Hi SSequence:
Yes.
If ##M## is a transitive model, then yes. For general models, the answer is no, see this stackexchange thread: https://math.stackexchange.com/questions/2118514/two-models-of-zfc-such-that-there-is-a-isomorphism-between-their-ordinals
Generally, forcing starts with a countable transitive model M of ZFC. Since it is countable, M will not even be close to containing all subsets of the naturals, so it is possible to add a new generic set G to M. G will not be in the L of M[G], since L^{M[G]}=L^M \subseteq M \subset M[G], but may...
Hmm... you and I have very different intuitions about what the most insightful solution is. I would consider mentioning that column rank = row rank to be far more insightful than offering a formula for the solution, since it tells us why there is a solution in the first place. Additionally, it...
The ##w_i##'s exist because each ##v_i## is in the range of ##X##. Saying that there is some ##w## such that ##Xw=v## is simply what it means for v to be in the range of ##X##. I'm really not sure what more I can say on that point.
As for ##Z##, I'm just using the fact that every linear...
All very true, but actually orthogonal to the point I was trying to make. What I was trying to get at is that if you are trying to find the limit of a rational function (or something which is effectively a rational function, since \sin (\theta) ~ \theta for small \theta) at the origin, the...
This limit does not exist. Note that the function isn't even defined on the path x=-y^2, and if you consider paths close to that path (e.g. x = -y^2 + y^4), the limit diverges to infinity.
Sure. Both players will almost surely be wearing at least one white hat. For i=1,2, let n_i be the position of the first white hat on player i's head. Then have player 1 point at hat n_2 and player 2 point at hat n_1. It is easy to see that both players win iff n_1 = n_2, which happens with...
No, this isn't true. Consider two fair coins flipped independently, let A be the event that the first coin comes up heads, B the event that the second coin comes up heads, and C be the event that at least one of the coins comes up heads. Then P(A) = P(B) = 1/2, P(A,B) = P(A)P(B) = 1/4, but...