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2d
revised Analysis in $R^n$
added 13 characters in body
2d
comment Analysis in $R^n$
@user262860 Yes, it is false.
Aug
29
comment Analysis in $R^n$
Personally, I feel that the linear-algebra tag would be more suitable to your question.
Aug
29
answered Analysis in $R^n$
Aug
29
awarded  Nice Question
Aug
29
comment Characterization of the weak topology
Where you wrote $M$ in your definitions should that be $A$?
Aug
28
comment Finding the order of permutations in $S_8$
This is a very nice answer. I allowed myself to correct a few typos, hope you don't mind.
Aug
28
revised Finding the order of permutations in $S_8$
deleted 2 characters in body
Aug
25
revised If $|xH|$ has order $n$, then there is an element $y$ with $|y|=n$ and $xH=yH$
edited tags
Aug
21
revised If equality of dual space of a Banach spaces implys the equality of pre-duals?
edited tags
Aug
20
comment $\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$
What happens if $x<0$?
Aug
20
comment $\arctan (x) + \arctan(1/x) = \frac{\pi}{2}$
What happens if $x < 0$?
Aug
20
answered Is the condition “the inverse image of a closed base set is closed?” sufficient for continuity?
Aug
20
answered Irreducibility Implication between $\mathbb{Q}[x]$, $\mathbb{Z}[x]$, and $\mathbb{Z_p}[x]$
Aug
19
answered Bijective Conformal Mapping onto the Open Unit Disc $\mathbb{D}$
Aug
19
comment If $f$ is one to one show that $f(a) \in \partial \Omega$
@zhw. THe argument I had in mind when I first read this post was that since $f$ is injective, if $f(a)$ was in $f(G-a)$ then $f(a) = f(b)$ for some $b$ in $G-a$ thereby contradicting injectivity of $f$.
Aug
19
comment If $f$ is one to one show that $f(a) \in \partial \Omega$
@tattwamasiamrutam It's not helpful to denote elements in $\Omega$ by $g$.
Aug
19
comment If $f$ is one to one show that $f(a) \in \partial \Omega$
@zhw. I am pinging on behalf of OP.
Aug
19
revised Open balls in the definition of a Euclidean submanifold
edited tags
Aug
19
comment Showing $\sup \{ \sin n \mid n\in \mathbb N \} =1$
Thank you for your reply. Is there a reason why you wrote $n \alpha$ instead? I find it confusing, especially because you also write $n/(2\pi)$ where I think it should also be the fractional part thereof. But I might be missing something.