Recent content by Thorn

I've been thinking...and am starting to think that I don't understand complex projective space...So, it's defined as ( Cn+1 \{0,0} / C\{0} ). Now, I think this is just the set of planes in 4 space that pass through the origin... and one can consider how they would all intersect a 3 sphere and...

I have a question about complex projective space... specifically CP1 which can be thought of as the action of C on C^2\{0} which gives rise to the equivalence classes of "lines" passing through the origin in C^2 (but not including the 0) Now, any vector in complex space, when multiplied by the...

If a sequence of measurable functions (real-valued) converges in measure, is it true that you can find a subsequence that converges almost uniformly? (This is obviously true if m*(domain) is finite...but in general is it?) If so, can someone outline a little why?

Say we have open sets from two topological spaces. A is an open subset of T1 and B is an open subset of T2. So for these two open sets, A, B. Is A X B open itself? I see this is the case in R x R, where R is the real number line. I am wanting to say that this is just true in general... If so...

Ha...well then it seems that EVERY non measurable set would be an example of m*(U Ei) < Σ m*(Ei)..

So I take it, then there isn't an example of this...even for non-measurable sets..?

yeah... m* is the Lebesgue outer measure. So, you can say things like m*(E) when E isn't even measurable!? I didn't think m would be defined from a non measurable set...

I was told that you can find a disjoint sequence of sets...say {Ei} such that m*(U Ei) < Σ m*(Ei).. That is the measure of the union of all these sets is less than the sum of the individual measure of each set... This is obvious if the sets aren't disjoint...But can someone give me an example...

Well...Actually, I could use Cauchy's theorem for Abelian Groups... "If G is a finite abelian group and p is a prime that divides |G|, prove that G contains an element of order p... we know that p divides |G|...and p2 divides the order of G...and so on and so on until we have pt divides...

Geez...I have no idea where to start... Ok, I was looking at a lemma... "Let G be an abelian group and a \in G and is an element of finite order. Then a = a1+a2+...+ak, with ai an element of G(pi), where p1,...,pk are the distinct positive primes that divide the order of a" ... now...

If G is an abelian group of order (p^t)m, and (p,m)=1, show that G(p) has order p^t and G(p) = {a e G| |a|=p^m where m is a natural number} any suggestions?

I told you to watch out for that 2x! ><

OH YEAH! DONT IGNORE THE ABSOLUTE VALUE...cause e^(-x) is a lot different that e^(x)...do you have limits on that integral?

Well, yeah...intergrate by parts! and try to reduce x^2! that is write e^(whatever) as the u'...so that when you apply the formula, you actually differentiate x^2...thus making it 2x...then do it again...and make it 2...then you will have nothing more than an integral with e(whatever that...

ha...tell me what you get