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Hand with Reflecting Sphere - M.C. Escher

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1d
awarded  Constituent
Dec
17
comment A real number $x$ such that $x^n$ and $(x+1)^n$ are rational is itself rational
I love it. This post should go to your list "Some fun answers of mine" in your profile page :)
Dec
16
comment Borel measurability is a local property
@PhoemueX: Ah! I finally get it :) Thank you! Feel free to post your last comment as an answer.
Dec
16
revised Borel measurability is a local property
added 238 characters in body
Dec
16
comment Borel measurability is a local property
@Ilya: If you could elaborate, I would appreciate it :)
Dec
16
comment Borel measurability is a local property
@PhoemueX: How do we know $\{y: g_{x}(y)>a\}$ is an open set? We only know it is a Borel set.
Dec
16
comment The set of differences for a set of positive Lebesgue measure
@JDH: Beautifully expressed. I wish more users had the same attitude.
Dec
16
comment A compact set, which is not closed.
Well, compact subsets of a Hausdorff space are closed… So the example you are looking for will come from a non-Hausdorff space.
Dec
16
comment A power series that converges for $|x| \leq 1$ and diverges otherwise.
Nice. Also, I love your hat choice!
Dec
16
revised Borel measurability is a local property
added 5 characters in body
Dec
16
accepted Typo in the book or am I going crazy?
Dec
16
asked Borel measurability is a local property
Dec
16
comment Irreducible in $\Bbb Q[x]$
1. Gauss's Lemma. 2. Zircht's Answer. 3. If a polynomial $f$ with coefficients in $\mathbb{Z}$ vanishes at infinitely many points, then $f=0$. 4. Apply step 3 to either $g(x)-1$ or $h(x)-1$ in Zircht's answer.
Dec
16
answered Typo in the book or am I going crazy?
Dec
16
comment A Pasting lemma for measurable functions
@Michael Would you mind posting that as an answer? So we can get this question out of "unanswered" list :)
Dec
15
comment Convergence of a series and sum
I see that you have not upvoted before. Please take a look at this page and considering voting answers to your questions :)
Dec
15
accepted Convergence in measure implies convergence almost everywhere (on a countable set!)
Dec
15
comment Convergence in measure implies convergence almost everywhere (on a countable set!)
Thank you, Davide! This is a very nice solution. I especially liked your edit, as it clarified where the assumption "$X$ is countable" was used.
Dec
14
asked Convergence in measure implies convergence almost everywhere (on a countable set!)
Dec
14
answered relation between units and non zero divisors in a ring