math.stackexchange.com/questions/3778134/… – I wonder if I should post an answer which simply points out that all non-discrete finite topological spaces are also non-Hausdorff.
Technically that doesn't answer the question at all; in order to have a complete answer to the question, you'd also need to have a "naturally occurring" example of a finite topological space which is not discrete.
@Lelouch It might be interesting to note what happens to quasi-isometries "in the limit". Suppose $(X, d_X)$ and $(Y, d_Y)$ are metric spaces and $f : (X, d_X) \to (Y, d_Y)$ is a $(K, \varepsilon)$-quasi-isometry, so $K^{-1} d_X(a, b) - \varepsilon \leq d_Y(f(a), f(b)) \leq K d_X(a, b) + \varepsilon$.
Scale the spaces $X$ and $Y$ by some large constant $\lambda >> 1$, so $d'_X = \lambda d_X$ and $d'_Y = \lambda d_Y$ are the new metrics. Then $f$ becomes a $(K, \varepsilon/\lambda)$-quasi-isometry with respect to this scaled metrics
If $\lambda \to \infty$, then supposing $(X, d_X)$ and $(Y, d_Y)$ converge in appropriate Gromov-Hausdorff sense to some limit $(X_\infty, d_X^\infty)$, $(Y, d_Y^\infty)$, $f$ induces a $K$-biLipschitz map between them
Eg scaled limit of the Cayley graph of $\Bbb Z^n$ is $\Bbb R^n$, so any quasi-isometry of $\Bbb Z^n$ gives a bilipschitz homeomorphism of $\Bbb R^n$ in the limit.
Heh I guess this gives a short proof that $\Bbb R^n$ is not quasi-isometric to $\Bbb R^m$ if $m \neq n$; both the spaces are stable under scaling
This is not a good perspective for hyperbolic space, because if you scale a $\delta$-hyperbolic space by $\lambda$ you get like a $\delta/\lambda$-hyperbolic space, so as $\lambda \to \infty$ you get a $0$-hyperbolic space... what is known as an $\Bbb R$-tree
Horrible objects in general
But anyway in the limit a quasigeodesic will be a bounded Lipschitz distortion of a tree path in this $\Bbb R$-tree, or something like this
For example you have a_n = (1/n) which has the limit of 0 as n approaches infinity. Ist ist enough to claim that a_n is convergent?
In order to prove that a sequence is convergent one should prove that it is 1. bounded and 2. monotonic. Does 'having a limit' contains both of these conditions in itself?
@Leaky Nun Convergence means: A function approaches a final and constant term as the input approaches infinity and limit is that constant value the the function approaches
Now I see that having a limit automatically means that the sequence converges
But then comes up another question: Why in math textbooks it is usually said, in order to prove convergence we should prove that it is 1. bounded and 2. monotonic? Why not simply say: If a_n has a limit then it converge?
@Yuvraj $f(x)$ is a constant function. Any continuous and differentiable function satisfying $f(x)=f(kx)$ where $k\in \mathbb R^+$ and $k\neq 1$ needs to be a constant function.
@Yuvraj $f(x)$ is a constant function. Any continuous and differentiable function satisfying $f(x)=f(kx)$ where $k\in \mathbb R^+$ and $k\neq 1$ needs to be a constant function.
@Yuvraj Note that this condition also implies that $f(x) = f(x/k)$ for all $x$ as well. (Why?) Therefore you can assume without loss of generality that $k < 1$. From there, you can consider $f(k^nx)$ for higher and higher $n$, and use the continuity of $f$, and I leave the rest to you.
Is there a good way of seeing that $\mathbb{R}[X,Y]/(X^2+Y^2+1,aX+bY+c)\cong\mathbb{C}$ (where $a,b$ aren't both zero) without doing the ugly computations
@BalarkaSen Nice, but at the risk of being wrong, doesn't this work to show that $\mathbb{R}^m$ and $\mathbb{R}^n$ are not quasi-isometric ? It suffices to show that $\mathbb{Z}^m$ and $\mathbb{Z}^n$ are not quasi-isometric. So basically, assume $m < n$. consider the ball of radious-$r$, in $\mathbb{Z}^n$. By the ifrst property of quasi-isometry, the image of $\mathbb{Z}^m$ inside this ball contains $\Theta(r^m)$ lattice points.
On the other hand, by the second properrty of quasi-isometry, this contains $\Theta(r^n)$ lattice points. Which is possible iff $m = n$
it's probably equivalent to what you said, I can't say becaues I don't understand how are you defining convergence of metric spaces
You say two compact metric spaces $X$ and $Y$ are $\varepsilon$-close if there is a map $f : X \to Y$ which is a $\varepsilon$-isometry, i.e., $d_Y(f(x), f(y))$ is bounded above/below by $(1 \pm \varepsilon) d_X(x, y) \pm \varepsilon$ and is $\varepsilon$-surjective, meaning every point of $Y$ is in $\varepsilon$ distance from $f(X)$
A sequence of compact metric spaces converge Gromov-Hausdorff to some metric space X if for any 1/n the tail of the sequence is 1/n close to X
Hello, is it generally true that a principal ideal of a ring (left, right, or two sided) are the same as cosets of the multiplicative group of the ring in itself?
I might not be thinking of the same thing you're thinking of, but the answer looks like "no": $\Bbb Z^\times = \{1,-1\}$, but principal ideals in $\Bbb Z$ are $n\Bbb Z$, which aren't cosets of $\{1,-1\}$.
https://math.stackexchange.com/questions/3780293/calculate-int-0-infty-frac-log-x-dxxaxb-using-contour-integra I have a question regarding taking the limit with residue
in this example at z=0, the residue is 0, so no problem with taking the limit when epsilon->0
on the selected answer @ (1) the author just has written that the integral vanishes as epsilon tends to 0.
my question, how to handle a case when the circle is not 0. I mean the limit won't be 0.
@hashman: No, $0$ is a branch point of the function, so residue doesn't even make sense. I don't understand your question. The whole reason the residue theorem works is that if you have a meromorphic function $f(z)$ with a pole at $a$, then the integral of $\int f(z)\,dz$ around a little circle centered at $a$ is precisely $2\pi i$ times the residue of $f$ at $a$.
Do you want to think about some topological dynamics with me? I don't know if I'm missing something or the book is missing an hypothesis (I assume it's the first)
So the setting is a continuous $G$ action on $X$, $G$ is a topological group and $X$ an Hausdorff space. The action is called (topologically) transitive if any open set has dense orbit, and point transitive if there is a point with dense orbit
It's obvious that point transitive implies transitive
And the book says that if $X$ is second countable then the converse also holds
I see that with some extra assumptions ($X$ compact, or more generally Baire)
But the book so far has been very careful in stating explicitly when $X$ is assumed compact and it's not an underlying assumption
The point is that if $Y\subseteq X$ is the set of transitive points (points with dense orbit), then $Y=\Bigcap UG$ where $U$ ranges over a basis for the topology of $X$, so if $X$ is second countable $Y$ is actually a comeager $G_\delta$ set, but I don't see why must it be nonempty without extra assumptions
@Alessandro Yeah I think this is just false. Take $f : S^1 \to S^1$, $f(z) = z^2$, and restrict the system to the dyadic rational points on $S^1$; then it's still a chaotic system but every orbit is periodic I believe
For semigroups you don't even have that point transitive implies topologically transitive, look at $\{0\}\cup\{1\n\mid n\in\Bbb N\}$ and $f(1/n)=1/(n+1)$
This image you will recognize as a re-indexing of the group law for $\Bbb{Z}_6$, together with a coloring that indicates the equal subgrids of the whole grid.
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