Luiz Cordeiro
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 36m comment Are holomorphic maps that “almost” preserve norm “almost” rotations? However, you could ask if the $f_n$ approximate rotations in the following sense: for every $\epsilon>0$, there exists $N$ such that for all $n>N$, there exists a rotation $T(n,\epsilon)$ with $\Vert T(n,\epsilon)-f_n\Vert_{\infty,\mathbb{D}}<\epsilon$, where $\Vert\cdot\Vert_{\infty,\mathbb{D}}$ is the uniform norm on the disc. This is interesting, and I don't have an answer, 41m comment Are holomorphic maps that “almost” preserve norm “almost” rotations? If $\theta$ is irrational and $f_n(z)=(1-1/n)e^{i\pi\theta n}z$, then $f_n$ satisfy the hypothesis you put but do not converge pointwise. 48m answered Confused about a neither statement and modular 1h comment Confused about a neither statement and modular I really don't think using logical proofs is the best way to proceed in this exercise, or even if it is feasible at all. I will try to explain better in an answer. 1h comment If $r(x)$ is an $n$th degree polynomial of the form $r(x) = a_0+a_1x+\cdots+a_nx^n$ with $n$ odd, prove that $r(x)$ has at least one real root You should at least state that $r(x)\to\operatorname{sign}\pm\infty$ as $x\to\infty$, where the sign depends on the sign of $a_n$, and similarly for $x\to-\infty$, and then conclude that there exists points $x_0$ and $x_1$ as you stated. I don't agree with using notation of order $O(x^{n-1})$ in a proof at this level (at most, just for some drafts), but if that is how you (and your professor) do in class then it is how you should do it. You also didn't explain what equality should hold or not!? 1h comment $\mathbb N\times\mathbb N$ is countable @AsafKaragila You should reference where this technique comes from: smbc-comics.com/index.php?id=1099 :) 1h comment Confused about a neither statement and modular Your negation of $q$ is correct. Assume $\lnot q$ is true. Now you can consider two cases: $3|a$ or $3\not |a$. In each case, prove that $a^2+2b^2$ is not congruent to $0$ modulo $3$. 1h comment Confused about a neither statement and modular I'm sorry, I misinterpreted how you wanted to proceed with the proof. You are right (but I think there shouldn't be a comma after "$a$"). 1h revised Confused about a neither statement and modular deleted 59 characters in body 8h comment find the gradient See that $\phi(|x|^2)$ is a number but $x(\nabla|x|^2)$ is a vector (although you are multiplying the vector $x$ by a number on the right, which is non-standard.), so you can't add them. This means you forgot something. Also, you should calculate $\nabla|x|^2$? 8h comment Hausdorff space with a additional property Look at preimages $f^{-1}(1/2,1]$ and $f^{-1}[0,1/2)$. Recall what normality means. 9h comment find the gradient You will need the product rule to separate $\phi(|x|^2)$ and $x$ (and calculate their derivatives), and the chain rule to calculate the derivative of $\phi(|x|^2)$ 11h answered Derivative of n x n Invertible Matrix 11h comment Derivative of n x n Invertible Matrix You can find the derivative of the maps $A\mapsto A^2$ and $A\mapsto A^{-1}$ and use the chain rule. (For the second one, use a formula for $X^{-1}$ as a series, whenever this makes sense). The things you get are similar to the real case. 11h comment The well-ordering principle implies Zorn's Lemma @akkarin Yes, it is. I just thought this should be mentioned because you asked for a direct proof of $WO\implies AL$. And Pedersen's proof is the same as in a paper by J. Lewin - A Short proof of Zorn's Lemma (JSTOR). 13h comment Show two norms are equivalent Try to look at the usual generating elements $e_n=(0,0,\ldots,1,0,\ldots)$ and compare their norms. 13h answered The well-ordering principle implies Zorn's Lemma 23h revised Isometric embedding of $\ell ^ 1$ in $\ell ^\infty$ in finite dimensions deleted 28 characters in body 23h comment Isometric embedding of $\ell ^ 1$ in $\ell ^\infty$ in finite dimensions @direct_limit There are $2^n$ ways of choosing $n$ numbers $a_1,\ldots,a_n$ in $\left\{\pm 1\right\}$, and each of these choices gives a mapping $\mathbb{R}^n$. Take the direct product of all the mappings obtained in this way, and you end up with a mapping $\mathbb{R}^n\to\mathbb{R}^{2^n}$, which ends up being isometric with the desired norms. 1d answered Isometric embedding of $\ell ^ 1$ in $\ell ^\infty$ in finite dimensions