Is there a collectionwise normal topological vector space which is not paracompact?

Vector space proof...?

• If V is a vector space with scalars, using the definition of a vector space show that 0 scalar * v = zero vector. How do I do this? Thanks!

• Answer:

  I do not provide "answers" that don't answer the question, and I will not post a solution if there is a correct solution already. Lucky for me, kb's proof in this question is TOTALLY INVALID -- it does not refer to the axioms. And it's also bad because it is excessively complicated by introducing arbitrary parameters like "a." Remember, kb, It is ABSOLUTELY NOT an axiom of vector spaces that -v = (-1)v. That requires proof. Worse yet, most proofs of this fact ASSUME this result, the one you are trying to prove. How terrible. Here is an actual proof. Let "z" denote the zero vector, so that we don't confuse it with the zero scalar which is "0". Use the compatibility condition of scalar multiplication. 0v + 0v = 1(0v) + 1(0v) [identity of scalar multiplication - axiom]         = (1+1)(0v) [distributive law - axiom]         = ( (1+1)(0) )v [ compatibility condition - axiom ]         = 0v [ what you know about scalars should tell you this* ]         = 0v + z [definition of zero vector] So 0v+0v = 0v+z. The vector 0v has some inverse u, and add that to both sides:         (0v + 0v) + u = (0v + z) + u Thus (commutative / associative laws):         0v + ( 0v + u ) = (0v + u) + z And since u is the additive inverse of 0v, you have:         0v + z = z + z Since z is defined by axiom as the vector so that         z + __ = __, we can remove it from both sides and conclude:         0v = z QED. Now THAT is a valid proof. * If you don't know this, consider a field F and 0 the zero scalar of F. We will show 0k=0 for any other scalar k in F. You know 0 = 0 + 0, so multiply both sides by any scalar k. Then 0k = (0+0)k = 0k + 0k. Thus 0k+0k = 0k. Subtract 0k from both sides to get 0k = 0. QED. That is what kb's proof is sort of like, except this works ONLY in a field, not in a vector space over that field!! But what would he know, he clearly is more interested in looking smart and getting points than providing valid solutions or having sportsmanship. What am I talking about? Well, I have a message for kb, someone I used to respect. Hey kb, how come every time I notice you are online I also notice that you get SIX votes for a question in which I provide a correct solution and you provided a crappy one? (You answered this question 30 minutes ago, which is exactly when your vote count shot up by SIX.) Why are you cheating? It is pathetic. I answered the question in full -- you didn't even bother to notice or look up the fact that part (b) was obviously relevant, and instead dismissed it as "unrelated." What nonsense. I am referring to: http://answers.yahoo.com/question/index?qid=20090927124235AAKUBb3 I am sick of seeing your answers be completely invalid -- like back when you said (Q>0,×) was a "divisible" group, because you can't tell the difference between (Q>0,×) and (Q,+): http://answers.yahoo.com/question/index?qid=20090927141823AA6VEMp Or when you give a patently false answer and then use fake accounts to vote yourself up: http://answers.yahoo.com/question/index?qid=20090922141840AAPHRC8 Don't kid yourself -- there are not six people on this entire site with brains enough to come, understand the permutation question, and yet not enough brains to avoid picking the answer that they can easily see is false. There is no way that you are not cheating, unless some benevolent cheater is cheating on your behalf without your knowledge, and just happens to always be on Yahoo!Answers exactly when you are. I expected better from someone I used to consider a worthwhile contributor to Yahoo and a good person. I always, always, ALWAYS delete my answer if there is a better answer, if I am wrong and someone else has shown a correct solution -- I have done this for you repeatedly, as a courtesy. And what do I get in return, for being honest, providing only good solutions, and showing you courtesy and respect? UPDATE: And now you are voting yourself up on an answer in which EVERYONE knows you are wrong because what you showed has NOTHING to do with growth rates of functions: http://answers.yahoo.com/question/index?qid=20090929085419AAArpFh You proved f(x) > g(x) when the kid EXPLICITLY TOLD YOU he did not want that. You have no understanding of the math involved, and yet here we are, with you cheating to get "best answer" when your answer is totally useless, and we all know it. Shame on you. So how many fake votes will you use on that question? How many are you going to use on this question? Are you going to ever stop cheating? Have you lost every scrap of integrity? Do you intend to just pretend you don't notice that we've caught on to your cheating? KB has denied that his proof is wrong, although he cannot deny cheating. KB, I challenge you then. Prove -v = -1 v in a vector space. Do not assume 0v = the zero vector. Cite only the axioms. Go on.