The major difference between 1. and 2. is that 1. is a sentence (with some truth value), while 2) is not.
When we say: "Let n be [something] satisfying P(n)" for some proposition P, we are not positing anything.
In a proof by induction, the goal is to prove A: P(1) and B: P(n) \Rightarrow...
Ok, let's take the most common form of Nakayama's lemma.
Statement 2: If M is a finitely-generated module over R, J(R) is the Jacobson radical of R, and J(R)M = M, then M = 0.
9a) Since R is local, the Jacobson radical (as the intersection of all maximal ideals) is simply the maximal ideal M...
9a) is just Nakayama's lemma for local rings, no?
9b) Suppose that r \in R is a non-zero, non-invertible element. Then the ideal (r) is contained in some maximal ideal m. But the Krull dimension of the integral domain R_m is zero, so the sequence of inclusions of prime ideals (0) \subseteq m...
I've corrected my question a bit, it now more accurately reflects the title. The complement of a dense set for the original question (where m had the fixed value n) could very well be the correct answer. I can't prove it right now, but I'll look into it. For general m however, it's not the right...
I was wondering if anyone knew of a name for such a set, namely a subset S \subseteq \mathbb{R}^n which at every point x \in S there exists no open subset U of \mathbb{R}^n containing x such that S \cap U is homeomorphic to either \mathbb{R}^m or the half-space \mathbb{H}^m = \{(y_1,...,y_m)...
One can consider a disk containing all 5 points on the interior, and deformation retract down to its boundary. Then draw 5 disks containing each point but none of the others, and deformation retract to their boundaries. Now you're left with a big closed disk with 5 holes in it (it is a closed...
The velocities are low, but so is the velocity differential. The difference in acceleration for two objects A and B with a tiny separation d is small, and I don't quite see why it should be sufficiently larger than any resulting relativistic effects.
I see, so I gather that "frame of...
The one true way to obtain a random chord, is to uniformly pick two points P,Q on the circle. These choices amounts to a uniform choice of angles \theta,\phi \in [0,2\pi). The angle POQ, where O is the center, is equal to \sqrt{2R^2-2R^2\cos(\theta-\phi)} = \sqrt{2}R\sqrt{1-\cos(\theta - \phi)}...
The space is, like WWGD explained, homotopy equivalent with a bouquet of 5 circles, and the fundamental group is the free product of 5 copies of \mathbb{Z}. Regarding the first homology group, this is the abelianization of \pi_1, hence the free abelian group \mathbb{Z}^5 (which can independently...
@PeterDonis Like I said, it was my motivation for my question, and did not involve black holes at all (S is not a black hole singularity in that example). Since I do not know general relativity, I merely attempted to look at what my (limited) knowledge of special relativity could tell me in such...
What do you mean by a Newtonian regime? I am not sure one can approximate the answer to that with Newtonian mechanics. Especially if the mass of S is large.
The question as presently stated does not require a third frame of reference, I believe. It refers exclusively to a system of two...
I'm not attempting to understand black holes from an SR perspective.
Assuming a non-black hole gravitational body: Let me pose the question as follows: A can meaningfully measure a set of distances to B in its own frame of reference, by regularly emitting and receiving photons being reflected...