757 reputation
212
bio website
location
age
visits member for 2 years, 6 months
seen Oct 14 at 2:01

Oct
12
comment Why König's lemma isn't “obvious”?
I don't think it's obvious at all. Intuitively, one can imagine building a tree, continually adding on longer and longer branches, but being careful never to extend any one branch too far. Indeed, this intuition is precisely how one constructs an unbounded, computable, finitely branching tree with no computable infinite branch. Or an infinitely branching counterexample. The assumptions are in some sense tailored to the proof, so I would say precisely the opposite: the proof is obvious (to someone who understands induction), but the lemma requires careful formulation.
Sep
30
awarded  Explainer
Sep
27
comment why have we chosen our number system to be decimal (base 10)
@J.M. The word digit is used to refer to the numerals below ten because you can count them on your fingers, yes, but using that as evidence for why we selected ten as a base in the first place is bogus. Using the multiple meaning of "digitus" as evidence that we originally selected base ten because of our fingers is like answering "Why do we use strings of numerals to assign unique identifiers to individual phone lines?" by pointing out that "digits" is common slang for "phone number". Certainly the slang came after the choice, not the choice to justify the slang.
Sep
27
comment why have we chosen our number system to be decimal (base 10)
@MichaelEdenfield The positional system was new, yes, and made arithmetic much easier. But the choice of ten as a base goes back to Mesopotamian civilizations, and this particular choice doesn't make arithmetic easier at all.
Sep
26
comment Weak convergence vs Convergence in Measure
@smiley06 $X_n$ have associated push-forward measures $\mu_{X_n}$ defined by $\mu_{X_n}(A) = P(X_n \in A)$.
Sep
20
comment Why some people don't like proofs by contradiction
In many cases, the direct proof actually establishes a stronger result and then applies modus ponens. For example, a useful way to prove that some particular set $A$ does not belong to a class $\mathcal{S}$ is by proving that every member of $\mathcal{S}$ contains an element not in $A$. If I did the argument by contradiction, I would assume $A\in \mathcal{S}$ and then deduce that $A$ contains an element not in $A$, a contradiction. The stronger statement--that every set in $\mathcal{S}$ contains elements outside of $A$--is obscured.
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 $[\epsilon, 2\pi-\epsilon]$.