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22:03
smooth and analytic are the same thing in complex analysis
$S$ isn't closed in $\mathbb{R}$, but it is closed in $X$
$S = [0, 1) = [0, 1] \cap (-1, 1) = [0, 1] \cap X$
$K = [0, 1],\ E = [0, 1)$ using your definition above
Joseph, I'm a little busy right now, but I'll take a look again in a few hours when I have more time
@JoeShmo I think the K should be [0,1] here
ok yes no prob
thanks for ur time
I'm having fun :-)
ooohhh
22:06
yeah, that is correct
sorry i meant
K should be [0,1)
care to share with us whats the excitement of the day @JoeShmo
0
A: Is direction of angular velocity just a definition or has a physical significance?

user76284Since the question mentioned higher-dimensional space, I wanted to give an answer that works in any-dimensional space, not just 3. I'll start with formal, mathematical definitions and then connect them to physical intuition. Rotations in $n$-dimensional space form a group. Specifically, they form...

Welcoming suggestions for improvement on my answer.
maximal functions, harmonic analysis
Does anyone know if is [0,1) closed in (-1,1)
yes, Joseph
if you take the induced topology from R
22:11
What does the equal sign with df over it mean? In the context of differential geometry/manifolds?
What about if [0,1] is closed in (-1,1)
df = definition?
@Joseph [0,1] is not a subset of (-1,1)
$[0,1]$ isn't a subset of $(-1,1)$
Thorgott beat me to it
but
$S$ isn't closed in $\mathbb{R}$, but it is closed in $X$
$S = [0, 1) = [0, 1] \cap (-1, 1) = [0, 1] \cap X$
$K = [0, 1],\ E = [0, 1)$ using your definition above
here you used [0,1]
I stand by how I used $[0,1]$ here
22:13
as the K by defn
but if you use [0,1] there
then the defn fails
no
read the definition carefully
because [0,1] is not closed anymore
ok i will read it again carefully
hmm X is (-1,1) here
and K is [0,1]
no, X isn't (-1,1)
Defn: E is relatively closed with respect to Y if and only if E = K ∩ Y for some set K ⊆ X which is closed in X. Example: Using the interval X:= (-1,1) , S=[0,1) and R. Is S closed in R here?
i wrote the X earlier on
what should the X be here if the textbook is wrong
I mean, what are you trying to do?
22:17
I am just trying to figure out
how to apply the defn
to my example
lol $X$ can be whatever the author wants it to be
S is not closed in R
if you say $X = (-1,1)$ then that's what it is
however, it is relatively close in X
Ok i see
so E, K , Y
i can define for myself accordingly
to whatever i wish as long as it fits?
like E= [0,1) so K= [0,1) Y = (-1,1)
will there be an intersection between K and Y?
22:22
what's the book?
and what's the page?
It is Terrence Tao- Analysis 2 page 16
is that his blog or his book?
its his book
2016, Springer
can you link it to me
ok should i link you the github
22:24
yeah thats even better
actually link me the exact page then
on git
howdy, @JoeShmo
Hi Ted! How are you
OK.. I guess I'll find it by myself
huh
no
can you see the link?
sorry, it seems the message didnt got sent across
version
i appreciate this a lot joe
23:21
𝑆 isn't closed in ℝ, but it is closed in 𝑋 𝑆=[0,1)=[0,1]∩(−1,1)=[0,1]∩𝑋 𝐾=[0,1], 𝐸=[0,1) using defn of relative topology , hey guys does anyone know if k here is closed in (-1,1)?
23:40
Consider the group generated by the two functions $x\mapsto \frac 1x$ and $x\mapsto -1-x$. It is isomorphic to $C_3$ and contains the following elements:

$$G=\left\{x,\frac 1x, -1-x,-\frac{1}{1+x},-\frac{x}{1+x},-\frac{1+x}{x}\right\}$$

I want to prove that for a prime $p$, when this group acts on an element $a$ of $\mathbb{Z}/p\mathbb{Z}$ such that $a^3\not\equiv 1\pmod p$ no pair of the group's actions maps $a$ to the same number. Assuming that $a$ and $a+1$ are invertible, I found by equating the identity to the other 5 transformations that there are collisions when $a\equiv -2^{-1}, -
hey sophie do you think [0,1) is closed in (-1,1)?
I don't know much topology
but I think so
I don't think $C_3$ should contain $6$ elements
thor u are back!
if [0,1) is closed in (-1,1) then is this wrong: 𝑆 isn't closed in ℝ, but it is closed in 𝑋 𝑆=[0,1)=[0,1]∩(−1,1)=[0,1]∩𝑋 𝐾=[0,1], 𝐸=[0,1) using your definition above
@Thorgott yeah you're right I meant $S_3$
23:43
thx soph
@Joseph no, that isn't wrong
but [0,1] is not closed in (-1,1)...
it's not even a subset
and in the above [0,1]∩(−1,1) is used
@Sophie but what if $S_3$ just acts as the identity with every element
what?
hi thor
basically i am trying to see how example 1.3.2
uses 1.3.4
You're considering a group action of $S_3$ on $\mathbb{Z}/p\mathbb{Z}$, no?
@Joseph it doesn't
huh
but why
doesnt 1.3.2 uses induced topology
@Thorgott Yes but I still don' get it
23:49
@Joseph yes, but Tao argues by definition
@Sophie one possible group action is given by $\sigma.x=x$ for all $\sigma\in S_3,x\in\mathbb{Z}/p\mathbb{Z}$
@Thorgott does this mean example 1.3.2 is not in any way related to 1.3.4?
I mean, of course you can also arrive at the conclusions of 1.3.2 using 1.3.4, as you've done above, but you don't have to
I don't understand why you would assume that an example relies on a proposition appearing later in the text when that is not in any way stated
yes but i am trying to see what is the KnY here
as it says K is a subset of X
and in the case Joe used K = [0,1]
and X=(-1,1)
@Thorgott I mean I want $g_1,g_2\in G$ so that $g_1(a)=g_2(a)$ and $g_1\neq g_2$
23:53
if you want to apply the proposition to the example, the X from the proposition should be R
erm but what should the K be
@Thorgott you are right, I didnt quite notice that, i just assumed it was all back and forth

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