I'd say that a set theory grad student should know pretty much all the first part of Jech, and at least half the second part. The third part is the advanced stuff.
I don't want to be a set theory researcher, frankly. I would like to know set theory to the point where I can understand the questions & answers like the one you gave me.
If you want a survey, I'm not sure which one is good. If you want a book then Jech is a very good book. The first part is about 200 pages, that's not THAT much.
Sheesh, ads for Facebook and Google+? I'd rather they used the ad space to mention any new cra- er, features they might be putting into the SE engine...
Okay, "she". :) Well, if she's now using the latter account, she won't be able to comment on posts (questions and answers) associated with the former account.
@JM Austria was great, a bit too cold for my taste, but the weather was nice enough for some walks at the Danube canal. Way too many tourists, though :)
@Srivatsan No it wasn't. I don't know how many times I was there (in Vienna most of the time). About a dozen times, I'd guess. Some very good friends of mine live there.
@Srivatsan I just saw your comment. Note that this doesn't help because the OP can't flag herself (you need 15 rep for that) and given the questions asked so far, it is unlikely that she will earn that anytime soon. In such obvious cases I usually flag for the mods to merge the old account into the new one.
@Gigili Your answer and the answer which OP copied from his book differ only in constant. Constant are unimporant in undefinite integrals. So both answers are in fact correct. (Both are primitive functions.)
That's what tards is explaining in the comment bellow yours.
@MartinSleziak Aha, I was trying to understand that comment but the constant 'C' we normally add after the anti-derivation process is something else, substituting x for 1+t^2 or t^2 would make the answer wrong, no?
@AsafKaragila I disagree. They're barking up trees in different forests. One is in the department called "we left that way behind us" and the other one is called "there's still a long way before we can even dream of tackling that one".
@Srivatsan Since the probability of drawing a white ball is the number of white balls over the total number of balls, and the probability of drawing a white ball is the expected probability over all cases, the probability of drawing a white ball is the expected number of white balls over the total number of balls.
Does that make sense, or should I expand on that some?
@Matt: there is no such thing as "the" compactification, there are many, and this one happens to be homeomorphic to [0,1]. The result asked about in that question is equivalent to Bolzano-Weierstrass, as Srivatsan's answer shows. The other direction is even easier.
@Srivatsan No. I think I went through all the questions and answers that Google yielded on site:math.stackexchange.com eom.springer -"link rot". Unfortunately, most of those roughly 50 results are in comments.
My first "answer" (that I edited before even posting) did all the bounding part by hand. Then it struck me that some of what I was doing must be known before.
It did not mention any of the words max-norm, operator norm, Weyl's inequality (perturbation). =)
@tb: I was looking for some paper from the 70's a few days ago and couldn't find it. Today I stayed at home, and I pulled a Buehler and found a .pdf scan :-)
By the way, I saw a similar file in treeware on my advisor's desk the other day. It was pretty neat. Typed out in a typewriter and the mathy stuff was filled by hand.
Yes, the famous filling in process. Halmos describes that somewhere. He dictated the typewriter stuff to his secretary and said what kind of symbols he would need to fill in later...
@tb and @Matt, is there a standard example of a sequence a_n such that: (a) a_n > 0, (b) a_n converges to 0, (c) \sum (-1)^n a_n does not converge. [To prove "tightness" of Leibniz test.]
I know it won't be nice simply because it is not monotone decreasing.
