haha, no it's good advice. i should be studying for the next exam but i can't focus...i'm too angry with myself. hence i am here in this chat procrastinating.
@akiva part of the problem is that surface isn't well-behaved along the axes. it might help to ask mathematica to instead plot $x^2y^2+x^2z^2+y^2z^2-xyz-\epsilon$ where $\epsilon$ is some small value
@Semiclassical Eh, the graph Wolfram Alpha gives is good enough. (By the way, it might look like it's bounded, but every point on the axes satisfies the equation. So, with a small epsilon, we end up with something approximating the bounded bit, plus six infinitely large tubes surrounding the axes.)
urgh, I don't really understand the definition of the simplicial circle (S^1 as a simplicial set) as a "quotient" of Delta^1 by its boundary. There are 3 2-simplices and 2 1-simplices. What are the three face maps from the former to the latter? I see two obvious ones, i.e. the only two non-decreasing maps that collapse a certain couple of adjacent cells into one, but I don't see a third one
@SohamChowdhury It's technically a mistake. The vertex set should be some $\mathbb R^k$ and the edges are just curves that pass through only elements in the vertex set. I couldn't find a notation for the class of continuous curves, just the class of continuous functions $C^0$.
Having a continuous curve on only the real-line that doesn't pass through itself would simply be an interval.
@AkivaWeinberger You're right. Subdividing $1 \to 2$The number of paths from $x$ to $4$ is the sum of the paths from $1 \to 4$ and $2 \to 4$ minus the paths that pass between $1 \leftrightarrow 2$
In this case, there are 4 such paths that pass between $1$ and $2$, so we know there are $5 + 5 - 4 = 6$. Subdivision does not preserve paths.
On the contrary, there exists a proof of correctness for truth-tables. That proof of correctness would show an equivalence between a series of well-formed formulas and a truth table. I don't know what it is, but it must exist. So we must be able to prove equivalence logically.
Truth tables are just easier for binary logical operations
I've read that two groups of order $n$ are isomorphic if $\gcd(n, \varphi(n)) = 1$. How come this doesn't work for $\mathbb{Z}_{4}^{\times}$ and $\mathbb{Z}_5$? We have $\gcd(4, \varphi(4)) = \gcd(4, 2) = 2.$ But these two groups are isomorphic and of the same order.
I meant $\mathbb{Z}_5^{\times}$ and $\mathbb{Z}_4$
Your criterion probably means "Any two (abelian!!) groups of order n are isomorphic iff gcd = 1". For n=4, you learned there are multiple abelian groups of order 4.
from the wikipedia article on hyperbolic geometry: There are at least two distinct lines through a point P not on a line R that do not intersect R, where all lie in one plane. This means that there are through P an infinite number of coplanar lines that do not intersect R.
I don't see how the latter statement follows from the former unless we say something like "two straight lines intersect in no more than one point" any hints?
@Simplex: For me, it is easiest to make the argument directly from any one of the models of hyperbolic geometry. But what they must intend is that all the lines through P "between" the two known lines not intersecting R will likewise not intersect R.
Ah, yes, that's a problem with literature, history, and philosophy courses ... or a good thing, depending on one's viewpoint. I actually loved writing literature papers, especially in French.
Ted: I was trying to derive the fact using only Euclid's first 4 axioms and the assumption that there are at least two such lines. I do see how it follows easily once the idea of betweeness is established though
@Simplex Elaborating on Ted's idea, perhaps construct a circle centered at P, intersect it with those two lines and iteratively take midpoints between two of the intersections (have to make the right choice here)
They actually prove as a corollary that for manifolds of dimension at most 3, the cotangent bundles are symplectomorphic iff the manifolds are diffeomorphic.
