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seen Apr 14 at 4:57

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awarded  Curious
May
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awarded  Good Question
Apr
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awarded  Yearling
Jan
6
revised when can you estimate curvature from finite information about two geodesics?
Adding a condition to the problem to avoid these counterexamples.
Jan
6
comment when can you estimate curvature from finite information about two geodesics?
Please see my comment on @PostNoBulls answer. I interpret your answer similarly as indicating that bounds on the side lengths are needed. This avoids sequences of triangles becoming degenerate as in your answer, or "too global" as in his answer. My intuition is that avoiding those and/or having enough homogeneity/symmetry will make it work. Constant curvature spaces are a bit silly of me to ask this question in. Maybe something like $\mathbb{H}^2 \times \mathbb{R}^2$ makes more sense as a potential example.
Jan
3
comment Soft Question:Is the following a Paradox?
@ABC I think the OP didn't have that sort of paradox in mind. After checking Quine's classification, I think he's asking if it's an antimony. I argue it isn't. Moreover, it can't be a falsidical paradox, since there is no demonstration, and hence no fallacious one. It's a single statement, and presumably the OP is asking whether it yields a paradox under some correct investigation.
Jan
3
comment Soft Question:Is the following a Paradox?
This isn't a question about English. It's ultimately more appropriate for math, but probably moreso philosophy until the problem is more formally stated. English is being used to convey the problem, but it's ultimately a problem about some kind of structure in the statement and whether there is a problem under some kind of evaluation. I.e. whether some function is well-defined. The issue here is that it isn't clear what that function is because we don't have a solid context.
Jan
3
answered Soft Question:Is the following a Paradox?
Jan
2
revised when can you estimate curvature from finite information about two geodesics?
Mostly, I think the title was unappealing.
Dec
31
comment $\forall$ is “distributiv”
For an analogy: it's like we're proving "for all x, x is red if and only if it is large" in a universe where all objects are large red balls. The proof wouldn't require any definition chasing to see how "red" and "large" relate to each other (and really, they may generally have nothing to do with each other), so it won't look like a math proof you're used to. You can simply observe that everything is red, and everything is large, so the $\leftrightarrow$ holds by just looking at $\leftrightarrow$'s truth table (it's true when both sides are true).
Dec
31
comment $\forall$ is “distributiv”
You're actually proving $\forall x (\phi \rightarrow \psi)$ using only $\forall x \psi$, which is a theorem. You could use $\forall x \psi$ to prove $\forall x (P \rightarrow \psi)$ for any P. The hypothesis part is irrelevant since the conclusion is true in the theory already.
Dec
31
comment $\forall$ is “distributiv”
Yes, I mean that $\forall x, y \phi(x,y)$ and $\forall x,y \psi(x,y)$ are true for these $\phi, \psi$ over the reals. So, you can prove $\forall x,y (\phi(x,y) \leftrightarrow \psi(x,y))$ like this: "Let $x,y\in\mathbb{R}$. Both $\phi(x,y)$ and $\psi(x,y)$ are true by the triangle inequality (noting $|x-y|=|x+(-y)|$ if you'd like), so $\phi(x,y)\leftrightarrow\psi(x,y)$ is true. Hence $\forall x,y (\phi(x,y)\leftrightarrow\psi(x,y))$." This is a valid proof, but it's silly...
Dec
31
comment $\forall$ is “distributiv”
It holds for sort of trivial reasons in your example as you've noticed, which is why you find proving it strange. Here both $\phi$ and $\psi$ are always true because of the triangle inequality (a property of the absolute value function). You don't have to use the hypothesis in a proof of an implication. This just indicates that stuff in the theory (here, the triangle ineq for $||$) is sufficient and you didn't need the hypothesis. Both $\phi$ and $\psi$ are always equivalent to "true" here. This is in contrast to something like "for all integers n, n is equal to 1 or 3 mod 4 iff n is odd".
Dec
30
answered Is it possible to change the step interval on a sum?
Dec
30
revised when can you estimate curvature from finite information about two geodesics?
adding another bounty, clarifying what I want.
Dec
30
awarded  Benefactor
Dec
30
answered $\forall$ is “distributiv”
Dec
28
comment when can you estimate curvature from finite information about two geodesics?
Ah, nice counterexample. Do you think it helps if we bound the lengths to avoid arbitrarily large or small triangles? For example, only look at side length 1 and compare at halfway along them? This is more what I had in mind. Or, do you have any insight into why it might be true for particularly nice spaces? Symmetric spaces in particular.
Dec
23
awarded  Promoter
Dec
23
answered Intuition, proof, one-sided group definition - Any set with Associativity, Left Identity, Left Inverse is a Group - Fraleigh p.49 4.38