Recent content by divergentgrad

\left( \begin{array}{cc} 0 & 1\\ -3 & 4\\ \end{array} \right) ?

Yes, thanks. As you'll see under number 2 above, that's exactly what I was considering doing.

Homework Statement I know that if a_n := n^{1/n} - 1, then \Sigma a_n is divergent. I know this (by the integral test) because the integral of 2^{1/n} - 1 from 1 to infinity is infinite. However, I want to avoid using non-elementary functions (here, the exponential integral) in my proof that...

I'm not sure if this question has any sense. Either way, hopefully someone can help me see either the right question or the right way of thinking about this. I don't have any special background in set theory, myself. A set is countably infinite if there is a bijection between it and the...

Ah, ah--I had those requirements mixed up. So it worked. Thanks for your help getting through the mire.

That looks so clear, but I think that what you brought up before is still an issue--unless one of the A_n is finite, I don't have a sequence of sets with limit \emptyset. EDIT: On the other hand it's still obvious that \bigcap^{\infty}_{n=1} A_n = Q - \bigcup^{\infty}_{n=1}B_n = Q - Q = \emptyset.

Back to the drawing board. Thanks for your help seeing the root of my misthought.

The original question, which appears in Robert Ash's Real Analysis and Probability, chapter 1.2: Let \mu be counting measure on \Omega, where \Omega is an infinite set. Show that there is a sequence of sets A_n \downarrow \emptyset with \lim_{n\to\infty} \mu(A_n) \neq 0.

No, no, I was addressing the second part of my question. I realize that the measure of the limit set is empty, because the limit set is the empty set. However, is the limit of the measure 0? (I realized from your post that I was just not thinking about it clearly. It is obvious that...

Thanks. I had thought up the An, Bn, believing it would work for the problem, until I puzzled over the limits. Now since B_n \subset Q, we have \mu(A_n) = \mu(Q-B_n) = \mu(Q) - \mu(B_n) = \infty - n. So then \lim_{n \to \infty} \mu(A_n) = \mu(Q) - \lim_{n \to \infty} \mu(B_n) = \infty...

It seems so obvious now. Thanks.

Homework Statement Suppose \Omega is an infinite set. If Q = \{x_1,x_2,...\} \subset \Omega is infinite and countable, and if B_n := \{x_1,x_2,...,x_n\}, A_n := Q - B_n , ... does A_n \downarrow \emptyset? If \mu is the counting measure on \Omega, is \lim_{n \to \infty} \mu (A_n) = 0?The...

Does every infinite set contain an infinite, countable subset? If so, why? (I'm assuming this depends on the Axiom of Choice).