@J.M. I had a question about accents. I understand the meaning and need for the accent in Poincaré. But what does it do in Bézout or Bézier? Should I elongate the vowel sound while pronouncing the word?
@JM Um, I tried the variations. But these distinctions just seem smaller than the least count of my tongue. =) // Thanks for the explanation though. :)
@JM I always wondered why English speakers are adding those "hints" of y to french e-aigu (they tend to exaggerate this a lot). To my (not quite native) ears it sounds way better without any such hint.
@HenningMakholm No, definitely not. Because it is usually quite long, the é may get a tiny little bit closer to an i towards the end of the pronunciation but hardly discernible, I think.
i hate most russian classics, they're written by neurotics for neurotics, and then interpreted by psychotics to mean whatever they fucking want them to mean.
Anyway, I was totally frustrated reading 'Golden Age' literature like Tolstoy, Dostoevsky etc. who were just chewing snots and encouraging 'Great Russian Revolution' instead of just saying people that without doing your job well you'll always be in a crap
He is deep, but the point is that this literature is about thinking and reflecting - as if it was missing in Russia where almost everybody preferred (and still prefers) to think and reflect guessing What would be if...
The closure of this L^\infty-question came surprisingly quick for me... It would have been nice if at least one of the voters had left an explanatory comment.
@AlexeiAverchenko I liked him as a person at least. Cliche to say, but among Russian writers I mostly like Bulgakov, not only because of MM but also because of White Guard
spelling reform is nice and all, but not when millions are dying at a war that nobody wants and the injustices of the land distributions are too numerous to count
@Srivatsan Let me try to explain you. There were Tsars who were autocrats and most of whom didn't care about people. There was the class called intellegence (people of a high culture) and there were young hot-blood empty-brain revolutioners
[contd.] this intellegence thought how to tell the Tsar that things should go in a liberal way and they were dreaming about making TheGreatRussianRevolution. Only thinking. Meanwhile revolutioners were bombing everything and everybody and killed the only smart Tsar
@Srivatsan the trick is that as a result the power was taken neither by infantile intellegence nor by revolutioners, but by bolsheviks who are suspected to receive money from Russian's enemies (the WW I was going on that time)
@AlexeiAverchenko this should be transparent. in the History textbook it should be written explicitly: Lenin was funded by Kaiser, so the whole October revolution was just a bullshit. Dot.
@Gortaur What's not in the history books? About the only thing everyone knows about Lenin is the sealed-train ride through Germany -- obviously it was in the German interest to support a Russian government that would agree to closing down the Eastern front.
@HenningMakholm 14-years old children do not necessary make such statements if it's not said explicitly
@HenningMakholm Nobody wanted to continue the war, but the war was to be finished with Russian to win. After October revolution Russia 1. betrayed France and others 2. lost the war
Why is complex projective n-sapce, defined by (C^(n+1)-{0})/C^*, Hausdorff? I looked in Munkres and he just says that in general, it's hard to find a condition that makes quotient spaces Hausdorff.
If it has a nice searchable title and description, I think a better solution would be to write an answer of your own, pointing to the MO thread and summarizing the solution.
Suppose a collection subsets U_\alpha of a set X is given. Define a topology on each U_\alpha. Define a topology on X by declaring a subset V open if and only if V intersect U_\alpha is open in U_\alpha for every \alpha. I have verified this is a topology. Question:
Suppose each U_alpha is connected. Form a graph with one vertex called v_\alpha for each U_\alpha, and with vertex v_\alpha connected to v_\beta iff U_\alpha intersect U_beta is nonempty. Prove or disprove: X is connected if and only if the graph is connected.
Could I get a hint for this? Not homework, just self study.
Martin: Oh, I took uniform measure over [0,1] and a step function. Your function is simpler (constant or identity?) but the measure is a combination of the Dirac measures. // Summarizing to myself basically.. :)
The uncountability is making this tricky. I want to say assume we have a separation U,V of X, and show this leads to a separation of the graph.
Suppose a collection subsets U_\alpha of a set X is given. Define a topology on each U_\alpha. Define a topology on X by declaring a subset V open if and only if V intersect U_\alpha is open in U_\alpha for every \alpha. I have verified this is a topology. Question:
Suppose each U_alpha is connected. Form a graph with one vertex called v_\alpha for each U_\alpha, and with vertex v_\alpha connected to v_\beta iff U_\alpha intersect U_beta is nonempty. Prove or disprove: X is connected if and only if the graph is connected.
I want to say take two of the U_\alpha, one in U, one in V, and show this leads to a contradiction, but I have a problem because the "path" of the graph joining of them my be uncountable.
Sorry for the interruption: can anyone tell me if this fails to be a counterexample and why? X = [1,2] union [3,4] with the usual topology. U1 = [1,2], U2 = [3,4], U3=X all with the usual topology. (Should verify that this is indeed ok.) (v1, v3) is an edge and so is (v2, v3). So the graph is connected. But, clearly the original space is not.
Ok so we take a connected component and wish to show the complement is open. The complement is open if and only if its intersection with every U_alpha is open. But this is now a trivial statement
Let our two vectors be w_1, w_2. Pick the one with smaller modulus, and let epsilon be half than that modulus. It suffices to show the ball around 0 with radius epsilon contains no point of the lattice.
New duplicate of the old question. But this one seems to be better worded (the new question states a lemma that can be used that is missing from the other questions).
Huh, scratch that. This question seems a little different.
