# Tag Info

4

Hint: if $S$ is a non-empty (bounded) set of real numbers, to show $\sup S \le b$, it is sufficient to show that for any $x \in S$, $x \le b$.

4

This is not always possible. Let $e_1=(1,0)$ and $e_2=(0,1)$ be the standard basis vectors in $\mathbb R^2$. Take $X=\{e_1,e_2,-e_1,-e_2\}\subseteq\mathbb R^2$, equipped with the $\infty$-metric, i.e. $$d(e_1,e_2)=d(e_1,-e_2)=d(e_2,-e_1)=d(-e_1,-e_2)=1$$ and $$d(e_1,-e_1)=d(e_2,-e_2)=2.$$ This space $X$ cannot be isometrically embedded into any $\mathbb ... 3 HINT: The pigeonhole argument sketched here shows that there is an$m\in\Bbb Z^+$such that$m\xi\bmod 1<\epsilon$. Choose$p\in\Bbb Z^+$large enough so that$p(m\xi\bmod 1)\ge 1$. Let$x\in\Bbb R$, and let$y=x\xi\bmod 1$; then $$\{(x+km)\xi\bmod 1:k=0,\ldots,p\}=\{(y+km\xi)\bmod 1:k=0,\ldots,p\}$$ is$\epsilon$-dense in$[0,1)$. Use this to show that ... 3 Note that the right hand side of the inequality$d(p, q) < 2\epsilon + \operatorname{diam}E$is a constant independent of$p$and$q$, so we see that$2\epsilon + \operatorname{diam}E$is an upper bound for$\{d(p, q) \mid p, q \in \overline{E}\}$. As such,$\operatorname{diam}\overline{E}$, the least upper bound for$\{d(p, q) \mid p, q \in ...

3

Every ball (open or closed) of $\Bbb Q_p$ looks like $a+p^n\Bbb Z_p$ for some $a\in\Bbb Q_p$ and $n\in\Bbb Z$. This is the image of $\Bbb Z_p$ under the map $x\mapsto a+p^nx$, and affine transformations send balls to balls... Note $\displaystyle\Bbb Z_p=\bigsqcup_{a=0}^{p-1} (a+p\Bbb Z_p)=\bigsqcup_{a=0}^{p-1}\bigsqcup_{b=0}^{p-1}(a+bp+p^2\Bbb ... 3 The metric on the subspace is just the restriction of the metric from the original space. So$d(2,5)$in the space you've written is$3$. There is not much you can say that is special about this situation from the metric point of view. One thing you can say is that if$A,B$are as above and$d(a,b)$is bounded below for$a \in A$and$b \in B$, then$X$is ... 2 No. Consider in the plane a "plus sign" that's one point thick, i.e., a union of line segments. It has no interior points, so your condition vacuously holds, but it's not convex. By "a plus sign", I mean $$([-1, 1] \times \{ 0 \} ) \cup ( \{ 0 \} \times [-1, 1]).$$ 2 No, it's not true. Let$A=\{(x,y)\mid 0<x<1\}$. Then$A$is convex and the closure of$A$minus the interior of$A$is$\{(x,y)\mid x=0\}\cup\{(x,y)\mid x=1\}$. Not connected. If you also require$A$to be compact then I have a feeling it may be true. 2 Assume for a contradiction that$x \in X$is a cluster point of the$x_n$. Since the$U_{\alpha}$cover$X$, there is some$\alpha$such that$x \in U_{\alpha}$, and since$U_{\alpha}$is open, for some$\epsilon > 0$,$B(x, \epsilon) \subseteq U_{\alpha}$. Now choose$N$such that$1/N < \epsilon/2$. As$x$is a cluster point of the$x_n$,$0 < ...

2

It might take a few lemmas to tighten this argument up, but I think the result pretty much follows from: What you've already proven about $\mathbb R$ $\mathbb R=\bigcup_n(-n,n)$ The metric $\delta_2$ is translation-invariant, so $(-n,n)$ represents all intervals. You shouldn't have to recapitulate the usage of Kronecker's Theorem, although that will work ...

2

We need to show that: $$d'(A,C)\le d'(A,B)+d'(B,C)\le\delta(A,B)+\delta(B,C).$$ Reversing the roles of $A$ and $C$ shows the other inequality in order to show the full triangle inequality for $\delta$. Now let's unravel all the suprema and infima. Goal: $d'(A,C)\le d'(A,B)+d'(B,C)$ Let $a\in A$, goal: $d(a,C)\le d'(A,B)+d'(B,C)$ $d(a,B)\le d'(A,B)$, thus ...

2

Just to cover both results: Assume that $(x_n)_{n\in\mathbb{N}}$ is a Cauchy sequence in a sequentially compact space $X$. Introduce a convergent subsequence $(x_{n_k})_{k\in\mathbb{N}}$ of $(x_n)$ with $x_{n_k}\to x.$ Let $\epsilon>0$ be given. Choose $N$ such that $\rho(x_i,x_j)<\epsilon/2,~i,j\geq N$. Choose $n_k>N$ such that ...

2

Hint: Recall that for metric spaces, compactness is equivalent to sequential compactness. Consider the sequence of sequences $(x_n)_m = 1$ if $n=m$ and $0$ if $n\not=m$. Every element has distance $1$ from the $0$ sequence. Can it have a convergent subsequence?

2

Hint: consider an open cover consisting of open balls of radius $1/3$ around every point of $B[x,1]$. Find infinitely many points whose pairwise distances are all $> 2/3$.

1

To show that $A = \{ x, x_1, x_2, \ldots \}$ is compact, you can do the following: Let $\{ U_i \}_{i \in I}$ be an open cover of $A$. Now there exists a $i_0 \in I$, such that $x \in U_{i_0}$. Since $x_k \to x$, there exists a $N \in \Bbb N^\times$, such that $x_k \in U_{i_0}$ for all $k \geq N$. For each $k \in \{1, \ldots, N-1\}$ we find a $i_k \in I$, ...

1

Here's one idea. Let $x_n$ be a sequence of points in $F$ which converges to $x \in X$ (and we do not yet know whether $x \in F$). Consider the set $K=\left \{ x_n : n \in \mathbb{N} \right \} \cup \{ x \}$. $K$ is compact (why?), so by assumption $F \cap K$ is closed. Hence $x \in F \cap K$ and hence $x \in F$. I think this is easier than trying to play ...

1

For the existence of an isometric embedding into a Euclidean space there's a necessary and sufficient characterization: the squares of the distances must be of negative type: specifically, given the $D_{ij} = d^2_{ij}$ values, then they must satisfy the inequality $$\sum_{i,j} b_i b_j D_{ij} \le 0$$ for all real $b_i$ such that $\sum_i b_i = 0$. ...

1

Suppose it is not bounded. Then you can, asuming $X$ is not empty (nor finite) so $x_0 \in X$, you construct this sequence : $x_1 \in B(x_0,1)$, $x_2 \in B(x_0,2)-B(x_0,1)$,... so on. Has this sequence a convergent subsequence?

1

Let $\epsilon>0$. If $X$ is finite the statement is obvious. If not, take some $x_1\in X$, then take $x_2\in X\setminus B(x_1,\epsilon)$,..., then take $x_{n+1}\in X \setminus\bigcup\limits_{k=1}^n B(x_k,\epsilon)$... (if you eventually run out of points you have proven the statement). Now use the fact that $\{x_n\}$ contains a subsequence that ...

1

Prove first that $X$ sequentially compact implies $X$ compact. With this result, take $\epsilon > 0$ and consider the open cover $\{ B(x,\epsilon) \}_{x \in X}$. By compactness you get: $$X \subset B(x_1, \epsilon)\cup \cdots \cup B(x_n, \epsilon),$$ and the quantity $n$ depends on how small $\epsilon$ is.

1

This is an easier version of Alex Ravsky’s idea. Let $C$ be the usual middle-thirds Cantor set, let $D$ be the set of midpoints of the open intervals deleted in the construction of $C$, and let $X=C\cup D$; then $D$ is a dense set of isolated points in $X$, so $X$ has no dense-in-itself open subset.

1

I assume that by "domain" we mean an open, connected subset of $\mathbb{R}^n$. Note, in particular, that if $D$ is a domain and $K\subseteq D$, then $K\cap \partial D=\varnothing$. Let's solve this using the contrapositives. First suppose that $f$ is not proper. Then there exists a compact set $K\subseteq D_2$ such that $f^{-1}(K)$ is not compact. But since ...

1

For the first exercise: write the metric $d$ as the sum of semi-metrics $$d_r(A,B)=\begin{cases}1,\quad \text{ if }r\in A\triangle B \\ 0 \quad\text{otherwise}\end{cases}$$ For each $r$, we have $$\sum_{i,j} b_ib_jd_r(A_i,A_j) = \sum_{r\in A_i,\ r\notin A_j} b_ib_j+ \sum_{r\notin A_i,\ r\in A_j} b_ib_j \\= 2 \sum_{r\in A_i}b_i \sum_{r\notin A_j} b_j = ... 1 Let X be a nonempty bounded convex subset of \mathbb R^n. If X has no interior points, then \partial X=\overline X is again convex and hence path connected. If X does have interior points we may assume wlog. (after translation) that 0 is an interior point. Consider the map$$\begin{align}f\colon \partial X&\to S^{n-1}\\ x&\mapsto \frac ...

1

Distance function is not differentiable in general. Consider the case $n=1$ - for $Q=\{0\}$, $d_Q(x)$ is same as the modulus function and is not even differentiable.

1

This is true for actions of countable groups, and in particular for $\mathbb{Z}$-actions. Let $C$ be a Cantor set and $G$ a countable group. Recall that $C^G$ is also a Cantor set. We let $\lambda:G\curvearrowright C^G$ denote the usual left action: $\lambda_g(f)(h)=f(g^{-1}h)$ for all $g,h\in G$ and $f\in C^G$. Now, suppose that $G$ acts freely by ...

1

Let $\{x(n)\}$ be any sequence in the intersection of compact subspaces of $X$. Then $\{x(n)\}$ is a member of every compact subspace individually. Compact spaces are closed in general hence every subspace contains the limit of the sequence. Now since arbitrary intersection of closed sets is closed this implies that the intersection is also closed. Hence $n$ ...

1

Here is a counterexample to the following claim: Let $X, Y$ be topological spaces with $x \in X$, and $f: X \rightarrow Y$ a function with the following property: if $M \subseteq Y$ is a connected set containing $y$, then there exists a connected set $N \subseteq X$ containing $x$, such that $M = f(N)$. Then $f$ is continuous at $x$. Let $X$ be the unit ...

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