it solves some nasty ...


2

I’ve not dealt much with the concept, but an example that I find helpful is $\Bbb R^2$ with the product partial order, $\langle x_0,y_0\rangle\le\langle x_1,y_1\rangle$ iff $x_0\le x_1$ and $y_0\le y_1$. Then it’s not hard to check that $\langle x_0,y_0\rangle\prec\langle x_1,y_1\rangle$ iff $x_0<x_1$ and $y_0<y_1$. Consider, for instance, the set ...


1

I cannot in good faith answer question 1, because I didn't see enough examples of that relation to get an informal idea of it. As for 2, first note that $q \succ p$ implies $q \geq p$ (as stated in the linked article, too), and here's an example where the two relations differ: Consider the set $S = \{a,b,c,d\}$, so that its power set $V$ is a complete ...


0

First, lets notice that $$E=(0,1]\backslash \left\{ \frac{1}{n} \right\}_{n\geq1}$$ By double contention: $$E\subseteq (0,1]\backslash \left\{ \frac{1}{n} \right\}_{n\geq1}$$ be $x\in (0,1)\backslash \left\{ \frac{1}{n} \right\}_{n\geq1}$, let be $n$ such as $\frac{1}{n+1}<x<\frac{1}{n}$ thus $x\in E$. $$(0,1]\backslash \left\{ \frac{1}{n} ...


1

As seems to be my habit, I'm going to say a lot of things here whose details I have not checked. Use at your own risk :). Given a metric space $(X,d)$, there is a geodesic remetrization $(X,d_G)$, equipped with a bijective short map $(X,d_G) \to (X,d)$. The geodesic metric is given by $d_G(p,q) = \sup_{\epsilon>0} \inf_{\{p = p_0,p_1,\dots,p_n=q ...


0

This is false. $(0,1)$ and $\mathbb{R}$ with their usual metrics are homeomorphic, but $(0,1)$ is totally bounded and $\mathbb{R}$ is not even bounded. To convert this into an example with two metrics on the same set, take $X = (0,1)$ and define the metric $d$ as usual by $d(x,y) =|x-y|$. Now let $f : (0,1) \to \mathbb{R}$ be your favorite homeomorphism, ...


0

Suppose that $Y$ is a countable set which is dense in $X$, then for any open Ball $U$ with $x\in U$, $U\cap Y\not=\emptyset$. for any open set $U$, there exists an open Ball $B(x,\epsilon) \subset U$, by the property, there is $y \in B(x,\epsilon) \subset U$, which shows that $Y$ is dense in $X$.


1

This has been stated in the comments. Consider $A=(1,2)\cup(3,4)$. $x\in A$ implies $x\in (1,2)$ or $(3,4)$. Suppose $x\in (1,2)$. As $(1,2)$ is open, $ \exists \epsilon>0 \text{ such that } B(x,\epsilon)\subset (1,2)\subset A$. Similarly, consider the case when $x\in (3,4)$. So, $A$ is open. In other words, open sets in $\mathbb{R}$ need not be in the ...


1

The set of open intervals of the type $(a,b)$ is a basis for the usual topology on $\mathbb{R}$. This means that all open sets of this topology can be written as countable unions or finite intersections of open intervals. As an exercise, you can take $a<b<c$ and search all the open sets that you can define from these three points; e.g. $(a,b) \cup ...


0

We define a primitive open set on the reals ($\mathbb{R}$) as: $$a,b,x\in \mathbb{R} : a<x<b$$ and we write $(a,b)$. A union of two open sets is also open. Let $A$ and $B$ be two open sets, their union is open whether $A\cap B=\emptyset$ or not. Any finite union of primitive open sets is open. Any countable union of open sets is also open, and it ...


0

take $X=(0,1]\cup (2,3)$ and take $E=(0,1]$ Then $E$ is closed in $X$ since its complement is open and take the continuous function $f:(0,1]\rightarrow [1,\infty)$ which is surjective but its image $[1,\infty)$ is not bounded


1

The extended real line is not a metric space. You are correct in noting that $d(0,+\infty)$ must be larger than any $d(0,x)$ for any $x\in\Bbb R$, and therefore cannot be a real number. However the extended real line is metrizable. This means that we can redefine the metric completely, while preserving the same topology (and therefore the same convergent ...


1

We know that $g \circ f$ is open, so for your open set $U \subseteq X$ we known that $V = g[f[U]]$ is open. We need $f[U]$ to be open. Show that $g^{-1}[V] = g^{-1}[g[f[U]] = f[U]$ because $g$ is injective. It then follows using continuity of $g$ that $f[U]$ is open in $Y$.


1

The fact that $g$ is injective implies that for an open $U\subset X$ we have$$f(U)=g^{-1}(g(f(U)))=g^{-1}(g\circ f(U)),$$and the claim follows from $g\circ f$ being open and $g$ being continuous. Note that the above equation does not hold for a general $g$.


1

I don't think there will be a nice property. Consider a space $X$ with the discrete metric. Then $X$ is bounded, but every sequence in $X$ is either divergent or eventually constant.


1

Yes. With that metric it's homeomorphic to $[0,1]$ with the usual metric, so it's compact, separable, and metrizable. A compact metrizable space is automatically complete in every compatible metric.


0

One possibility is the following: you can define the metric in $\mathbb{N}\cup\infty$ as you suggest: $d(x,y)=|\frac{1}{x}-\frac{1}{y}|$, $d(x,\infty)=|\frac{1}{x}|$. Suppose you have a function $f:\mathbb{N}\longrightarrow Y$. Then The function $f$ is extensible continuously to $\bar{f}:\mathbb{N}\cup\infty\longrightarrow Y$ if and only if the sequence ...


2

(i) The interior of the set $(a,b]$ with respect to $[a,b]$ is $(a, b]$. Perhaps you mistyped the left parenthesis? Your reasoning for why there exists an open ball around $b$ which is full contained within $(a,b]$ is correct. But $a$ cannot be in the interior, since $a$ is not even in $(a,b]$ in the first place. (ii) The closure of $[a,b)$ with respect to ...


0

Let $X$ be a compact metric space and let $\{p_n\}$ be a Cauchy sequence in $X$. Then define $E_N$ as $\{p_N, p_{N+1}, p_{N+2}, \ldots\}$. Let $\overline{E_N}$ be the closure of $E_N$. Since it is a closed subset of compact metric space, it is compact as well. By definition of Cauchy sequence, we have $\lim_{N\to\infty} \text{diam } E_N = \lim_{N\to\infty} ...


0

Let $X$ be $\Bbb Q$, the rationals, with the usual metric, and let $A$ be the set of dyadic rationals, rationals that can be written as fractions whose denominators are powers of $2$. Then $A$ and $X\setminus A$ are both dense in $X$, and $X$ itself is only countable. HINT for the proof that $(A')'=A'$: If $A$ is dense, then $\operatorname{cl}A=X$, and that ...


0

think of the relative topology on A in connection with the disjoint partition: $$ A = (A\cap B) \cup (A \cap \bar B) $$ where $\bar B$, the complement of $B$, is also both open and closed. recall the definition of connected


2

Observe that $A\cap B\neq \emptyset$, and we have that $A\cap B\subseteq A$. Since $B$ is open and closed in $X$ then by the subspace topology $A\cap B$ is going to be open and closed in $A$. Since this set is nonempty and $A$ is connected we must have that $A\cap B = A$, thus $A \subseteq B$.


1

Suppose $A \subset S$ is complete. To prove that $A$ is closed, it suffices to prove that if $(x_n)$ is a sequence of points of $A$ which converges to $x \in S$, then $x \in A$. So let $x_n$ be such a sequence. Since $x_n$ converges in $S$, it is a Cauchy sequence in $S$. Therefore $(x_n)$ is also a Cauchy sequence in $A$, so by completeness of $A$, there ...


0

It seems the following. Put $$\hat X=\{(a_i)\in F^\Bbb N: \sum_{i=1}^{\infty} |a_i|<\infty\}.$$ Define on $\hat X$ a norm $$\|\mathbf{a}\|’ = \sum_{i=1}^{\infty} |a_i|.$$ This norm induces a metric on $\hat X$, which makes it a complete metric space. Moreover, $\|\cdot\|’|_X=\|\cdot\|$ and $X$ is a dense subspace of the space $\hat X$. So the ...


0

Any sequence of points in a compact metric space has a convergent subsequence. We know that a sequence that has a convergent subsequence is in fact convergent. So this easily shows us that a compact metric space is complete.


4

We have the following result in Set Topology: Prop. Let $X$ be a space with the two properties that: (i) each point has a neighborhood basis of closed neighborhoods ($X$ is regular), and (ii) every continuous image of it in a Hausdorff space is closed. Then $X$ is compact. Proof. Suppose first $X$ is Hausdorff (the case of metric spaces). Let $\{U_i\}$ an ...


0

Because $\verphi$ is proper ,there exists $ u\in X$ SUCH THAT $\verphi(u)<\infty$. let $ C={x\in X :d(u,x)\leq\varphi(u)-\varphi(x)}$.\ then C is a nonempty closed subset of X . we show that C in invariant under f. For each $x\in C$, we have \begin{center} $d(u,x)\leq\varphi(u)-\varphi(x)}$\\ \end{center} and hence from ...


1

Yes, this is correct: $X$ is not homeomorphic to any of the choices, and your reasoning is correct. You can also rule out the rectangular hyperbola by noting that it is not connected, while $X$ is connected, and you can also rule out the circle by noting that it’s compact, while $X$ is not.


0

The simplest argument is that $X$ is regular, Hausdorff, and second countable and therefore metrizable by the Uryson metrization theorem. It’s second countable, since the base used to define it is countable. Each of the basic open sets $a+b\Bbb Z$ is clopen, so it’s regular (indeed, completely regular): if $n\notin a+b\Bbb Z$, then $n+b\Bbb Z$ is an open ...


1

I believe that you need to have $\varphi:X\to[0,\to)$ (or at least have its range bounded below). I don’t immediately see a substantially different argument, but I can get rid of the nets for you. Let $X,d,\varphi$, and $f$ be as given in the statement of the theorem. Define a partial order $\preceq$ on $X$ by setting $x\preceq y$ iff ...


1

Notice that there is no Riemann tensor for an inner product space. Moreover, at one chosen point $p \in M$ the metric inner product in $T_p M$ can always be made Euclidean. In other words, the condition for a metric to be flat is local, not pointwise. To see what is going on, one may look at the Taylor expansion of the metric. In special ("geodesic" or ...


0

We recall the definition of open ball in a metric space $(X,d)$. $B(x,\delta) = \lbrace y \in X$ such that $d(y,x) <\delta \rbrace$. Now, you are asked to find the open ball $B(0,1)$. $B(0,1) = \rbrace a \in \mathbf{R}$ such that $d(0,a)<1 \rbrace$. Next step : $d(0,a)=0$ if $a=0$ and else $d(0,a)= 1+\vert a-1 \vert+2\vert a \vert$. So your open ball ...


0

Your proof is virtually complete as it stands (apologies for writing it out again with slightly different notation in an earlier version of this answer). To show that $X \cup Y \cup Z$ is connected where $X$, $Y$ and $Z$ are connected, you do not need to show that $X \cap Y \cap Z$ is non-empty. All you need is to show that two of $X\cap Y$, $X \cap Z$ and ...


1

To show $V$ is connected use the following facts: The set in question $V = \displaystyle\bigcup_{a \in A}B(a, \varepsilon)$ where $B(a, \varepsilon)$ is the ball in Euclidean space with centre $a$ and radius $\varepsilon$. Each $B(a, \varepsilon)$ is connected. If two connected sets (for example $A$ and some $B(a, \varepsilon)$) overlap their union is ...


2

To adapt and complete the OP's initial line of attack, instead of using ordinary unit balls it will make life easier to instead use "polar coordinate basis sets". For each $x = (\cos(\theta_0),\sin(\theta_0)) \in S^1$ there exists $\delta_x > 0$ such that the polar coordinate basis set $$B_x = \{(r \cos(\theta), r \sin(\theta) \bigm| 1-\delta_x < r ...


1

At each point $x$ on the sphere, let $\rho(x)$ be the largest number for which the open ball of radius $\rho(x)$ centered at $x$ is a subset of $\bigcup_\alpha U_\alpha$. The $\rho(x)$ must be finite unless $\bigcup_\alpha U_\alpha=\mathbb R^n$, and if that happens then it's easy to answer the question. If I'm not mistaken $\rho$ is continuous. I'd try to ...


3

If that weren't the case you could find $z_n = (x_n,y_n)$ with $1 - {1 \over n} \leq |z_n| \leq 1 + {1 \over n}$ with $z_n \in (\bigcup_\alpha U_\alpha)^c$. Taking a convergent subsequence, you'd have some $z_{n_j}$ converging to a $z$ with $|z| = 1$, but it would also be in $(\bigcup_\alpha U_\alpha)^c$ since the latter set is closed. Hence you have a ...


1

Let $p \in U$, and let $m=f(p)$. Then $\{ m \}$ is a closed set in $\mathbb{R}$, so the preimage $f^{-1}(\{m\})$ is a closed subset of $U$, by continuity of $f$. Let $a \in f^{-1}(\{m\})$. Choose $r>0$ such that $B_r(a) \subset U$. Let $y \in B_r(a)$, and define $\gamma: [0,1] \to B_r(a)$ as $t \mapsto a+(y-a)t$. So $\gamma$ parametrizes a line segment ...


2

Hint: Calculate the Euclidean distance between $Z_1$ and $Z_2$ then express it in terms of $z_1$ and $z_2$.


1

This set $S$ is a branch of the unit hyperbola. It can be parameterized as the set of points $(\cosh t,\sinh t)$ for $t$ real. This map $f:\mathbb R\to S$ is a bijection, so points $a,b\in S$ can be written as $a=f(t_a),b=f(t_b)$ for unique $t_a,t_b\in \mathbb R$. The desired path from $a$ to $b$ is $p:[0,1]\to S$ with $p(t)=f(t_a+(t_b-t_a)t)$.



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