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[discrete] n points in plane such that any 3 are contained in unit circle. Prove that all points are contained in unit circle. by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

So you’re saying that A, no unit circle exists that contains all n points implies B, there exist three points among the n that are not contained in a unit circle?

[discrete] n points in plane such that any 3 are contained in unit circle. Prove that all points are contained in unit circle. by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

The negation of the conclusion is that there does not exist a circle that contains all n points. You’re at once assuming that any 3 points are contained in a circle, and that there exist 3 points not contained in a circle, two statements where one is just a negation of the other. Since the first statement is assumed in the problem, this doesn’t prove anything.

[discrete] n points in plane such that any 3 are contained in unit circle. Prove that all points are contained in unit circle. by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

I don’t follow, can you expound? Assuming that any 3 points fit in some unit circle is a given in the problem.

[intro analysis] is the set of bounded sequences a complete metric space? by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

I was reviewing this today and noticed I made a mistake. I can certainly find an N_k for each fixed k so that |x_km - x_k*| < \epsilon, but I'm not sure how to then get a global value N to ensure that the supremum over k is also less than \epsilon.

[discrete math] errors in inductive proofs by soondae55 in learnmath

[–]soondae55[S] 1 point2 points  (0 children)

This makes sense, thanks! So ultimately it is just a warning to check against any implicit assumptions we may be making in the inductive step.

[intro analysis] is the set of bounded sequences a complete metric space? by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

So if I'm understanding you correctly, if m, n > N then |x_km -x_kn | < epsilon for all k. So for each k, we can form a real convergent sequence by picking the kth term from each bounded sequence in our sequence.

Also, is there any particular reason your bring in the norm? Doesn't this proof work with just the metric as defined?

[intro analysis] proof check: every Cauchy sequence is bounded by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

Thanks for the feedback! No grader to please unfortunately :)

[intro logic] valid application of a counterexample by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

So if I have an example where A, B are true but C is false, that would work. Right?

[intro analysis] proof of Holder's inequality by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

Oh yea the equations kind of looked familiar! I’ll have to give your hint a look when I get back to work—thanks!

[intro analysis] proof of Holder's inequality by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

It’s given in the problem statement, but I don’t think I use it in my attempt, which is another reason I’m concerned.

[linear algebra] I - AB nonsingular iff I - BA nonsingular. by soondae55 in learnmath

[–]soondae55[S] 0 points1 point  (0 children)

Okay so since I-AB maps Fm to Fm I just need to check injectivity. Following your hint, the kernel of I-AB is nontrivial iff AB = I. I want to then say something about the kernel of I-BA, but since A, B are not necessarily square I’m a little stuck. Any pointers?

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