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Jan
17
awarded  Popular Question
Dec
5
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Nov
24
comment Applications of the Hahn Banach theorem for normed spaces?
Yes, this works. This space is simple so your function that extends should be simple too, but it's a fair question.
Nov
20
comment Proving two matrices are equal
Your intuition for why it seems true is natural though and motivated by your familiarity with the $1\times 1$ real valued matrices.
Nov
19
comment Complex integration and Gauss mean value theorem
Thank you .... It was stupid easy and now I feel silly having asked.
Nov
19
accepted Complex integration and Gauss mean value theorem
Nov
19
comment Complex integration and Gauss mean value theorem
If $|a|> 1$ it can be shown the integral you wrote is equal to $\log |a|$ by the Gauss MVT
Nov
19
asked Complex integration and Gauss mean value theorem
Nov
17
comment Computing the integral using cauchy's theorem
If this is on the real line then you are just integrating at -1 and 1, so the integral is zero. I assume this is a complex integral? If so, it's more customary to use $z$ than $x$.
Nov
16
comment Infinite dimension of a polynomial ring as a vector space
Even the subspace $k[x_1]$ has infinite dimension since $1,x_1^1,x_1^2,\ldots$ is linearly independent.
Nov
12
revised Riemann Lebesgue Lemma for locally compact ableian groups
deleted 6 characters in body
Nov
12
asked Riemann Lebesgue Lemma for locally compact ableian groups
Nov
11
comment on the boundary of analytic functions
How precisely is the dominated convergence used? I was thinking of using $g_n= \chi_{\Delta(\frac{n}{n+1})} g$ but it's not quite clear....
Nov
6
accepted Order of pole and evaluating residue
Nov
5
awarded  Popular Question
Nov
2
awarded  Notable Question
Oct
29
revised Rationals are not locally compact and compactness
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Oct
28
comment How to prove this result involving the quotient maps and connectedness?
Thank you! That makes sense.
Oct
28
comment How to prove this result involving the quotient maps and connectedness?
You wrote that $U=f^{-1}(f(U))$, this is true if $f$ is injective... however we are assuming $f$ is surjective (since it's a quotient map), the analogous result is that $U=f(f^{-1}(U))$....so this needs some cleaning up.
Oct
27
awarded  Popular Question