somewhere near the beginning.

Bolzano-Weierstrass theorem

Filed under: Mathematics — Alex @ 12:08 pm 2/6/2005

As I mentioned before, I’m taking a complex analysis class that is being taught via the Moore method: we are given definitions, and then asked to prove propositions and defend them against the class; this is the entirety of the class. One of the stipulations is that we are not to reference outside sources, or discuss the solution to problems among ourselves. Unfortunately, I’m taking the second half of a Real Analysis course, so I have to reference outside sources. Usually, this is not a problem— I’ve made the decision to recuse myself from doing any problems that I feel I have had an unfair advantage on.

Now I have a conundrum. I spent a week and a half worrying about one of the propositions:

If s_0, s_1, \ldots is a bounded sequence of complex numbers, and |z|<1, then  \sum_{n=0}^\infty s_n z^n converges.

I finally got it, using a subdivision argument that I thought was very unique. In the process, I proved the triangle inequality and the nested interval theorem, which are tools that I think would really speed up the progress of the class. For instance, the last proposition— to prove that if a series converges, then the associated sequence converges to zero— is trivial to prove once you have the triangle inequality. But the guy doing it either didn’t think of using the triangle inequality, or didn’t think he could prove it. Instead, he took a circuitous, although ingenious, roundabout method of proof that took two classes to lay out, and he’s still not finished!

The problem is, in preparation for a test this Friday in my real analysis class, I started reading a supplementary book on real analysis, and yesterday I came across the Bolzano-Weierstrass theorem:

If x_n \in \R^n and there exists a K such that |x_n|\leq K for all n, then we can find  n(1)<n (2)<\ldots and x\in\R such that x_{n(j)} as j\rightarrow \infty.

Turns out that the proof I have is very similar to the proof the book gives for the theorem. So should I recuse myself? Given the people in the class, I think if I don’t prove this one, we’re going to end up skipping it.

Possibly relevant posts:

No Comments »

No comments yet.

RSS feed for comments on this post. TrackBack URL

Leave a comment