1:23 AM
@BalarkaSen hmm I don't know, I am guessing its because for each of the finite points, you get to throw countably many darts at each such point and some probability following poisson random distribution for all the points near those selected points, thus it is possible that the infinite is hiding within the semingly finite number of dirac deltas
so heuristically, the equation may actually read (random finite)*(random finite that is hiding an infinite) = (random infinite)

4 hours later…
5:13 AM
How do all infinite algebraic extensions of $\Bbb{Q}$ which are not equal to $\bar{\Bbb{Q}}$ look like?

adjoin any infinite proper subset of all the algebraic numbers

Ah nice, yes

6:18 AM
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@SayanChattopadhyay infinite Galois theory

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 $N$?

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…
1:58 PM
Is there a name for a basis that block-diagonalises a reducible matrix representation of a group?

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?

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

3:37 PM
hi all
3

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…
5:13 PM
0

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...

0

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...

2 hours later…
7:19 PM
question
what is the identity element of the poincare group

7:34 PM
1

okay that's what I thought
for $H$ which maps hyperbola to hyperbola, non-involutively, $H=\lbrace \frac{s}{x} : s \in \Bbb R \rbrace$ is it a one parameter Lie group?
the point of confusion for me, is that the transformation is not sending points along the hyperbola, but it's sending the family of hyperbola to itself

1 hour later…
9:10 PM
How is in pure mathematics the novelty of research judged? In applied mathematics i see often that there is a "real-world application", which you can then solve faster/understand better and by this is the math judged. But if there is no application what differs a mediocre theorem from a great theorem?

Hi @Balarka

9:27 PM
Hey

Any interesting math to share?

9:50 PM
anyone want to talk about math
even number theory
wait I just realized that my professor introduced group actions and never told us about isotropy subgroups