# Matt N.

less info
reputation
43087
bio website tancast.com/wp-content/… location age member for 3 years, 2 months seen 7 hours ago profile views 5,045

 7h comment $f$ is one-to-one with domain $\mathbb{R} - \{ a\}$ implies range $\mathbb{R} - \{ b\}$? @Unwisdom Thank you. I had a long day, could you please elaborate a bit? I know that continuous maps map connected sets to connected sets but it's not obvious to me this very second why this implies that they also have to map disconnected sets to disconnected sets. 7h comment $f$ is one-to-one with domain $\mathbb{R} - \{ a\}$ implies range $\mathbb{R} - \{ b\}$? Is it also possible if $f$ is required to be continuous in addition to being bijective? Mar1 comment What is the order of $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ and is it cyclic? Ooh, you are right, I missed lots of copies of $\mathbb Z$. Thank you very much for your patient comments. I do not fully understand your comment somehow but it gave me the idea on how to prove that the image is $\mathbb Z \oplus \mathbb Z_2$: define a map $\mathbb Z\oplus\mathbb Z \to \mathbb Z\oplus\mathbb Z$ that maps $\langle (0,2) \rangle$ to $(0,0)$ and whose image is $\mathbb Z \oplus \mathbb Z_2$. Like e.g. $f((n,k)) = (n, k \mod 2)$. My only problem now is that I don't know how to figure out what the image looks like. How did you figure out it's $\mathbb Z \oplus \mathbb Z_2$? Feb28 comment What is the order of $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ and is it cyclic? Just a quick question to see if I am "seeing" the groups in this question correctly: The group $\mathbb Z \oplus \mathbb Z$ contains $3$ copies of $\mathbb Z$ -- one looks like $(k,0)$, one like $(0,k)$ and one like $(k,k)$. Then taking the quotient by $\langle (2,2) \rangle$ eliminates half of the copy $(k,k)$ (namely the even numbers). So what remains is all elements of $\mathbb Z \oplus \mathbb Z$ minus all elements of the form $(2n,2n)$? Feb24 comment Counting roots of polynomial inside $S^1$ @DanielFischer Yes, thanks. I was 99% sure it wasn't but... you (=I) never know. : ) I think I understand: It looks like if $a_n$ are the zeros of $p$ then the fraction ${p' \over p}$ expands as a sum $\sum_n {k_n \over z-a}$. Computing the integral of the sum yields $\sum_n k_n$ because $\oint {1 \over z-a}dz$ around $a$ is $1$. It's pretty neat. Feb24 comment Counting roots of polynomial inside $S^1$ @DanielFischer That $q$ in your first comment is not the same $q$ as in the question, right? Feb22 comment Concatenation with continuous function is entire Thank you for your comment. I'm still a bit confused. I thought $\varphi$ was the linear operator $A_x$ here and $A_x$ is the derivative of $f$ at $x$. But then we'd have $\varphi = A_x = f'$ at $x$.Which of the equalities here I thought were true does not hold? Feb22 comment Concatenation with continuous function is entire Did you mean $f'(a) =\varphi$ for all $a \in A$? (in the last sentence before the third quotation) Or perhaps $f'(a) =\varphi(a)$? Feb21 comment Continuity of the derivative @Etienne Thank you for your comment. Indeed a seemingly not very thought out comment of mine, I'm sorry about that. Especially if one thinks about real functions where the linear map that is the derivative becomes the tangent! Feb21 comment Continuity of the derivative @Etienne I am not sure if I understand your question correctly. If one considers $f: \mathbb C \to \mathbb C$ as a function from $f: \mathbb R^2 \to \mathbb R^2$ then if $f$ is (complex) differentiable at a point $x$ its derivative at $x$ is given by the Jacobian of $f$. But in finite dimension every linear map is continuous which would yield what I understand you are asking. But I think I misunderstand your question or I am probably making a fundamental mistake in what I wrote. Feb19 comment A closed ideal in a commutative Banach algebra $C(X)$ @DanielFischer I went ahead and made your comments into an answer. Hope it's ok. Feb18 comment Cycling Digits puzzle @DavidConrad Yes the number is assumed to be in base $10$. Feb18 comment A closed ideal in a commutative Banach algebra $C(X)$ @DanielFischer To me it looks as if for any $f$ not in $I(x)$ you can just find an epsilon ball in the $\sup$ norm to show the complement of $I(x)$ is open. But then all the additional assumptions are not needed (like "natural" and that $K$ is Hausdorff and the thing about the characters). Feb16 comment Why are isometries continuous? Like this: $$\|f(x) - f(y)\| = \|x - y\| < \delta = \varepsilon$$ Feb16 comment Why are isometries continuous? If $f$ is an isometry (=distance preserving) then you can use $\delta = \varepsilon$ to prove $f$ is continuous. Jan29 comment O'Neill Formula in terms of Exterior Derivative of Killing Form I couldn't agree more with your profile: Hatcher is not a great book, even before cohomology. At least according to my taste. Enjoy Kreyszig, I have not finished it but I found the parts that I did read very enjoyable and enlightening. Jan28 comment Seifert matrices — Figure 8 knot Thank you for your comment. I will have to look into it when I have more time. Nov19 comment Prove Continuous functions are borel functions I don't know off the top of my head. You might want to post your comment as a question. Nov15 comment Roadway and book recommendations to math study. @AlexanderGruber While the article in the paper is an interesting piece of research could you please elaborate on what it has to do with the question? The question seems to be of the sort "I would like to understand A, B and C. My background is X. Could you please give me help on how to understand A, B and C, given my background" and I can't work out how an article on self-definition pertains. Sep25 comment Relation between primary ideal and prime ideal @leducquang Can you give me a list of all prime ideals containing $q$?