Recent content by Kocur

Let us assume that X has Bernoulli distribution, with P(X = 1) = p and P(X = 0) = q = 1 - p. Of course, E(X) = p and Var(X) = pq. Now, since pq < 1, standard deviation is bigger than variance. I have got the following question: Does this fact make standard deviations and theorems based on...

Thank you for help guys. I think I got it at last :biggrin: . By the definition of union of sets, any point x of a union belongs to at least one of the contributing intervals. Since the respective intervals are open, each interval containing x must also contain a neigbourhood of x. A...

I think I got it: By the definition, any point x of a union belongs to at least one of the contributing intervals. Since the respective intervals are open, each interval containing x must also contain a neigbourhood of x. By the definition of union of sets, the union contains some...

Matt, I am clearly very resistant to the reasoning :blushing: . I think it goes like that: By the definition, any point of a union belongs to at least one of the contributing intervals. An open interval is an interval which does not contain its borders. Combining these two above we...

Matt, I think I that I do not quite get it. It is clear that any point of the union belongs to at least one of the contributing intervals (by the definition). Unfortunately, the second part escapes me.

I am a computer scientist that only recently got interested in topology. I have got the following question: How to prove, in an elegant way, that an arbitrary sum of open intervals (a, b), where a is in R and b is in R, is open? I have got some kind of a proof, but I am not sure it is...

I think that you are right AKG. Thank you for helping me out of confusion. Kocur.

AKG wrote: Huh? If your operation is associative, then no, the result will not be different. What does: x + (y + z) may be different from x + (z + y) and x + (y + z) may be different from (y + z) + x and x + (y + z) may be different from (z + y) + x. have to do with whether or not (x...

How much commutativity in associativity? Please correct me if I am wrong. By the definition, binary operation "+" on set S is associative if and only if, for all elements x, y, and z from S, the following holds: x + (y + z) = (x + y) + z. In other words, the order of operation is...

Well, you might consider my remark to be really silly :smile:. I was thinking about subtraction of reals and the following ideas came to my mind: From one side, we may consider subtraction to be addition of the inverse of an element, for example: (5 - 3) - 2 = (5 + (-3)) + (-2). This...

Well, I would like to thank you all for your input. Now, I have a lot of stuff to think about again it seems:).

I have been thinking about bases of vector spaces and have come to some strange results. I mean, we cannot present vectors without giving a base first. That is okay :smile:. But which base shall we use for presenting the base used? How can we identify "the real" elements (vectors) behind our...