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6

Consider the subset of $\mathbb R$ $$\{ \frac{1}{n} | n\in \mathbb N^*\} \cup \{0\}\;$$ which inherits a metric structure from the standard metric of $\mathbb R$. The sets $\{1/n\}$ are countably infinitely many clopen subsets.


6

This is true. In particular, consider any sequence $x_n$ such that all continuous functions $f:S\rightarrow \mathbb R$ converge. We will start by showing that if $x_n$ has a limit point $y$, then it converges to $x$. In particular, the condition of being a limit point may be expressed as $$\liminf_{n\rightarrow\infty}d(x_n,y)=0$$ however, since $x\mapsto ...


6

$X=(0,1)$ with Euclidean metric. Then $X$ itself is closed and bounded yet not compact.


5

If $k = 1/3$, then $f(x) = k(x+\tfrac{1}{x})$ is not a function from $X = [1,\infty)$ into $X$. For instance, $f(1) = \dfrac{1}{3}(1+\tfrac{1}{1}) = \dfrac{2}{3} \not\in [1,\infty)$.


5

Of course the question is whether there's a countable $\epsilon$-cover for every $\epsilon>0$. And we really can't tell, because we don't know whether $\aleph_2>c$. The answer is no if $\aleph_2>c$ (and certainly yes if $\aleph_2\le c$): If $M$ is a metric space and there is a countable $\epsilon$-cover for every $\epsilon>0$ then $|M|\le c$. ...


3

Since $n_0\in\mathbb N$, you can't choose $n_0=\frac{1}{\varepsilon}$. What you can choose is $n_0=\lfloor\frac{1}{\varepsilon}\rfloor +1$. But you don't need to explicit the $n_0$. Let $\varepsilon>0$. Since $$\lim_{n\to \infty }\frac{1}{n}=0,$$ there is a $n_0\in\mathbb N$ s.t. $\frac{1}{n}<\varepsilon$ when $n\geq n_0$. Therefore, if $n\geq n_0$, ...


3

Let $n \in \mathbb{N}$, and define $ f_n(x) = \left\{\def\arraystretch{1.2}% \begin{array}{@{}c@{\quad}l@{}} 0 & \text{if } 0 \leq x \leq 1/2\\ 2nx - n & \text{if } 1/2 < x \leq (n + 1)/(2n)\\ 1 & \text{if } (n + 1)/(2n) < x \leq 1\\ \end{array}\right.$ Then $\{f_n\}$ is a Cauchy sequence whose limit is $ f(x) = ...


3

Real numbers with the discrete metric. I.e. $d(x,y) = 1$ if $x \neq y$.


3

You need to do this for the standard metric on the reals. If not, it should have been specified, as Cauchy-ness depends on the metric used, and it can fail for other choices of the metric on the reals. So you need to show that for every $\varepsilon > 0$ there exists some $N$ such that for all $n,m \ge N$ we have $$\left| d(x_n, y_n) - d(x_m,y_m) ...


3

The topology doesn't care about exact distance values in the metric; it only cares about what points are close to what others in a broad sense. So for a question where the exact distance values are important, you wouldn't expect the question to say the same under different metrics even if they induce the same topology.


3

If you are familiar with degrees, one can argue as follows. The identity selfmap $z\mapsto z$ of the circle has degree $1$. The squaring map $z\mapsto z^2$ has degree $2$. The function $(g(z))^2$ is the composition of $g$ and $z\mapsto z^2$. Under composition the degree multiplies, i.e., the degree of $g$ would have to be $\frac{1}{2}$ in order for the ...


3

Hint: $S_k=[-k,k]$ [edit after the question was changed] If you want $S_k\subset\tilde{S}$ for some compact $\tilde{S}$ then let $S_k$ be the unit line segment intersecting the origin of the plane at angle $\theta=2\pi/k$.


3

Your original question "Construct a metric such that the resulting metric topology is the same as the standard topology" makes no sense since, as Donkey_2009 was trying to tell you, you have to have a metric to begin with in order to define "radius" and "ball". I suspect your problem was to construct another metric which gives the same topology. That's ...


3

Any metric space is a topological space. So topological terms generally have the same meaning as in a general topological space. In particular a metric space is separable if it has a countable dense set.


3

Show two inclusions, based on your definition of closure as the union of interior and boundary: If $x \in \operatorname{Int}(A)$ then in particular $x \in A$ and so $d(x,A)$ is the infimum of a set that contains $d(x,x) = 0$ in particular, so $d(x,A) = 0$. If $x \in \partial A$, pick $r>0$ arbitrarily. Then $B(x,r)$ intersects $A$ by the definition of ...


3

No. Consider that this problem is equivalent to finding the roots of $f$ (if you knew the distance from $x$ to the nearest root, you could check only two points to find the root). If you know the roots and they are countable, you can of course simply sort them and quickly find the one closest to $x$. As for finding the distance in practice, you can ...


3

The result is not true. Let $M=\mathbb R$ and $f(x)=\max(x,0)$. Then $A=(0,\infty)$ and $\partial A=\{0\}$, while $\{x:f(x)=0\}=(-\infty,0]$. What is true is $\partial A\subset\{x:f(x)=0\}$.


2

Yes, this is a metric space. It is (isomorphic to) the subspace of $\mathbb R^2$ consisting of the four points $(0, 0), (0, 1), (1, 0), (1, 1)$, with the metric inherited from the metric on $\mathbb R^2$. 2,3. Nothing particular that I can think of. Most texts for a first class in real analysis should give an overview of metric space theory. I ...


2

You are given a metric $$ds_z:={|dz|\over y}\qquad(z\in H)$$ and a metric $$ds_w:={|dw|\over 1-|w|^2}\qquad(w\in D)\ .$$ In addition a map $$W: \quad H\to D, \qquad z\mapsto w:= {z-i\over z+i}$$ is considered. One computes $$1-|w|^2=1-w\bar w={(z+i)(\bar z-i)-(z-i)(\bar z+i)\over|z+i|^2}={-2i(z-\bar z)\over|z+i|^2}$$ and $$W'(z)={2i\over(z+i)^2}\ .$$ It ...


2

The conclusion I want to jump to is that the infinite union is compact. But, that is not true. Suppose $S_n = \{1,\frac1n\}$ and $\bar S = \{\frac1n: \forall n \in \mathbb N\} \cup \{0\}$ $\bar S$ is compact. $\bigcap S_k = \{1\}$ i.e. non-empty. There exists a sequence $\in\bigcup S_k$ that converges to $0.$ $0 \notin \bigcup S_k$ $\bigcup S_k$ is ...


2

Why don't you linearly do it: interval length should be expaded/shrunk, and $a$ must map to $0$, so $$f:[a,b] \longrightarrow [0,1]$$ $$\ \ \ \ \ \ \ \ \ x\longrightarrow \frac{x-a}{b-a}$$ ? It has a continuous inverse.


2

For any $k$, $$ x_k\neq y_k\quad \Rightarrow\quad x_k\neq z_k \quad \vee\quad y_k\neq z_k. $$ Then $$ \min\{k:x_k\neq y_k\}\geq \min\{k:x_k\neq z_k\} \quad \vee\quad \min\{k:x_k\neq y_k\}\geq \min\{k:y_k\neq z_k\} $$ hence $$ d(x,y)\leq d(x,z) \quad \vee\quad d(x,y)\leq d(y,z) $$


2

You're a little muddled about what open and closed mean. In a metric space, open and closed sets are defined using the concept of a ball. This proof does not deal with metric spaces, but with topological spaces, which are more general, so there is no such thing as a ball here. Here, to know whether the two halves of the separation are open, you just need ...


2

So this doesn't quite work but you are definitely on the right track. The only problem with your argument is that we don't know a priori that $\frac{1}{\epsilon}$ is a natural number. However, try repeating your argument with any natural number $k$ such that $k<\frac{1}{\epsilon}$. This will solve the problem, because given $\epsilon>0$, we have ...


2

Consider $X = \mathbb{Z}$ with a metric $d$ such that $d(m, n) = 1$, if $m \neq n$ and $d(n, n) = 0$. $d$ is a metric. Take any infinite subset $A$ (for example, $\mathbb{N}$). Then $A$ is open (and closed) and it is also bounded (because the maximum distance is 1). Moreover, $A$ is not compact, because the family of sets defined by $\{\{x\} \mid x \in ...


2

Let $(x_n)$ be a convergent sequence in $M$ with limit $a$. Then we can consider the sequence $x_1, a, x_2, a, x_3, a, \ldots$ which intertwines $(x_n)$ with the constant sequence $(a)$. This sequence converges to $a$ as well, so by assumption, the sequence $f(x_1), f(a), f(x_2), f(a), \ldots$ converges in $N$. Since every other term is $f(a)$, the only ...


1

HINT: Show that $d$ is equivalent to the usual metric on $\Bbb R$, meaning that it generates the same topology on $\Bbb R$. Compactness is a topological property, not a metric property, so as long as two metrics generate the same topology, a set compact in one is compact in the other. Completeness then comes for free. It’s also possible to show directly ...


1

There are at least two reasonable approaches. One is to construct a complete metric equivalent to the usual one; I’ll get to that in a moment. The other is to prove the result that a $G_\delta$ set in a complete metric space is completely metrizable; since $[0,1)$ is certainly a $G_\delta$ in $\Bbb R$, that theorem yields the desired result. To get a ...


1

HINT: You don’t need to find $N$ explicitly. Fix $\epsilon>0$. Since $\sum_{k\ge 0}\frac1{k!}=e$, you know that there is an $N\in\Bbb N$ such that $$\left|e-\sum_{k=0}^N\frac1{k!}\right|=e-\sum_{k=0}^N\frac1{k!}=\sum_{k\ge N}\frac1{k!}<\frac{\epsilon}2\;.$$ The terms of the series are positive, so for each $n\ge N$ we have ...


1

$p_i^{-1}(A_i)$ is the set of all things mapped by the projection onto $A_i$. This is called the pre image of $A_i$ under the projection onto the $i^{th}$ coordinate and it is precisely $M_1\times \dots M_{i-1} \times A_i \times M_{i+1} \times \dots M_n$. So then what is the intersection $\bigcap_{i=1}^n p^{-1}(A_i)$? Well, its the intersection of all of the ...



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