# Tag Info

19

It is easy to see the answer is no. Take the unit circle in the plane, with the metric inherited from the Euclidian metric in the plane. Then remove the "top" point (coordinates $(0,1)$). Then a "ball" (small disk) around a point close to "the top", restricted to the "circle with the top removed," will be disconnected. Here the metric space is the circle ...

16

No. The Knaster-Kuratowski fan is a connected subspace of the plane that becomes totally disconnected when a certain point is removed, so open balls centred at the other points cannot be connected if they are small enough to exclude the explosion point.

7

If $f_1$ and $f_2$ differ in values just at let us say in the middle of the interval, thus at a point, this would not change the integral value, it would vanish, but $f_1$ and $f_2$ would be considered different.

6

$d(x,\emptyset)=\inf\emptyset=\infty.$ That makes sense because in general, if $A\subseteq B,$ then $d(x,A)\ge d(x,B);$ so $d(x,\emptyset)\ge d(x,B)$ for all $B.$ If the infimum of a set $S$ is attained, that is, if $\inf S\in S,$ then $\min S=\inf S;$ otherwise, $\min S$ does not exist. Thus $\min\emptyset$ does not exist, just as ...

4

Start with an arbitrary metric space (with metric $d$) that is not totally bounded, and take the new metric $\overline{d}(x,y) = \min(1, d(x,y))$.

4

Just take a plane without a segment, and consider a small ball near the center of the segment. There's no need to get complicated stuff.

4

Assume by contradiction that $A$ is unbounded. Let $a \in A$. Then for each $n$ there exists some $x_n$ such that $$d(x_n,a) >n$$ Now, $x_n$ has a converging subsequence $x_{k_n} \to b$. Now, for $\epsilon=1$, since $x_{k_n} \to b$ there exists some $N$ so that for all $n >N$ we have $$d(x_{k_n}, b) <1$$ Then, for all $n >N$ we have $$d(a,b) ... 4 I think that when he says f is continuous from E to B. he intends to mean that the important part which was justified was: (..) to B. and assumes continuity is a trivial matter (which is, since it is division by \frac{1}{1+\Vert x \Vert }. Since the norm is continuous, multiplication by scalar is continuous, inversion is continuous in ... 3 It depends on what you want for your "metric". It makes perfect sense as far as the definition goes to replace \mathbb{R} by any totally ordered abelian group. But of course the properties won't be the same... First note that when you suggest to interpret |x|_p = p^{-v_p(x)} as an element of \mathbb{Q}_p, this has the pretty annoying inconvenient that ... 3 Yes, it is separable - in fact, it has a countable dense subset consisting entirely of finite sets! Exercise. Let C be a closed bounded set, and N\subseteq C such that for each c\in C, there is an n\in N such that d(n, c)<\epsilon. Then the Hausdorff distance between N and C is <\epsilon. Such an N is sometimes called an ... 3 The first part of your analysis is fine, but the second is wrong. Consider the set of sequences that are eventually 1. Let S_n be the set of these sequences that are 1 for all but at most the first n terms. There is an easy bijection between S_n and \Bbb N^n: just throw away all of the ones after the n-th term. But \Bbb N^n is countable for ... 3 d_1 and d_2 are equivalent metrics if and only if the identity map is a homeomorphism (X,d_1)\to(X,d_1). However, it is possible for (X,d_1) and (X,d_2) to be homeomorphic through a non-identity map even though d_1 and d_2 are not equivalent. For example, take X=\mathbb R and define$$ f(x) = \begin{cases} -x & \text{when }|x|<1 \\ x ...

3

Perhaps I will write something wrong? (This seems easy to me). First of all, I think you mean there can't exist an injective function $f:I_m\to I_n$, for if $n<m$ there actually exists an injective $I_n\to I_m$ (the inclusion). Suppose $n<m$ and $f:I_m\to I_n$ injective. Then $f:I_m\to f(I_m)$ is bijective, where $f(I_m)=\{f(1),...,f(m)\}$. Clearly ...

2

I have not checked if $d(x,y)$ is indeed a metric, but to show that $d$ induces the product topology $\tau$ on $X$ it is enough to show that for every non-empty open $V\in \tau$ there is a basic ball $B(a,r)\subseteq V$, and that for every ball $B(a,r)$ there is an open $V\neq \emptyset$ such that $V\subseteq B(a,r)$. Suppose $V$ is a basic open in the ...

2

If $f:X\rightarrow Y$ is a bijective isometry, $d(f(f^{-1}(x)),(f^{-1}(y)))=d(f^{-1}(x),f^{-1}(y))=d(x,y)$, thus $f^{-1}$ is an isometry and $f^{-1}(B(x,r))=B(f^{-1}(x),r)$ thus $f^{-1}$ is continuous.

2

You want to prove that if $x_n \to x$ in $S$, then $d(x_n,x_0) \to d(x,x_0)$ in $\Bbb R$. One does this as follows: let $\epsilon > 0$. By convergence of $(x_n)_{n\geq 1}$ there is $n_0$ large enough such that $d(x_n,x)< \epsilon$ if $n \geq n_0$. Then: $$|d(x_n,x_0)-d(x,x_0)| \color{red}{\leq} d(x_n,x) < \epsilon$$for all $n \geq n_0$, and so ...

2

Let us denote the metric of $X$ by $d$. Let $(x_n)_{n=1}^\infty$ be a Cauchy sequence in $X$. For each positive integer $n$, there is some $y_n \in S$ with $d(x_n,y_n) < 1/n$. Hence $$d(y_m,y_n) \leqq d(y_m,x_m) + d(x_m,x_n) + d(x_n,y_n) < 1/m + d(x_m,x_n) + 1/n,$$ for every pair of positive integers $m$ and $n$, whence $(y_n)_{n=1}^\infty$ is also a ...

2

The thing is, by definition, $F^n$ is the set of $n$-uple $(x_1,\dots,x_n)$ with $x_i\in F$. That's the very definition of the symbol $F^n$. And indeed $F^n$ is a vector space over $F$. Now if you start with an "abstract" vector space $V$, then what you want to prove is that $V$ is isomorphic to $F^n$ for some $n$ (which is the dimension of $V$). The ...

2

Let $S$ be the set of Cauchy sequences of natural numbers. As you pointed out, any such sequence is eventually constant, hence we can write $$S=\bigcup_{n=1}^{\infty}S_n$$ where $S_n$ is the set of sequences $\{a_k\}$ of natural numbers such that $a_k=a_n$ for all $k\geq n$. Now there is a bijection $\mathbb{N}^n\to S_n$, hence $S_n$ is countable. So $S$ ...

2

There are two definitions useful in this context: A crystallographic group of the euclidean $n$-space is a discrete subgroup $\Gamma$ of the isometry group $Isom(E^n)$ of $E^n$, such that $E^n/\Gamma$ is compact. A discrete subgroup $Isom(E^n)$. You also have to decide on the notion of equivalence between such subgroups: Conjugation in $Isom(E^n)$, ...

2

The first thing to do is to take a Cauchy sequence $\{f_n\}$. Notice that for every $x \in [a,b]$, $\{f_n(x)\}$ is a Cauchy sequence in $\mathbb{R}$, a complete metric space! Then for every $x$ we can define $f(x)$ as the limit of $f_n(x)$. Now that we have a pointwise limit there are two things left to do: Show that $f_n \to f$ in the metric $d$ Show that ...

2

At heart, this question depends entirely on what "countable" means. Here are two options: A set $A$ is said to be countable if there is a one-to-one function from $A$ into the set of natural numbers. A set $A$ is said to be countable if there is a one-to-one function from $A$ into and onto the set of natural numbers. As far as I'm aware, all textbooks ...

1

Let $(f_{n})$ be a Cauchy sequence in $C^0[a,b]$. For each $x\in [a,b]$, $(f_{n}(x))$ is a Cauchy sequence in $\mathbb{R}$, and hence converges to some limit $p_{x}$. Define $f:[a,b]\to\mathbb{R}$ by $f(x)=p_{x}$. We claim that $f\in C^{0}[a,b]$ and that $(f_{n})$ converges to $f$ in $C^{0}[a,b]$. Note that the first assertion will follow from the second ...

1

Yes, translation by $x$ is a homeomorphism $\Bbb R^n\to\Bbb R^n$ (and an isometry too).

1

The next thing that one would try is to prove that these two topologies are equivalent. But they are not. If they were, we would also have that $x_n \to x$ using $d_2$ implies $x_n \to x$ using $d_1$, but we don't have this. Hence, I believe that $\tau_2 \subseteq \tau_1$ is the best you can get.

1

Your counterexample is fine. Here's how you can get more counterexamples, which might be considered more natural. Plainly, "metrizable by a bounded metric" is a topological property; if two topological spaces are homeomorphic, and if one is metrizable by a bounded metric, so is the other. The usual metric on $\mathbb R$ is unbounded. Since a finite open ...

1

Let $V$ be a vector space of dimension $n$ over $\mathbb{F}$. Let $\{v_1,\cdots,v_n\}$ be a basis for $V$. Let $u \in V$. Since the $v$'s form a basis, there are scalars $a_1,\cdots,a_n \in \mathbb{F}$ such that $u = a_1v_1 + \cdots + a_nv_n$. Let $T$ be the tuples $(a_1,\cdots,a_n)$ then $T$ is a representation of $u$ as a tuple.

1

No, find one fixed $\varepsilon$, that you can choose and show that for every $\delta > 0$ we can find a point $x_\delta$ with $d(x_\delta,0) < \delta$ and $d(f(x_\delta),f(0)) \ge \varepsilon$. The latter implies indeed that $f[B_\delta(0)] \nsubseteq B_\varepsilon(f(0))$. Note that $d(x,1) \ge 1$ for any $x \neq 1$. Also note that near $0$ we have ...

1

Let $(x,y) \in \mathbb{E}^2$, then $(y,x)\in \mathbb{E}^2$ and $f(y,x)=(x,y)$, so $f$ is surjective. Let $(x_1,y_1),(x_2,y_2) \in \mathbb{E}^2$, then $f(x_1,y_1)=(y_1,x_1)$ and $f(x_2,y_2)=(y_2,x_2)$ and $$d((x_1,y_1),(x_2,y_2))= \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$$ $$d(f(x_1,y_1),f(x_2,y_2))= d((y_1,x_1),(y_2,x_2))=\sqrt{(y_1-y_2)^2 + (x_1-x_2)^2}$$ ...

1

I think that answer is revealed in Lebesgue's number lemma. In this case $B_x(d)$ is ball of radius $d$ centered at $x$. $\mu(d)$ behaves as part of elements in $B_x(d)$ and $0 \le \mu(d) \le 1$.

Only top voted, non community-wiki answers of a minimum length are eligible