Oh, I met Deven and we discussed you briefly. He was one of the people running our training session.
Really cool thing that came up as an exploration: Characterize all numbers that can be written as the sum of at least two consecutive positive integers. [I'd never seen this before.] Hint: They're called trapezoidal numbers. @Balarka @Leaky @Akiva @anyone else
No, it doesn't, @MikeM. I figured out easily why the name. But only last night did I give a proof characterizing the necessary and sufficient condition for an integer to be trapezoidal.
hey, guys. Quick question: if I take a finite subset out of $\mathbb{Q}$, is the result still dense in $\mathbb{R}$?; I think it is, because a finite amount of elements does not affect the convergence of a sequence.
But Deven had a good laugh at my expense, because we (and, in particular, I) had talked about the importance of reading carefully, and my head forgot the word "consecutive" ...
I have ignored some users in chat by clicking the ignore button, but there is no button on the main site, so I will have to manually ignore them there.
@Will It's your favorite, but I'd be rather cautious about saying it's truly the best, like Ireland/Rosen and Niven/Zuckerman are supposed to fantastic
@TedShifrin Steamy just pointed out something that should have been obvious. We don't have to worry about negative terms at all, since if they occur, we can cancel symmetrically around the term that's zero
Well, I plan to give a few of my own exercises, but I think most of the exercises are more thought-provoking than any high school or college precalculus text's.
I never knew where you guys got negative numbers in the first place. I certainly didn't think about the problem with any negative numbers.
Sort of the way Leaky's went. But I gave the odd numbers explicitly in the form $2k+3$, so take one row of $k+1$ dots and one row of $k+2$ dots. There's your trapezoid.
Otherwise, I broke things into two cases.
Probably not the most elegant, but it was a proof.
So... first a sanity check, if $E\subseteq\Bbb R^n$ has Hausdorff dimension stricly smaller than $n$ then $E^c$ must have Hausdorff dimension $n$, right?
So if we call $H^d$ the $d$-dimensional Hausdorff outer measure (working in some $\Bbb R^n$ with $n>d$) and $M_d$ the $\sigma$-algebra of $H^d$-measurable sets what's the relation between $M_r$ and $M_s$ when $n>s>r$?
@TedShifrin How do you define the Hausdorff measure? In my course we defined the outer measure first and then got an actual measure when restricting to its $\sigma$-algebra of measurable sets, but I don't see a straightforward way to do it without going through the outer measure
$M_r\subseteq M_s$ since all $H^r$-measurable sets have measure $0$ with respect $H^s$ and if we have an outer measure the sets with measure $0$ are measurable
Ah, actually I'm not sure. $M_r$ is full of sets that don't have Hausdorff dimension $r$
Hi. Let $2^A$ denote the power set of a set $A$, and let $\Omega=[0,1]$. It is true that $(\Omega,2^{[0,1]})$ is a measurable space, but we typically consider, e.g., the Borel $\sigma$-algebra of $[0,1]$ rather than $2^{[0,1]}$ because no reasonable measure $\nu$ can be defined so that $(\Omega,2^{[0,1]},\nu)$ would be a measure space?
That is, I understand in what way $2^{[0,1]}$ is too big, but am I right that it is still a valid $\sigma$-algebra and the problem appears only when looking for a ''reasonable'' measure?
Very Secret Number Theory Study Group
Very Secret Number Theory Study Group
