Recent content by D_Miller

I have a problem I can't quite figure out: I have a first order system S, and an interpretation I of S. I have to show that a closed well formed formula B is true in I if and only if there exists a valuation in I which satisfies B. I've done one of the two implications, but I still have...

I have a problem in my logic course which I can't get my head around: I have to show that there is a well formed formula \mathcal{A}(x_1) in the formal first order system for arithmetics, \mathcal{N}, with precisely one free variable x_1, such that \mathcal{A}(0^{(n)}) is a theorem in...

The only non-obvious prerequisite for the cosmology course (which is actually about more than just cosmology) is a course on Riemann geometry and Einstein metrics - a course I'm taking as we speak. This is another reason why I'd prefer that the other course has something to do with a branch of...

I have almost finished my bachelor's degree in mathematics, with a minor in physics. A while ago I decided not to continue with physics on my master's degree, but to focus on pure mathematics while learning a bit of computer science as well. However, after planning out the two semesters of...

I may be missing something completely obvious, but why is this true? Thanks for your time, guys.

I'm terribly sorry to kick this thread up the board when others probably have more urgent problems, but I'm very curious as to why/if micromass' previous argument (post #16) is enough to prove the stated problem. I'd appreciate it if anyone took the time to elaborate...

Yes, but how does this eliminate the possibility that there are simply multiple vectors with the smallest norm? The task is to show that M doesn't contain any such vectors at all.

Hmm. I do have a bad habit of over-complicating mathematics. Thanks for your help though. There is a second part to the problem, which I'm not supposed to do. Apparently it's relevant to different course, but just for the heck of it, it asks me to show that M does not contain a vector with...

Yeah, it stands to reason that a sequence f_n:M\rightarrow \mathbb{R} converges uniformly to f:M\rightarrow \mathbb{R} iff \lim_{n\rightarrow\infty}\| f_n-f\|_S=0. But is it enough to write that \int_{0}^{\frac{1}{2}}f(x)dx-\int_{\frac{1}{2}}^{1}f(x)dx=\lim_{n\rightarrow...

Uniform convergence would be in \mathbb{R}, but the norm is stated in the assignment itself to be the sup-norm.

Perhaps I've overlooked something, but how can you be sure that the convergence is uniform? I was thinking about the convergence theorems of Lebesgue integration, but if the convergence is uniform, we can interchange the limit and the integral for Riemann integrals as well.

I'm sorry but these kind of proofs always confuse me. I take it I'm supposed to use the result that a set is closed iff it contains all its limit points and then look at \int_{0}^{\frac{1}{2}}f_t(x)dx-\int_{\frac{1}{2}}^{1}f_t(x)dx=1 and prove it by letting limits pass through the...

Of course. That makes sense and I have now done it. Any idea as to which of my closed set definitions would be easiest to use?

Yeah, if I'm not mistaken I have to prove that for f,g\in M and 0<t<1, the fuction h_t=(1-t)f+tg also lies in M. Then I have to show that the complement of M is open or, equivalently, that the closure of M is equal to M itself.

I am taking a self-study individual course on convex analysis but I'm having some troubles with the basics as I'm trying to do the exercises in my notes. I'm asked to consider the space C[0,1] of continuous, complex-valued functions on [0,1], equipped with the supremum norm \|\cdot\|_{\infty}...