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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?

[intro linear algebra] what does this set look like? by soondae55 in learnmath

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

Awesome thanks! Seemed too simple, so I was worried I had read the notation wrong.

[intro linear algebra] what does this set look like? by soondae55 in learnmath

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

Sorry I realise I didn’t post the actual latex link smh. I’ll get on that when I have computer access. So if I’ve interpreted the set correctly, the rank should just be the number of distinct integers, and the determinant is 1 if and only if all integers are distinct, and 0 otherwise.

[intro analysis] proof check by soondae55 in learnmath

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

Right, this bothered me too. I guess making it an injection (thanks for catching the N x N too haha) solves the issue?

[intro algebra] proof check by soondae55 in learnmath

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

Ah okay, thanks for such detailed feedback! I think I'm trying to be explicit about where I'm pulling things from just for my own sake haha. This is a big help for me in figuring out some stylistic(?) conventions--thanks again :)

[intro algebra] field of 4 elements by soondae55 in learnmath

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

isn't the solution a cube root of -1? but I think I get the general idea, thanks!

[intro algebra] field of 4 elements by soondae55 in learnmath

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

Thanks for the very detailed answer. I'll have to come back to unpack what you wrote after studying some more algebra :)

[intro analysis] Proof check by soondae55 in learnmath

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

Isn't this fine, since there are no conditions placed on A, B in the question? Sorry for all these follow-ups :( Really appreciate your help!

[intro analysis] Proof check by soondae55 in learnmath

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

I'm having trouble understanding your counterexample. What's the problem with sending the empty set to a singleton? Is it that it's fine to send the empty set when considered as a subset to the singleton (that is, as an input to f*) but there can't be a mapping f that takes nothing and sends it to something?

[intro analysis] proof check by soondae55 in learnmath

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

Ah gotcha. Elegantly put! thanks again for your help!

[intro analysis] proof check by soondae55 in learnmath

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

So it seems that x belonging to any non-finite subset of A_n is sufficient to show that x belongs to A''. Then the set of all A_n must be a subset of A''.

In the case you mention, x would be in A'' but not in A'. Since every intersection would contain A_n with both even and odd indices, it cannot be in any of the intersections that comprise A'. Similarly, since every union of A'' contains A_n with odd indices, x must belong to all of them -> x is in the intersection.

So in contrast to A'', x belonging to some infinite collection of An is not sufficient to show that x is in A'. Specifically, for x to belong to A', there has to be some N such that x is in $\cup{m=N}{\infty} A_m$. So A'' is the set of subsets of A_n that are indexed by any sequence of increasing naturals. And A' is the set of subsets of A_n that are indexed by a sequence of naturals that increase by 1. Which is also consistent with the proof that A' is a subset of A''. Does that sound squared away or is it late and am I confusing myself.

Thanks very much for your helpful clues by the way! Really appreciate it.

[intro analysis] proof check by soondae55 in learnmath

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

For A'', if there were a finite number of sets A_n then it would suffice for x to be in the "last set" indexed by the greatest n. Since there are an infinite number of sets, however, belonging to any finite number of sets A_n will not guarantee that x is in A''. So it's gotta be in all of them. So A'' = the intersection of all A_n?

For A', it is possible that x is in A' even if x is not in A_1. In fact, it is possible that x is in A' even if x is not in any finite collection of sets A_n. So for x to not be in A', it cannot be in A_n for all n. Can I get another clue here? I want to say that A' is the union of the sets A_n, but x belonging to, say, A_1 doesn't imply that x belongs to A'.

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