Timeline for answer to What does infinity in complex analysis even mean? by José Carlos Santos
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| Jun 1, 2020 at 14:14 | history | edited | José Carlos Santos | CC BY-SA 4.0 |
Fixing typos
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| Oct 8, 2017 at 16:32 | audit | First posts | |||
| Oct 8, 2017 at 16:32 | |||||
| Sep 13, 2017 at 13:41 | comment | added | David C. Ullrich | @EricTowers Yes, there are various compactifications of $\mathbb C$. But the question was about the meaning of $\infty$ in complex analysis. In complex analysis the standard meaning of $\infty$ is the added point in the one-point compactification. (There are good reasons for that. For example, in the compactification you seem to have in mind $\lim_{z\to0}1/z$ does not exist.) | |
| Sep 11, 2017 at 5:26 | comment | added | gen-ℤ ready to perish | Why add extra confusion by distinguishing between $\infty$ and $+\infty$? Why not leave the limits at $\lim_{n\to\infty}|z_n|=\infty$ and $\lim_{|z|\to\infty}f(z)=0$? | |
| Sep 10, 2017 at 16:20 | comment | added | José Carlos Santos | @EricTowers Interesting! (+1) | |
| Sep 10, 2017 at 16:17 | comment | added | Eric Towers | @JoséCarlosSantos : There is nothing wrong with the one-point (Alexandroff) compactification of $\mathbb{C}$. But there are other compactifications. See my answer, if you are interested. | |
| Sep 10, 2017 at 15:53 | comment | added | José Carlos Santos | @GCab Why not in maths? If I wrote a thing, is there a reason why I shouldn't later remind that I wrote it? | |
| Sep 10, 2017 at 15:52 | comment | added | G Cab | "as I wrote .. there is only one" looks as an answer in .. subjects other than maths! | |
| Sep 10, 2017 at 15:52 | comment | added | José Carlos Santos | @EricTowers I will rephrase my answer then. The way I defined it, there is only only $\infty$ in $\mathbb C$. | |
| Sep 10, 2017 at 15:50 | comment | added | G Cab | @EricTowers: thanks, that is exactly what I meant. | |
| Sep 10, 2017 at 15:36 | comment | added | Eric Towers | @GCab : Yes. See directed infinity. | |
| Sep 10, 2017 at 15:32 | comment | added | José Carlos Santos | @GCab No. As I wrote in my answer, in $\mathbb C$ there is only one $\infty$. | |
| Sep 10, 2017 at 15:23 | comment | added | G Cab | for the modulus of $z$, that' s clear. But for $z$ as complex number can't we say that $lim_{n\to\infty}z_n=(1+i)\infty$ ? | |
| Sep 10, 2017 at 15:19 | history | edited | José Carlos Santos | CC BY-SA 3.0 |
Improving notation
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| Sep 10, 2017 at 15:18 | comment | added | José Carlos Santos | @GCab $\lim_{n\to\infty}z_n=\infty$, because $(\forall n\in\mathbb{N}):|n+in|=n\sqrt2$ and $\lim_{n\to\infty}n\sqrt2=+\infty$. | |
| Sep 10, 2017 at 15:17 | comment | added | G Cab | ok, but what about , e.g., the sequence $z_n=(n+in)$ ? | |
| Sep 10, 2017 at 14:48 | history | answered | José Carlos Santos | CC BY-SA 3.0 |