@JM I plea guilty of that. I gotta go for a while, maybe we'll talk again tomorrow morning (for you), afterwards I'll be gone for about a week. So, see you later or next week.
@AsafKaragila "Since we defined cardinality as equivalence classes, if we show that (A^B)^C is equinumerous with A^(B×C) for arbitrary sets." Perhaps you meant something like "...as equivalence classes, it suffices to show that.."
Sure, $$\operatorname{dom}A \cong \operatorname{ker}A \oplus \operatorname{im}A,$$ but it's not an equality, the equality is actually $$\operatorname{dom}A \cong \operatorname{ker}A \oplus \operatorname{im}{}^t\!\!A.$$
[I like to do things abstractly, but you can just pretend that in the followin...
It did. You pointed out an error after which the author deleted it, commenting: thanks. I'll try it one last time, if I still don't get it right I'll stop posting on mathSE after having had one beer...
Yes, unless one can come up with some really clever combinatorial argument.
OTOH, it's not obvious that the OP is asking for an elementary closed form -- just an algorithm to compute it might well satisfy him.
(Like: set up a Markov chain counting number of consecutive successes, with a trap state if the limit is reached. Write down its transition matrix. Raise to the Yth power).
To quote Arnold Miller's paper's footnote: "When I was a graduate student in the early 1970's and puzzles over whether the ground model must satisfy AC, Jack Silver told me, "But forcing has nothing to do with the Axiom of Choice.""
I probably could, but I am trying to remember the proof my friend gave me for some interesting AC related topic which came up on the site. I may succeed and then write an answer.
I am glad to see that my new meta thread is getting good support (+7 at the moment), as I was feared it would be downvoted.
Well, I lose a point for downvoting, so a sufficiently keen observer should be able to deduce that I've downvoted something. Which total rep change do you see for me on Nov 21?
Does it require rep to see the downvotes on the user profile?
Also, everybody can see the total number of downvotes I've ever cast. Someone who polls my user page regularly can figure out roughly when I downvoted.
Except ... either I'm a complete misanthrope compared to the rest of you, or only some of a user's downvotes are tallied when someone else views their user page. How many up/downvotes do you guys see on my user page?
@JonasTeuwen: Like I said, I don't push the button. If I think someone says something wrong then I can write a comment. And if a question doesn't belong here it's more constructive to vote to close rather than to down vote.
On the front page of the users-by-votes-this-month, it seems I'm bested (worsted?) only by cosine666 (5.2 up per down), Austin Mohr (5.7 up per down) and Benjamin Lim (6.7 up per down).
Are there keyboards that type an apostrophe as ´ rather than '? If you want the context, the OP of this question always uses the former kind of aphostrophe marks, and I am not too fond of it.
Keyboard layouts for languages that use accents typically have dead-key accents somewhere, which are used to type, say, è. If the key you press after the dead key is not something that can be composed with the accent (in the keyboard lay-out designer's charset), the dead-key is converted to a free-floating accent such as ` or ´.
My impression is that some users habitually use this method for typing apostrophes, because they have never noticed that there's a difference between an appostrophe and a free accent.
I am going to upgrade bash, kernel, xorg stuff and the nvidia driver. I should return in 10 minutes or so. When I do, I demand to have 15 or more reputation points.
@JonasTeuwen First, I have no idea how to upgrade a kernel without a reboot. Not to mention that some modules require a recompile with dkms after a kernel upgrade; second, I have to unload the nvidia module to upgrade it (or at least to reload the new one) my experience shows that upgrading without rebooting soon after may heat in the fan stopping and everything heating to near melting point.
Interesting. FF9 should have a significant JavaScript performance improvement.
So this is a super basic question, but I'm going to ask it anyway: I'm working through Rick Miranda's Riemann Surface/Algebraic curves book. Why is the emphasis on projective space and not affine space? Why do algebraic geometers care so much about projective space?
I had a downvote for some hours which was retracted after it had already qualified another upvote to count partially. I expect it to be rectified at the next rep recalc.
So I was looking for Ahlfors again in preparation for my final, and there's a problem in the very first chapter in the section about stereographic projection that says "find the images of the vertices of a regular tetrahedron inscribed in the sphere (that we are projecting from) under stereographic projection." (Problem 3, page 20 in the 3rd ed.). What is the purpose of this problem? I think you just get a really messy, unenlightening answer.
It can be made simpler by choosing the right orientation of the tetrahedron. Inscribe a cube in the sphere such that the phases are parallel to the axes. Take every second vertex of the cube to be vertices of the tetrahedron.
Hmm. I wonder whether there's anything better to do then than to compose the projection formula with rotation formulas for the tetrahedron, and call that a day.
In other news, I am one problem away from completing my complex analysis problem set, but there is one problem that still resists me. It's going to occupy me during my Thanksgiving dinner :(
@J.M. I figure I'd rather catch the "obvious" thing after 2 hours then not catch the "obvious" trick after two hours and feel dumb when someone shows me the solution.
I'm hoping this change in my work habits is going to have a positive long term impact on my mathematics.
That's another thing you'll want to address. You can and will make dumb mistakes from time to time. You have to be upfront, and be careful not to repeat those mistakes though.
