Tag Info

New answers tagged

2

A non principal ultrafilter on the set of natural numbers is a tail set but it is far from Borel. What you are missing is that you need to make use of the fact that your set is Borel (or Lebesgue measurable) so that you can apply something like Lebesgue density theorem.


0

The simplest example of a limited metric is the discret metric, given by $d_0: X\times X\rightarrow X$ s.t. $$d_0(x,y)=\left\{\begin{array}{lr}1&\text{if $x\neq y$} \\ 0 & \text{if $x=y$}\end{array}\right.$$ You can see that $d(x,y)\leq 1$ for all $x,y\in X$. Another way is to construct a bounded metric from any given metric is: Given a metric $d: ...


1

Started to write a comment, turned out too long. $[0,5]\notin\tau$ in the Euclidean topology; $[0,5]^c=(−∞,0)\cup(5,∞)\in\tau$. $\tau$ denotes the collection of open sets; closed sets are not included in $\tau$. This axiomatic characterization of topological spaces should be thought of as the definition of a topology. In metric spaces, we're dealing with ...


1

This is an enlarged comment. You seem to have a misunderstanding. What is the definition of a topology? Let's look at your first bullet. You have that $X\in\tau_X$. So, if $X=[0,5]$, then $[0,5]\in\tau_{[0,5]}$. Why would a singleton not be an open set? Let $X$ be a topological space endowed with the discrete topology, and let $x\in X$. Then, since ...


1

There are multiple ways to define a valid topology on a given set. You are talking about all these things such as '$[0,5]$ cannot be an open set' in the context of standard topology on $\mathbb{R}$, which includes as 'open sets' all subsets that can be written as unions of disjoint open intervals, but if you define the topology on $\mathbb{R}$ otherwise ...


0

Here is the abstract construction of the topology induced by a metric $d$ on some space $X$. There is a base of open sets called open balls of radius $r$ centered at some point $x$ which is defined as the set $\{y \in X : d(x, y) < r\}.$ We allow $r$ to vary over the entire positive reals and $x$ to vary over all of $X$ to get a collection of open sets ...


0

Rudin is defining the concept of a topology on a set. For any given set, there may be several different topologies on it. On a metric space $(X,d)$, one can define a topology on it by using the open balls as a basis. This will give one possible topology on $X$, the metric topology given by $d$ which you are probably familiar with. However there are other ...


0

A metric space $X$ here isn't just a space that happens to have both a topology and a metric $d$; it's one in which the topology is generated by the open balls $B_r(x) = \{y\in X:\, d(x, y) < r\}$. Thus, for example, $\mathbb{R}$ with the usual topology and the metric $d(x, y) = |x - y|$ is a metric space; $\mathbb{R}$ with the finite-complement topology ...


0

Hint: If $d:X\times X\rightarrow \mathbb R$ is a metric on $X$ then so is $d':X\times X\rightarrow \mathbb R$ prescribed by $(x,y)\mapsto d(x,y)$ if $d(x,y)<1$ and $(x,y)\mapsto 1$ otherwise. Note that $d'$ is a bounded metric. Metrics $d$ and $d'$ induce the same topology on $X$.


-1

Yes. Suppose that the topologies are not the same and let $U$ be open in $(X,d_1)$ and not open in $(X,d_2)$. Then $\overline{U}/U\neq\emptyset$ in $(X,d_2)$. Let $u\in\overline{U}/U$. Then for each nonnegative integer $$\left\{ x\in U\mid d_{2}\left(x,u\right)<\frac{1}{n}\right\} \neq\emptyset$$ Let $x_{n}$ be an element of this set. Then ...


1

Many important quotient spaces are not metric, nor even Hausdorff. This happens quite commonly for orbit spaces of group actions, which are very important in geometric group theory, in dynamical systems, etc.


0

Following an argument of Wobst, we can prove Lemma 2 without proving that $r\mapsto n(r)$ is continuous on $(0,r_{0})$. Instead, we prove the inequality $$s_{r;r_{0}}:=\inf_{r\leq\left\|x\right|\leq r_{0}}\dfrac{\left\|2x\right\|}{\left\|x\right\|}>1,$$ where of course $0<r<r_{0}$, by using the concavity of $t\mapsto\left\|tx\right\|$ and the ...


2

1) No standard name. 2) You speak of the symmetric distance but you write $A\backslash B \cup B\backslash A$ which is the symmetric difference. It is not a metric and does not turn $S$ into a metric space. 3) Let $A$ and $B$ be any two distinct straight line segments. Their symmetric difference has $0$ area. 4) Is $S$ complete? well, you'll have to ...


0

Due to a misunderstanding on my behalf my original answer was flawed. My new argument does not fully work (I have indicated the gap), but as I think this is better than nothing, I'll leave it. Let $r_0$ be as in the cited paper. Then every nonzero vector in $X$ has a multiple with norm at least $2r_0$. It suffices to show (for the purposes of the paper) ...


0

Every homogeneous metric induces a norm via: $$\|x\|:=d(x,0)$$ while every norm induces a homogeneous and translation-invariant metric: $$d(x,y):=\|x-y\|$$ The clue herein lies in wether the induced norm really represents the metric as: $$d(\cdot,\cdot)\to\|\cdot\|\to d(\cdot,\cdot)$$ which is the case iff the metric was also translation-invariant: ...


3

The first definition is that of compactness while the second definition is for sequential compactness. For a general topological space, the two notions do not coincide, but for metric spaces they do. A proof of the equivalence between the two definitions in this setting can be found in most topology books (for example, Theorem 28.2 of Munkres' Topology, ...


4

If one excludes the trivial close set $\Bbb Z$, then the topology one's looking for is the cofinite topology. This topology is not Hausdoff, hence not a metric topology. So the answer is still "NO".


5

No, $\mathbb{Z}$ is closed (and not finite) in every topology on $\mathbb{Z}$. If you meant to say "the closed sets other than $\mathbb{Z}$ are finite", then the answer is still no. If the finite sets and $\mathbb{Z}$ are the only closed sets, then the only open sets are the cofinite ones (i.e. complement of a finite set) and $\varnothing$. This is known ...


2

This is already false for the sphere $S^3$ with the standard metric. The group G is isomorphic to $SU(2) \times R$. Edit: Here is a simplified version of the above example, which assumes that you only know linear algebra (and general topology). Verify that the unitary group $U(2)< GL(2, {\mathbb C})$ acts isometrically on ${\mathbb C}^2={\mathbb R}^4$ ...


4

Ordinal spaces occur naturally, and any uncountable ordinal space is not metrizable. You can also talk about Moore spaces, Stone-Cech compactification of $\Bbb N$ (which is compact but has is too big to be metrizable), there are Zariski topologies which are often non-metrizable, and there are plenty of Cantor cubes which are naturally occurring in set ...


0

A very simple way in which the quotient can fail to be metrizable is if there are equivalence classes that are not closed. Take your original space to be $\mathbb{R}$ and let $\sim$ have as its equivalence classes $(0,1)$ and all singletons $\{x\}$ with $x\notin(0,1)$. In the quotient topology, you can not separate $(0,1)$ and $1$ by open sets and the ...


1

You have it right, much better than the other solutions, minus some notational discrepancies. Don't take the one you have checked, you need to use the compactness of $X$ for a proper proof. Clearly $X=\cup_{r \in \mathbb{R}} \, G_r$. Since $G_r=f^{-1}((-\infty, r))$ is open in $X$ for each $r \in \mathbb{R}$, you have that $\{G_r\}_{r \in \mathbb{R}}$ is an ...


4

If $x\in X$, then $f(x)<|f(x)|+1=r$, hence $x\in G_r$.


5

Assume that $\{ G_r \}_{r \in \mathbb{R}}$ is not an open cover of $X$, then there exists $x \in X$ such that $f(x) > r$ for all $r \in \mathbb{R}$ i.e. $f(x) = \infty$. But, we can't have a point $x$ that takes infinite value, since the range is the reals.


2

By the closed map lemma (http://en.wikipedia.org/wiki/Open_map), you can answer by the affirmative in the case where the group $G$ itself is compact. This is because metric spaces are always Hausdorff, so the inclusion map $i : G \to \mathrm{Isom}(M)$ is continuous, hence a closed map (by the closed map lemma ; note that it is not that hard to prove, you ...


1

Maybe the defintion on faithful I found was a bad definition but it seems from the definition of faithful every point $g$ of $G$ is an isolated point. Now from this it would seem that the big question is if $G$ has finitely many elements, since single point sets are closed sets and closed sets are closed under finite unions. If you are leaning towards $G$ ...


0

Both are correct. For the first, if you assume that it is not a singleton or the empty set, you can use the "Hausdorffness" to create for every $x$ of $C$ an open set that separates $x$ from the rest of $C$ because $C$ is finite. By this you create a finite separation of $C$, and this is a contradiction to its contentedness. For the second, every point in ...


2

The result is not true. Take $A=\{\pi+n+1\colon n\in\mathbb N\}$, and $ B=\left \{-n-1+\frac1 {n+2}\colon n\in \mathbb N\right\} $. Both sets are closed, $\pi $ is a limit point of their sum, but is not on their sum. If both $ A, B $ are compact, so is their sum (Since $ A+B $ is the image of $ A\times B $ under the continuous function $(x, y)\mapsto x+y ...


9

Let $A$ be the set of negative integers. Let $B$ be the set of all $n+\frac{1}{2^n}$ where $n$ ranges over the positive integers. Then $A$ and $B$ are closed. But $A+B$ is not closed, since it contains numbers arbitrarily close to $0$ but does not contain $0$.


3

It can't be a vector space because scalar multiplication doesn't behave as required. For instance, given your algebraic operations, we'd have $$\frac{1}{2} \star (2 \star \pi) = \frac{1}{2} \star 0 = 0$$ but $$\left(\frac{1}{2} \cdot 2\right) \star \pi = 1 \star \pi = \pi \ne 0$$ so scalar multiplication doesn't satisfy the necessary conditions. To ...


1

You can certainly define a 'scalar multiplication' as you have. But that doesn't make it a vector space. Why? Because 1) if it were a vector space, it would obviously be 1-dimensional because any element in $(0, 2\pi)$ is a generator under your scalar multiplication but 2) the additive group of the angles is not isomorphic to $\mathbb{R}$, because it ...


0

1) and 2) $C$ is not a vector space over $\mathbb{R}$, since if $x\in C$ such that $x\neq 0$, then $1\cdot x = \frac{x}{2\pi} \cdot (\frac{2\pi}{x}\cdot x) = 0$. 3) $C$ inherits a metric from the euclidean plane when viewing $C$ as the unit circle.


0

An isometry $f$ on the line satisfies $|f(x) - f(y)| = |x - y|$ for any two real numbers $x$ and $y$. If $f(x_0) = x_0$, you get $|f(x) - x_0| = |x - x_0|$. This leads to only two possible isometries: either $f(x) = x$ or $f(x) = 2x_0 - x$ which are the two you describe.


2

If $X$ is a standard flat torus and the equivalence relation declares two points to be equivalent if they lie on the same line with a (fixed) irrational slope, the quotient is clearly not metrizable.


-1

If $(X,d)$ is a pseudo-metric space, and if you define: $$x \sim y \iff d(x,y) = 0$$ then $(X/_\sim, d^*)$ is a metric space, with: $$d^*([x],[y]) = d(x,y)$$ It has to be checked that $d^*$ is well-defined. The space $(X/_\sim, d^*)$ is called the metric identification of $(X,d)$. If I remember well, this is an exercise in Stephen Willard's General Topology. ...


0

Not an answer, but too long for a comment: Given a complete Riemannian manifold $M$ such that every two points are joined by a unique minimizing geodesic it follows that for a fixed $p \in M$ every point $q \in M$ different from $p$ is a regular point for the distance function $d(p,.)$. As a consequence $M$ is diffeomorphic to $\mathbb R^n$. So if there is ...


1

If $d$ is a metric, then $\lambda\cdot d$ is an equivalent metric to $d$ for any $\lambda\in\mathbf R$. Assume that $X=\{x_1,\dots,x_n\}$. If $d$ is a metric on $X$, define $c_{i,j}:=d(x_i,x_j)$. The set of all possible metrics on $X$ is equipotent to a subset of $\mathbf R^{n^2}$, which is equipotent to $\mathbf R$.


4

Let $x$ be an element of $A$ and let $y$ be an element of $B$. We know that $\|x-y\|$ is a real number $r$ and so $d(A,B)$ must be at most $r$ by the definition of $d(A,B)$, hence $d(A,B)$ is finite.


11

The distance between two sets of the same metric space is defined as: $$d(A,B) = \inf_{x\in A,\ y\in B} d(x,y)$$ That means that if $x\in A$ and $y\in B$ then $d(x,y) \geq d(A,B)$. Now, $d(x,y)$ is always finite in a metric space so $d(A,B)$ must be too.


1

If the space consists of binary $m\times n$ matrices for fixed integers $m,n \gt 0$, then the number of such matrices is finite. Any finite metric space has a discrete topology, so there is only one metric topology up to topological equivalence. Indeed strong equivalence of metrics must also hold in this finite setting. Let $d_1(x,y)$ and $d_2(x,y)$ be two ...


1

Thanks wspin. However, let me elaborate a bit, some detail is still not clear to me. For any $s \in (-\epsilon,\epsilon)$, let $\gamma_s(t)$ be a minimal geodesic connecting $\gamma(0)$ with $\gamma(t+s)$. Let $V(t) :=\partial_s \gamma_s(t)|_{s=0}$ be the associated vector field along the original geodesic $\gamma$. By construction $V(0) = 0$ and $V(1) = ...


1

Assume there exists precisely one geodesic segment from $p$ to $q$, say $\gamma : [0,l] \to M$, parametrized by arc length. By compactness we can extend $\gamma$ to the intervall $[0,\infty[$. Now consider a sequence $\gamma_n : [0,l_n] \to M$ of minimal arc length geodesics from $p$ to $\gamma (l + \frac 1 n)$. Since limits of minimal geodesics are minimal ...


0

I want to mainly mention a point that hasn't been made in the other answers so far. If $X$ is any topological space and $x \in X$ has a finite neighborhood basis, then it has neighborhood basis of size one. This is because the intersection of any finite family of neighborhoods of $x$ is also a neighborhood of $x$, and is a subset of each of those ...


2

Each of your finitely many $U_i$ is a neighbourhood of $x$, hence by definiiton contains an open ball $B_{r_i}(x)$ for some positive number $r_i$. Pick one for each $i$. Let $r$ be the minimum of these finitely many positive real numbers, so $r$ is itself a positive real number. Consider $B_{r/2}(x)$. As $B_r(x)\setminus B_{r/2}(x)$ is nonempty, we have thus ...


2

The characterization of the topology on $M$ that you have written (which, I see, is the characterization on the wikipedia page for pseudometrics) has an error. It should be: "a set $A$ in $M$ is closed (open) iff $A$ is saturated and $h(A)$ is closed (open) in $M^*$".


0

If you look at $\mathbb Q$, $\mathbb Z$ and $\mathbb N$ as metric spaces disregarding the ambient real axis, you can distinguish them. The space $\mathbb Q$ is the only one of them without isolated points (the others have only isolated points). The space $\mathbb N$ contains such an element that no two other elements have the same distance to it; this is ...


1

If $X$ is a countably infinite set, then there exists a bijection $f:X\to \Bbb Q$. Now what about the metric $d(x,y)=|f(x)-f(y)|$.


1

How about $d(x,y)=\max\{1,|y-x|\}$ or $d(x,y)=\frac{|y-x|}{1+|y-x|}$?


2

1. Metrics that satisfy $d(x+r,y+r)=d(x,y)$: There are many metrics of this kind, as remarked by other users. Also the sum of any such metrics still has the property. 2. Metrics that satisfy $d(xr,yr)=d(x,y)$: There is such a metric on $(0,\infty)$, namely $d(x,y)=|\log(x)-\log(y)|$. There are also such metrics on $\mathbb R$, and something can be said ...


0

The logic of notation in the book is that $u,v $ are vectors and $s,t$ are scalars. So the composition $F^2(y+su+tv)$ is a function of two scalar variables $s,t$. We take second mixed partial of that function. What you wrote in the second equation may be familiar to you, but I do not recognize "partial derivative with respect to a vector" as a precise ...



Top 50 recent answers are included