37

No. Let $x$ and $y$ be connected by an edge and let's use the graph distance as our metric.
At $x$, connect paths of length $n$ for each $n\in\mathbb N$. At $y$, do the same, but also connect an infinite path.

answered Oct 6 '14 at 17:15

Bjørn Kjos-Hanssen

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36

There is no such continuum. See
Z. Waraszkiewicz, Sur un problème de M.H. Hahn, Fund. Math. 22 (1934) 180–205.
Waraszkiewicz constructed an uncountable family $W$ of continua in the plane called Waraszkiewicz spirals so that no continuum can be mapped continuously onto every continuum in $W$. Start with the space $X=S^1\cup [1,2]$ and consider the maps $f,...

27

There is no hope of making any such metric continuous with respect to $\theta$ at $s=0$ without violating the triangle inequality even in the plane. Consider parallel lines $L_1$ and $L_2$ which are some huge distance $s$ apart and a third line $L_3$ which intersects both at a very tiny angle $\theta$. Here's a picture:
In order to make $d(L_1,L_2) = s$ ...

22

Well, if $M$ has a metric basis $\{b_1, \ldots, b_{n+1}\}$ then the map $$x \mapsto (d(x,b_1), \ldots, d(x,b_{n+1}))$$ is a continuous injection from $M$ into $\mathbb{R}^{n+1}$. So, for example, no infinite dimensional Banach space $E$ could have this property for any $n \in \mathbb{N}$: letting $K$ be the closed unit ball of an $n+2$ dimensional subspace ...

19

There is a 5-point example $\ X := \{x\ y\ a\ b\ c\},\ $ with (symmetric) metrics $\ d\ $ as follows:
$$d(x\ y) = d(a\ b) = 1$$
$$d(x\ a) = d(y\ b) = 2$$
$$d(x\ b) = d(y\ a) = 3$$
$$d(x\ c) = d(y\ c) = 6$$
$$d(a\ c) = 5\qquad\qquad d(b\ c) = 4$$
Oviously, $\ B_r(x)\ $ and $\ B_r(y)\ $ are isometric for every $\ r>0,\ $ while there is no isometry $\ f:...

answered Oct 6 '14 at 22:03

Włodzimierz Holsztyński

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16

Consider $(\mathbf{Z}\times[0,1])/(\mathbf{Z}\times\{0\})$, infinitely many spokes attached at a single point. Given a finite basis, the only points uniquely characterized by distances from it are the points on the same spokes as the basis. So this space is infinite-dimensional.

14

Let us improve slightly on Joel's answer by avoiding not only choice but also excluded middle (which is used in assuming that the minima $n_x$ exist). In passing we also generalize to an arbitrary metric codomain. Since the various notions of compactness are not equivalent intuitionistically, we have to specify which one we mean. We mean by "compact" the ...

14

If $\text{NMSC}$ is consistent, then so is $\text{NMSC}+\text{"there are no strongly inaccessible cardinals"}$.
This is because if $V \models \text{NMSC}$, then $V_\kappa \models \text{NMSC}$ for any inaccessible cardinal $\kappa$. (A proof sketch is given below.) In particular, if $\kappa$ is the first strongly inaccessible cardinal, then $V_\kappa \...

13

Unfortunately, my original answer below was completely misguided (i.e., wrong), and the people who up-voted it should feel free to reverse their votes! I'm leaving the answer below so that people can see what it was, but I'm prefacing it with a correction. I'm not deleting it because then maybe people won't be able to see it and know that my answer was ...

answered Jul 20 '13 at 15:21

Robert Bryant

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13

$d_2$ is indeed a metric. Abbreviating $\gcd(m,n)$ to $(m,n)$, we need to show that
\begin{align*}
1-\frac{2(a,c)}{a+c} &\le 1-\frac{2(a,b)}{a+b} + 1-\frac{2(b,c)}{b+c}
\end{align*}
or equivalently
\begin{align*}
\frac{2(a,b)}{a+b} + \frac{2(b,c)}{b+c} &\le 1 + \frac{2(a,c)}{a+c}.
\end{align*}
Furthermore, we may assume that $\gcd(a,b,c)=1$, since we ...

12

It seems to me that this is provable without using the axiom of
choice.
Suppose that $X$ is a compact metric space and $f:X\to\mathbb{R}$
is continuous. Let's show it is uniformly continuous. Fix any
$\epsilon\gt 0$. For each point $x\in X$, there is a small ball
$B$ centered at $x$ such that $f(y)$ is within $\epsilon/2$ of
$f(x)$ for all $y\in B$, and we ...

answered Feb 17 '13 at 15:48

Joel David Hamkins

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12

Let $X=[0, \infty)$ with the metric $d(x,y)=|x-y|$.
The equivalence relation is $x\sim y$ iff $x=y$ or $xy=1$.
In the Hausdorff metric on $X/{\sim}$, the open ball with radius $1/2$ around the equivalence class $\{0\}$ contains only $\{0\}$.
However $\{0\}$ is not open in $X$, so $\{\{0\}\}$ is not open in the quotient topology on $X/{\sim}$.

12

Pretty much the papers to read are
Nikolaev, I.G., Smoothness of the metric of spaces with two-sided bounded Aleksandrov curvature, Sib. Math. J. 24, 247-263 (1983); translation from Sib. Mat. Zh. 24, No.2(138), 114-132 (1983). ZBL0547.53011.
Nikolaev, I.G., A metric characterization of Riemannian spaces, Sib. Adv. Math. 9, No.4, 1-58 (1999). ZBL0956....

12

The issue here is that a metric space might not have non-trivial (read: not eventually constant) Cauchy sequences. For example, if the underlying space is a Dedekind finite set.
Indeed it is consistent that there is a dense subset of $[0,1]$ which is Dedekind finite. As a space with the inherited metric it is complete already and totally bounded, but it is ...

answered Jan 16 '18 at 17:28

Asaf Karagila

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12

Metrics are strongly equivalent if the identity mapping $Id:(X,d_1)\to (X,d_2)$ is bi-Lipschitz. They preserve the class of Lipschitz mappings.
Roughly speaking classical topology deals with notions that are invariant under homeomorphisms. In particular, changing metric to the one that gives equivalent topology does not make much harm. But the situation ...

11

In other words, you are interested in the spaces where isometric subsets are congruent.
If the metric space is locally compact and intrinsic and simply connected then you get only spheres, Euclidean spaces and hyperbolic spaces.
If not simply-conected, then in addition you get real projective spaces.
If the metric is not intrinsic you get discrete spaces ...

11

Okay, so $X$ is a finite metric space and $D(X)$ is the positive part of the unit sphere of $l^1(X)$. We can consider $X$ as sitting inside $D(X)$ by identifying a point $x \in X$ with the function that is $1$ at $x$ and $0$ elsewhere.
The literal question you have asked is whether the mass transport metric on $D(X)$ is the only metric on $D(X)$ whose ...

11

The answer is no if $M$ is assumed to be complete:
Let $x, y \in M$ such that for all $r > 0$ the balls $B_r(x)$ and $B_r(y)$ are isometric via
$$f_r : B_r(x) \to B_r(y).$$
Set $R := \inf \{ r \mid B_r(x) = M\}$ (here $R = \infty$ if $M$ is unbounded).
First I claim that
For all $0 < r < R$ we have $f_r(x) = y$:
Assume on the contrary that $...

11

Functional analysis abounds with such examples. There are many complete locally convex spaces which are countable unions of closed subspaces with empty interior. One such is the space of smooth functions on the line which have compact support, the test functions of L. Schwartz. An even simpler one is the space of finite sequences with the locally convex ...

11

Having infinite metric dimension is not bi-Lipschitz-invariant.
On the real line, consider the two metrics
$$
d(x,y)=\min(|x-y|,1)\quad\text{and}\\
\delta(x,y)=\arctan(|x-y|).
$$
The two metrics both give the usual topology, they are bi-Lipschitz to each other, but $\dim(\mathbb R,d)=\infty$ (as observed by Benoît Kloeckner) and $\dim(\mathbb R,\delta)=1$ (...

answered Jul 16 '17 at 14:32

Joonas Ilmavirta

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10

The result does appear in Dunford/Schwartz, Linear Operators Part I (page 437), but is only stated as an exercise.
Edit after @JosephVanName' comment:
Conway's Functional Analysis has the result for completely regular spaces as Theorem 6.6 (page 140).

10

Since you haven't given a distribution, let me make an observation giving the right form of the answer in the case where the $D_{xy}$ are independent uniform $[0,1]$ random variables.
I want to claim that the probability, $p$, that the matrix defines a metric satisfies
$\alpha^{n^2}\le p\le \beta^{n^2}$ for constants $\alpha$ and $\beta$.
First, notice ...

10

No, in order for a subspace of a complete metric space to be completely metrizable it is necessary and sufficient for it to be $G_\delta$. There are only $2^{\aleph_0}$ many $G_\delta$ subsets of $\mathbb{R}^2$ but there are $2^{2^{\aleph_0}}$ sets in $$\{X \subseteq \mathbb{R}^2 : (0,1)\times(0,1) \subseteq X \subseteq [0,1]\times[0,1]\},$$ each of which is ...

answered Dec 1 '13 at 15:53

10

Your answer is correct, except that in the example you gave, one has to adjoin far more numbers than you described in order to get to that ordered Pythagorean not-Euclidean field K.
That example is described on p.594 of the fourth edition of my book Euclidean and Non-Euclidean Geometries: Development and History (W.H. Freeman, 2008). There I called such a ...

10

To provide some context the subsets of a Euclidean space that can be approximated by affine planes on every scale are known as Reifenberg-flat sets after E. R. Reifenberg who proved in the 1960s that such sets are bi-Holder to a Euclidean space. There is a substantial literature on the subject (search on "Reifenberg-flat").
Now regarding your specific ...

answered Mar 14 '17 at 1:48

Igor Belegradek

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10

If I understood the question correctly, what you're asking is related to the maximum distance of binary codes with large minimum distance $d\geq s$.
In coding theory, $A_q(n,d)$ is defined as the maximum cardinality of a $q-$ary code with length $n$ and minimum distance $d.$
You can never have more than 2 codewords at full distance by binary geometry. In ...

9

As to the compact-open topology of $C(X,Y)$, it is metrizable if and only if $Y$ is metrizable, and $X$ is hemicompact.

9

Let $\varphi$ be a continuous function supported on $[0,1]$. Then continuum many combinations $\sum c_k \varphi(x+k)$, $c_k\in \{0,1\}$ are separated in our space.

answered Feb 10 '15 at 12:57

Fedor Petrov

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9

More elementary than Ascoli:
If it was normable, it would mean that there exists a norm $n$ which is continuous, hence for some $k$, $n(\phi) \leqslant C\ p_k(\phi)$ and which defines the topology, i.e. such that for all $k$, $p_k(\phi) \leqslant C_k\ n(\phi)$.
This would imply that all the norms $p_k$ are equivalent for $k$ large enough, which is not the ...

9

Here's a related positive result:
Alan Dow and KP Hart proved in 1998 that the (non-metrizable) continuum $\beta[0,1)\setminus [0,1)$ maps onto every continuum of weight $\leq \omega_1$, and therefore onto every metrizable continuum. See https://arxiv.org/pdf/math/9805008.pdf.

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