Reputation
1,839
Next privilege 2,000 Rep.
Edit questions and answers
Badges
5 21
Impact
~40k people reached

  • 0 posts edited
  • 1 helpful flag
  • 186 votes cast
Oct
29
revised Let $r_n$ be an enumeration of the rationals in $[0,1]$, does the sequence $\{B_{\frac{1}{n}}(r_n)\}_{n=k}^m$ cover $[0,1]$ for $m-k$ finite?
edited body
Oct
29
asked Let $r_n$ be an enumeration of the rationals in $[0,1]$, does the sequence $\{B_{\frac{1}{n}}(r_n)\}_{n=k}^m$ cover $[0,1]$ for $m-k$ finite?
Oct
29
asked Sequence of continuous fuctions $f_n:[0,1]\rightarrow [0,1]$ s.t. $\lim_{n\rightarrow\infty}m(E_n(\varepsilon)) = 0$ but…
Oct
29
comment Trying to do a measure theory proof concerning almost everywhere convergence
Thanks! this makes perfect sense.
Oct
29
accepted Trying to do a measure theory proof concerning almost everywhere convergence
Oct
28
revised Trying to do a measure theory proof concerning almost everywhere convergence
deleted 7 characters in body
Oct
28
revised Trying to do a measure theory proof concerning almost everywhere convergence
added 63 characters in body
Oct
28
asked Trying to do a measure theory proof concerning almost everywhere convergence
Oct
24
comment Explaining projective space to master students
I'm actually a masters student who is just now learning projective space on my own, and I found the introduction to projective space given in the book Ideals, Quotients, and Algorithms to be really good and eminently understandable, you might take a look at it and write some notes up based on it, on your own.
Oct
21
accepted What does arithmetic actually mean (as an adjective)
Oct
21
comment What does arithmetic actually mean (as an adjective)
Ok thanks André, so by number-theoretic you mean dealing with the problem of finding the subset of a polynomial's roots which happen to lie in your non-algebraically closed field? Also, are there algebraically closed fields which do not contain the algebraic numbers?
Oct
21
asked What does arithmetic actually mean (as an adjective)
Oct
21
accepted Prove that any subgroup of a metabelian group is metabelian
Oct
21
comment Prove that any subgroup of a metabelian group is metabelian
nm, it's fine thanks.
Oct
20
comment Prove that any subgroup of a metabelian group is metabelian
This is the second isomorphism theorem, so this should be provable with out it, I'm not sure if I can use it or not.
Oct
20
comment Prove that any subgroup of a metabelian group is metabelian
I dunno just to be thorough I guess, it is isn't it?
Oct
19
asked Prove that any subgroup of a metabelian group is metabelian
Oct
13
accepted Prove that in $\mathbb{R}$, if $|a-b|>\alpha$ for all $a\in A$ and $b\in B$, then outer measure $m^*(A\cup B)=m^*(A)\cup m^*(B)$
Oct
13
comment Prove that in $\mathbb{R}$, if $|a-b|>\alpha$ for all $a\in A$ and $b\in B$, then outer measure $m^*(A\cup B)=m^*(A)\cup m^*(B)$
hmm ok I'll play around with that, thanks.
Oct
13
comment Prove that in $\mathbb{R}$, if $|a-b|>\alpha$ for all $a\in A$ and $b\in B$, then outer measure $m^*(A\cup B)=m^*(A)\cup m^*(B)$
Ok ya I kinda went down that path too, but there aren't any theorems about the outer measure of the countable union of disjoint intervals being equal to the sum of there lengths.