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Jul
31
comment What are some conceptualizations that work in mathematics but are not strictly true?
It seems like you're trying too hard. Your way doesn't have to be the "pedagogically most efficient" way to be justified. It could be the fourth, fifth, or thousandth most efficient way and it would still be a damn good way.
Jul
31
comment What are some conceptualizations that work in mathematics but are not strictly true?
Outer measure is a perfectly reasonable definition of "area", and agrees with the Lebesgue definition on the measurable sets. The existence of so-called "non-measurable sets" (terrible name: you can easily measure them using their outer measure, or their inner measure for that matter) just shows that the most obvious definitions of "area" fail to be sigma-additive. Which is not surprising, considering sigma-additivity is a ridiculously powerful thing to ask of a reasonable notion of area, certainly when you also want to have the Axiom of Choice.
Jul
29
comment Integral test for convergence?
This would actually be an example where the conclusion of the integral test holds, although the hypotheses don't. Both the series from 1 to infinity and the integral from 1 to infinity of $\cos(\pi x)$ fail to exist.
Jul
27
revised What is the proof to the fact that all prime numbers are 1 above or below a 6 multiple?
added 1 character in body
Jul
27
revised What is the proof to the fact that all prime numbers are 1 above or below a 6 multiple?
added 95 characters in body
Jul
27
answered What is the proof to the fact that all prime numbers are 1 above or below a 6 multiple?
Jul
27
comment A Result About Sequences And Series
While this hint is not directly applicable (for example, $a_n=\frac{1}{n\log n}$ doesn't converge in sum), the proof of her hint using Cauchy criterion can be adapted to prove your result. Here is the proof of her hint, see if you can adapt it to your problem: math.stackexchange.com/a/369824/29892 this one is from @robjohn. There are of course other proofs of her hint (one nice proof combines Cauchy condensation test and the squeeze theorem), but the one I linked to is more easily adaptable to your problem.
Jul
27
suggested suggested edit on $\lim_{n\to\infty}a_n$ and Cauchy condensation
Jul
27
comment Are there any well known mathematicians who published very little?
I'm not an historian, but I only know of four publications from Baire. He struggled with his health all his life, which kept him from publishing as much as he might have otherwise.
Jul
27
comment Countable or uncountable set 8 signs
@DavidBevan Disjoint figure-eights cannot fill the plane. If the figure 8s consist of circles, any line will only intersect each figure 8 in finitely many points. If the figure 8s consist of rectangles, any circle will only intersect each figure 8 in finitely many points. Lines and circles are uncountable, but the plane only fits countably many figure 8s.
Jul
26
comment Sum of two closed sets in $\mathbb R$ is closed?
Perhaps it's too trivial to mention, but if both are compact, then of course their sum will be compact (the continuous image of a compact set is compact).
Jul
24
awarded  Critic
Jul
24
comment Lebesgue's Density Theorem - intuition and weaker forms
Catchy slogan: "The edge of a measurable set is negligible."
Jul
22
comment $S(x)=\sum_{n=1}^{\infty}a_n \sin(nx) $, $a_n$ is monotonic decreasing $a_n\to 0$: Show uniformly converges within $[\epsilon, 2\pi - \epsilon]$
Interesting to see how Dirichlet did it. Another way: $\sin(kx)=Im(e^{ikx})$ so $\Sigma_{k=1}^N sin(kx)= Im \left(\sum_{k=1}^N e^{ikx} \right)$ which by Geometric Series formula is $Im ( \frac{1-e^{ix(N+1)}}{1-e^{ix}} )$ which, maximizing the top, is $\le \frac{2}{|1-e^{ix}|}$. Now $|1-e^{ix}|$ is continuous with zeros only at the endpoints of $[0, 2\pi]$. So on compact subsets, it attains a uniform minimum $>0$, which gives us a uniform finite bound $M_\epsilon$ on $[\epsilon, 2\pi-\epsilon]$.
Jul
22
comment Showing that rationals have Lebesgue measure zero.
Using the countable additivity of Lebesgue measure doesn't seem in the spirit of the problem, or the example proofs given. The originally approach does in fact work, with slight modification. Notice we run into the problem that $\sum_{n=1}^\infty 2\epsilon =\infty$. We need a convergent sum. The fix clearly presents itself: simply take the intervals to be $(q_n-\epsilon_n, q_n+\epsilon_n)$ where $\epsilon_n=\epsilon\cdot 2^{-n}$. Then $\sum_{n=1}^\infty \epsilon 2^{-n}=4\epsilon$, which now goes to $0$ with $\epsilon$.
Jul
2
awarded  Curious
Jun
9
awarded  Popular Question
May
9
comment How does Hilbert's Nullstellensatz generalize the “fundamental theorem of algebra”?
It is usually the "weak Nullstellensatz" that is referred to as a multivariable analogue of FTA. For $k$ an algebraically closed field the "weak Nullstellensatz" can be stated as: "For any proper ideal $I$ in $k\left[X_1, X_2, \ldots, X_n\right]$, $V(I)$ is non-empty." Since ideals in $k[X]$ are principle, the $1D$ analogue contains precisely the same content as FTA. A proper ideal $I$ in $k[X]$ is given by a non-constant polynomial with zeros $V(I)$.
May
9
revised What is the completion of this space?
deleted 22 characters in body
May
9
comment Advice on taking a Course in Logic.
It might be worth replacing the word "logic" everywhere with "mathematical logic". As you can see from this comment section, there is a big difference. For example people who study plain "logic" are capable of saying things like "set theory really isn't logic".