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Jan
24
comment Prove the uniformity of the Cantor/Lebesgue function defined on $A^c$ where $A$ is a Cantor set on $[0,1]$
Thank you for a lucid explanation. Is this a stylistic choice on the author's part that is reasonable to expect most readers to understand or would most writers make the distinction between $A$ and $A_k$?
Jan
24
accepted Prove the uniformity of the Cantor/Lebesgue function defined on $A^c$ where $A$ is a Cantor set on $[0,1]$
Jan
23
asked Prove the uniformity of the Cantor/Lebesgue function defined on $A^c$ where $A$ is a Cantor set on $[0,1]$
Jan
7
asked Why do we construct the Lebesgue measure with finite measure sets before sets of arbitrary measure?
Jan
7
comment Why do we usually construct the Lebesgue Measure on finite outer measure sets before arbitrary measurable sets?
I'm still a little confused as to why that's an issue. Can you elaborate?
Jan
7
comment Why do we usually construct the Lebesgue Measure on finite outer measure sets before arbitrary measurable sets?
Surely the first method does not imply that every set had measure $\infty$, only those that have inner measure $\infty$ have measure $\infty$, which is what we want, correct?
Jan
7
asked Why do we usually construct the Lebesgue Measure on finite outer measure sets before arbitrary measurable sets?
Dec
24
revised Prove that every nonempty open subset $G$ of $\mathbb{R}^n$ can be expressed as a countable union of nonoverlapping closed rectangles.
added 285 characters in body
Dec
24
comment Prove that every nonempty open subset $G$ of $\mathbb{R}^n$ can be expressed as a countable union of nonoverlapping closed rectangles.
@bof, I suppose the problem could be wrong, but I am quite sure that I am quoting the problem correctly. It is from the book Lebesgue Integration on Euclidean Space by Frank Jones. It is Chapter 2 Section A Problem 9.
Dec
23
comment Prove that every nonempty open subset $G$ of $\mathbb{R}^n$ can be expressed as a countable union of nonoverlapping closed rectangles.
Why does the density of $\{x_n\}$ mean that all points in $G$ must have been covered? Aren't the points in $G$ uncountable? Maybe I have some trouble with the terminology "countably dense" which I have not encountered before.
Dec
23
comment Prove that every nonempty open subset $G$ of $\mathbb{R}^n$ can be expressed as a countable union of nonoverlapping closed rectangles.
For this questions, the rectangles must not overlap and they are $n$-dimensional. They could also be cubes.
Dec
23
asked Prove that every nonempty open subset $G$ of $\mathbb{R}^n$ can be expressed as a countable union of nonoverlapping closed rectangles.
Dec
20
awarded  Constituent
Dec
16
accepted Calculate $I_m = \int_{-\infty}^\infty \frac{dx}{1+x+x^2+\cdots+x^{2m}}$ using complex variables
Dec
16
comment Calculate $I_m = \int_{-\infty}^\infty \frac{dx}{1+x+x^2+\cdots+x^{2m}}$ using complex variables
Lightbulb moment. Of course. Thank you!
Dec
16
comment Calculate $I_m = \int_{-\infty}^\infty \frac{dx}{1+x+x^2+\cdots+x^{2m}}$ using complex variables
How do you get from $\left(e^{-i8\pi/5} - e^{-i6\pi/5} + e^{-i16\pi/5} - e^{-i12\pi/5}\right)$ to $\left(e^{i2\pi/5} - e^{i4\pi/5} + e^{i4\pi/5} - e^{-i2\pi/5}\right)$?
Dec
16
asked Calculate $I_m = \int_{-\infty}^\infty \frac{dx}{1+x+x^2+\cdots+x^{2m}}$ using complex variables
Dec
15
accepted What conditions are necessary on $a,b,c,d$ so that the Mobius transformation $w=\frac{az-b}{cz-d}$ has only one fixed point?
Dec
15
comment What conditions are necessary on $a,b,c,d$ so that the Mobius transformation $w=\frac{az-b}{cz-d}$ has only one fixed point?
@DanielFischer, on your last question, no. I am only assuming that $ad-bc \neq 0$. I thought the "standard" definition of a Mobius transform implied that $a,b,c,d \in \mathbb{R}$ -- I guess I'm mistaken on this point. Also, I would be happy to upvote your comments if you were interested in submitting them as an answer (since you are addressing my question directly).
Dec
15
comment What conditions are necessary on $a,b,c,d$ so that the Mobius transformation $w=\frac{az-b}{cz-d}$ has only one fixed point?
@DanielFischer The reason I believe it should lie in $\mathbb{R}$ is because $a,b,c,d \in \mathbb{R}$. If there is only one fixed point, then (if I'm right) the discriminant is zero and what remains is in $\mathbb{R}$. I think the only way for a complex number to "arise" is if the discriminant < 0. I could be mistaken.