Mathematics - 2020-09-20

5:13 AM

How do all infinite algebraic extensions of $\Bbb{Q}$ which are not equal to $\bar{\Bbb{Q}}$ look like?

Ted Shifrin

adjoin any infinite proper subset of all the algebraic numbers

Ah nice, yes

skullpatrol

6:18 AM

Abandon all hope, ye who enter here.

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@SayanChattopadhyay infinite Galois theory

skullpatrol

Welcome @Spectre

@LeakyNun Hmm never heard of it

Any references?

Stacks Project tags 0BMI, 0BML

e.g. $\operatorname{Gal}(\Bbb Q(\zeta_{2^n}) / \Bbb Q) = (\Bbb Z/2^n\Bbb Z)^\times$, so $\operatorname{Gal}(\bigcup \Bbb Q(\zeta_{2^n}) / \Bbb Q) = \varprojlim (\Bbb Z/2^n\Bbb Z)^\times = (\Bbb Z_2)^\times = 1 + 2\Bbb Z_2$

and $\Bbb Z_2$ looks like the Cantor set

6:49 AM

That's a lot to unpack

@BalarkaSen ok

@SayanChattopadhyay this is probably much basic than what you're looking for, but math.brown.edu/~abrmovic/MA/f1415/253/Zijian.pdf is a quick and good read.

7:06 AM

That's a nice reference, thanks! @Lelouch

I'm trying to prove (for $f:X \mapsto Y$ where $X, Y$ are topological spaces) that $x \in \bar A \implies f(x) \in \overline{f(A)}$ for $A \subseteq X$ means that $f$ is continuous at $x$, i.e. for each neighbourhood $N$ of $f(x)$ in $Y$, $f^{-1}(N)$ is a neighbourhood of $x$ in $X$.

This would be simple if the implication was $f(x) \in \overline{f(A)} \implies x \in \bar A$, but because the implication goes the other way, I am stuck on the possibility that $f(x) \in \overline{f(A)}$ but $x \not \in A$. Can anybody help?

@SayanChattopadhyay Keith Conrad is always the best explainer youtube.com/…

Yeah his notes have helped me a lot before.

@remana Doesn't it follow almost immidiately that $f^{-1}(O)$ is open as well ?

What you've written expands to "for each neighbourhood of $y$ in $Y$, the preimage is a neighbourhood of $f^{-1}(y)$"

@remana let $B \subseteq Y$ closed. with $A = f^{-1}(B)$ we get $x \in \overline{f^{-1}(B)} \implies f(x) \in \overline{f(f^{-1}(B))} \subseteq B$, i.e. $f^{-1}(B)$ is closed

7:28 AM

@Lelouch That's what I'm trying to prove, I'm not given that the preimage of a neighbourhood of $y \in Y$ is a neighbourhood of $f^{-1}(y) \in X$

@LeakyNun That's brilliant, thank you

7:41 AM

@remana Well, if for some neighbourhood $N$ of $y$, if $y = f(x)$ and there's no neighbourhood of $x$ contained in $f^{-1}(N)$, then $x \in \overline{f^{-1}(N)^c} = \overline{f^{-1}(N^c)}$, which forces $y = f(x) \in f(f^{-1}(N^c)) \subset N^c$, which is a contradiction

7:53 AM

@Lelouch I think I understand, I'm assuming your $N^c$ notation indicates the closure of $N$?

Stupid question inc

8:05 AM

hej!

is there any good video lecture which covers full analysis course ?

@Lelouch Actually I'm still confused. Why does it force $f(f^{-1}(N^c)) \subset N^c$?

@LeakyNun Looking back over your statement, how do we know that for all neighbourhoods $B$ of $f(x)$ in $Y$, that $x \in f^{-1}(B)$? You just assume that $x \in f^{-1}(B)$

when did I say neighbourhood

You didn't, that's why I'm asking how do we know that.. since that was my original question

8:20 AM

@remana N^c is complement, bar denotes closure

@remana f(f^{-1}(A)) is a subset of A for set theoretical reasons

@Lelouch Ahh, that makes more sense. I understand now, thank you!

@Lelouch Since $f(x)$ is forced into $\overline{f(f^{-1}(N)^c)}$, there isn't a contradiction if $f(x) \in \bar N$ but $f(x) \not \in N$ though, right?

Ah, but $N$ has to be open because $Y$ is a topological space. I see. Nevermind, thank you for your help!

5 hours later…

Charlie

1:58 PM

Is there a name for a basis that block-diagonalises a reducible matrix representation of a group?

Alessandro Codenotti

2:40 PM

Weird topology fact of the day: Čech-complete spaces are Baire, a product of (even two) Baire spaces need not be Baire, a product of Čech-complete spaces need not be Čech-complete, however a product of Čech-complete spaces is always Baire

3:07 PM

are 4-to-1 maps generally not desirable?

and just to check my understanding, a 4-to-1 map means that for every point, one has 4 values attached to that point?

Q: Julia set of $x_n = \frac{ x_{n-1}^2 - 1}{n}$

3:19 PM

okay I think I understand it now. If you take $\Bbb R^2$ (minus the x and y axes) and restrict it to $(0,1)^2$ then you get a 4-to-1 map

and this restriction of $\Bbb R^2$ would not be very useful because I don't think you could have a well-defined metric. So I conclude that 4-to-1 maps are not good

mick

3:37 PM

hi all

Consider the following iterations : $x_0 = z$ Where $z$ is complex. $x_n = \frac{ x_{n-1}^2 - 1}{n}$ It is well known that for real $z > 3$ the sequence grows double exponentially. It is known that for $z = 3$ the sequence grows linear ; in fact like $3,4,5,6,7,...$. In fact When considering...

anyone feel like attacking that or plotting it ?

2 hours later…

mick

5:13 PM

Let $n$ be a positive integer. Let $b = 2 n - 1$. Let $x$ be a positive integer. Define $f(x)$ as : $$ f(x) = \prod_{i = 1}^x ( \sin( i \frac{\pi}{n}) + \frac{5}{4}) $$ Then it appears that $$ f(b) = \frac{4}{5} + C(n)$$ And $C(n)$ is close to zero. In fact $$ lim_{n \to \infty} C(n) = 0 $$ How d...

real-analysis limits trigonometry asymptotics products

Q: $ R(t) = R_0 - \int_0^t v(T) dT , v(t) = \tanh(\int_0^t \frac{C dT}{R^2(T)})$ ??

Let $R_0,C$ be positive real constants and consider the following differential equation : $$ R(t) = R_0 - \int_0^t v(T) dT $$ $$ v(t) = \tanh(\int_0^t \frac{C dT}{R^2(T)})$$ How to solve this ? How to simplify it or change it in a simpler differential equation ? How to solve it numerically ? I co...

real-analysis ordinary-differential-equations numerical-methods closed-form inverse-function

2 hours later…

7:19 PM

question

what is the identity element of the poincare group

Thorgott

7:34 PM