04:00 - 18:0018:00 - 22:00

4:49 AM
@Koro yes

5:12 AM
why do people say conditional "expectation" even though it's a random variable not a value

1 hour later…
6:28 AM
@Jakobian Once you wrote the argument, I was able to understand the rest:)
My biggest problem is that I get too nervous during exams, and start doubting obvious things. Like there was this question yesterday. There exists a metric space with exactly $2024$ open sets(True/false). I checked the question at least $5$ times to get sure it was written $2024$ and not $2048$.

7:20 AM
@onepotatotwopotato the undergraduate version is a number, i presume know that

yeah so people just abuse the term I guess
conditional random variable is more suited

It is a pity the same term is used, it is very confusing for those making the transition

15 hours ago, by Sine of the Time
@ThomasFinley It's still valid and I answered it assuming $e_3=(0,1,1)$
@SineoftheTime Thanks a lot!
4

For intuition, you might be interested in reading Why does Newton's method work? and Math Insight. As for this specific example, we cannot find a nice closed-form solution, so we are stuck using numerical methods instead and will choose Newton's Method, but many other root finding methods will wo...

I think tge iterations are impossible to do manually unless some technology is implemented.

there is this thing called a compoopoo

2 hours later…
9:39 AM
@Jakobian Let $\Lambda\subset\Bbb S^2$ be a closed subset such that $\Bbb S^2\setminus\Lambda = \Omega_1\coprod\Omega_2$ where $\Omega_1$ and $\Omega_2$ are connected, simply connected open subsets of $\Bbb S^2$. Then $\Lambda\cong \Bbb S^1$?

0

For $d \in \Bbb{N}, d \geq 1$, define for $x \in \Bbb{R}$: $$[d|x] := \begin{cases} 1, \ \text{ if } x \in d\Bbb{Z}; \ \\ 0, \ \text{ otherwise} \end{cases}$$ defines a family of linearly independent functions $B = \{[d \mid \cdot ] : [1, b] \to \{0,1\}; d \in I\}$ over any field $K$, whenever ...

I guess topologist sine curve works

10:08 AM
the topologist's sine curve has connected complement, but the "Warsaw circle" (topologist's sine curve plus an arc joining both ends) should work

10:48 AM
@onepotatotwopotato I mean you can take $\Lambda$ to be a thicker strip is all. Unless you meant homotopy equivalent?
@Thorgott people from Warsaw clearly have never seen a circle

11:09 AM
Can someone help me to understand why is it trivially claimed that $KerJ^k=ImJ^k$? math.stackexchange.com/questions/2269721/…
I struggle to see it
Or is there an equivalent way to describe this so it looks more intuitive?

11:20 AM
@Jakobian or you've never been to Warsaw

Let $X$ be a topological space. If $X$ is T1, then for each $x \in X$ the set $\{x\}$ is closed. My attempt to the proof: let $x \in X$ and $y \in X \setminus \{x\}$ be arbitrary. Since $X$ is T1, there exists a open neighborhood $V_y$ of $y$ such that $x \notin V_y$. This means that $\{x\} \subseteq X \setminus V_y$, and so taking the complement $V_y \subseteq X\setminus \{x\}$.
This shows that each $y \in X\setminus \{x\}$ has a open neighborhood $V_y$ contained in $X \setminus \{x\}$, that is $X \setminus \{x\}$ is open and so $\{x\}$ is closed. My lecturer proves this differently, by noticing that $X\setminus \{x\}=\bigcup_{y \in X\setminus\{x\}} V_y$. Is my proof correct as well?

1 hour later…
12:23 PM
Hello! I am an electrical engineering student and I want someone to explain some concept’s related to vector spaces, measure and change of variables. No need for formal proofs, I will ask my own questions. I want the intuition for now but I don’t have the time to study topology and functional analysis. I can pay for it. Thanks

1:15 PM
"A topological space has the fixed-point property if and only if its identity map is universal."
What is the definition of universal map?

12

Let $X,Y$ be topological spaces. We call a continuous map $u:X\to Y$ universal if for every continous map $f:X\to Y$ there is $x\in X$ such that $f(x) = u(x)$. If $u:X\to Y$ and $v:Y\to Z$ are universal, is $v\circ u: X\to Z$ universal? (I am thinking about this question because of an inspirati...

@ZaWarudo Sure its correct

Okay so the statement holds as for any map $f$ we have a point $x$ such that $f(x)=id(x)=x$

Yep. Its literally what it means for $X$ to have fixed point property

1:34 PM
And fixed point property is topologically invariant so this is another way of checking whether two spaces are homeomorphic or not

Sure but checking the FPP for a space is often nontrivial

Just one crazy paper that relates PDE and AG

2 hours later…
3:26 PM
So no help I take it
Not that I deserve it in some way but its sad that the world wide math space with so many people over the world gives no answer

:-)

@giannisl9 its not the whole world and people prefer to help people for free
without any sort of obligations coming from being paid

I said I could pay
Not that I will
@Jakobian I just don't have the knowledge and I would like someone to help me because I don't want to go in depth, I want to get a feeling from someone who understands it for now, because I don't have the time
@Jakobian and because its not one question and certainly not a question to ask in the main questions
@Jakobian I just want to chat with someone one-on-one
@Jakobian that takes time and so if someone wishes one can get some compensation for his time
that's all

3:42 PM

I'd love to chat with someone one-to-one that would explain my doubts about everything, sadly that's not possible

it's possible in UG courses.
but not in PG courses like algebraic topology etc.

well yeah, the accessibility of people of knowledge starts to dim after some point

I say it's all demand and supply.

I will try

3:44 PM
There are so many books on real analysis.

And just remember I want the intuition

I mean most people here want to learn on their own and discuss something potentially interesting

I am an electrical engineer student

ask a question on mse on 'how to compute this limit?' and see the upvotes and answers.

late in my studies
and don't have the time to study topology
rn
and so on

3:45 PM

I know somewhat
General idea of vector space
Functions as members of vector space
Domain
Set, Subset
Topology, topological space
L^2 Norm, Inner product
I want to know better
Change of variables in integrals
Measure
change of measure -> change of space?
Field vs Space
How to go from topological space to vector space for functions
Notation

Example

∫_1^4▒〖dx=3〗

Here the function is 1 the set is (1, 4) and the measure is x?
Is area 3 a characteristic of the function regardless of space/transform? Is the area related to how we measure it? ie x?

Measuring the same set with x^2 so in terms of x^2 is (1, 2)

u(x)=x^2

∫_1^2▒du=1

∫_1^2▒〖2dx=2〗

Now I am lost..

I think u is the measure in another space where the same object (1,4) is something else?

1 min ago, by Koro
ask a question on mse on 'how to compute this limit?' and see the upvotes and answers.

are you on PC? See the channel description on how to use MathJax in chat

3:46 PM
add to it 'do the limit without L'Hopital', that's more response.

Can you relate some terms I know to the change of variables case?

I mean for measure spaces?

But ask a question on Algebraic topology etc. and wait for it to be marked as duplicate or be closed.
or deleted

there's no such thing as measure is "x" too

I want an explanation
Not just there is no such thing
and so on

3:47 PM
there are only a few books on AT unlike real analysis. This is purely due to low demand.

what is x?

the standard measure on $\mathbb{R}^n$ is called Lebesgue measure, the one for which $\lambda([a_1, b_1]\times ...\times [a_n, b_n]) = (b_1-a_1)\cdot ...\cdot (b_n-a_n)$ and which we call the $n$-dimensional area

profit from 1 book is say 1$and if no. of students studying AT is only say 1000 in the world, then that's only 1000$ with taxes.

If you want to talk about integral $\int_a^b f(x)\mathrm{d}x$ then one way to interpret it is as integration over the Lebesgue measure $\int_a^b f(x)\mathrm{d}\lambda(x)$
but you can also interpret it without measure theory

but in case of real analysis the profit is much more due to huge demand.
that's why plethora of real analysis books.

3:52 PM
Can you relate that to the example in more simple language?
If we change variables does that mean that we change space?

see chat description on how to see latex in chat
its better if you write things in latex

@Koro Also one of the best books in AT is free so why would someone bother to write new books in AT

@giannisl9 we change the area of integration
do you know integration on $\mathbb{R}$?

@SoumikMukherjee that's propaganda. It sure is not one of the 'best'.

Hatcher fans will surely argue: But its free, and this is for a certain audience!

3:55 PM
@Jakobian Rieman integral of a function?

for example, yeah

@Jakobian most of it
you also said on R
you also said on $\mathbb{R}$

I mean details of how it works are irrelevant, but the point is that if you change variables, you are integrating on a different domain, the transformation maps some set to some other set you integrate on

what is meant by on $\mathbb{R}$?

it works in multi-dimensional case how it works in $\mathbb{R}$ is my point

3:58 PM
@Jakobian the domain is part of the definition of the function. is the domain the same as the set for which the topological space is defined?
@Jakobian does different domain mean different space?

Why are topological spaces relevant here?

I am trying to think this way
are they not relevant?

Maybe its better if you just told me what problem you're thinking of
topological spaces are relevant to integration and measure theory if we want to mix them in together
but the two things can be discussed independently
that usually comes up with when we're trying to talk about Borel and Radon measures

Let me walk you through how those questions came up then
Maybe you get more understanding

alright

4:01 PM
I studied the Fourier series in one of my courses
And I liked the idea of basis functions and expansion and two different representations of the same thing
so
I thought where are the complex exponential defined
is it the circle
is it time
Is it a different space angle and time
\int_{X}fdμ(x)
$\int_{X}fdμ(x)$

Complex exponential is defined on complex numbers

@Jakobian how you mean
by on you mean the domain?

yes
$\exp(z) = \sum_{n=0}^\infty \frac{z^n}{n!}$ is the definition of the exponential
by theorems from real analysis, this defines a continuous function on complex numbers

is time some sort of parametrisation?

what do you mean?

4:07 PM
I mean one sees $e^{j2\pi t}$

thats just a certain volue of the exponential
but $\exp(z) = e^z$ is defined for all complex numbers $z$

for example, we can talk about it for $z = j2\pi t$

lets see it as a function

arguably its the most important function in complex analysis

4:08 PM
$f(t)=e^{j2\pi t}$
so one question that comes up is

for $t\in [0, 1]$?

@Jakobian ok
Now $[0,1]$ is what?
we say L^2 on [0,1]

$[0, 1] = \{x\in\mathbb{R} : 0\leq x \leq 1\}$
its an interval

lol 😂 I like you
okay
I rephrase
$L^2$ on $[0,1]$ means square integrable l functions that have domain $[0,1]$

So are we discussing Fourier transform or something? You've studied Fourier series and discovered Fourier transform?

4:11 PM
?
now just the Fourier series
is my statement okay

$L^2[0, 1]$ is the set of equivalence classes of functions $f\equiv g$ iff $f = g$ everywhere except for a set of measure zero, such that $|f|^2$ is Lebesgue integrable

@Koro there's a lot

@Jakobian that doesn't help me at all
at least intuitively
is it okay to say that?
all the square integrable functions on that interval?

Alright so first define $\mathcal{L}^2([0, 1]) = \{f:[0, 1]\to\mathbb{C} : f^2\text{ is Lebesgue integrable}\}$

@Jakobian okay

4:14 PM
Where the latter means that $f$ is measurable, and $\int |f|^2 < \infty$
where the integral is a Lebesgue integral

@Jakobian ok
now my question is what is the measure? you didn't include it in the integral

now we want to define a "norm" $\|f\|_2 = \sqrt{\int |f|^2}$

I am ok with norm

but the issue is that $\|\cdot \|_2$ is not a norm on $\mathcal{L}^2$
$\|f\|_2 = 0$ if and only if $f = 0$ everywhere except for a set of measure zero
thats why we need to identify functions which are equal everywhere except for a set thats irrelevant

wait let me think of that for a moment

4:17 PM
Thus we define $L^2 = \mathcal{L}^2/\equiv$, the set of equivalence classes, where $f\equiv g$ iff $f = g$ everywhere except for a set of measure zero
this makes $L^2$ into a normed vector space
For example, if you modify the zero function, $f(x) = 0$ for $x\in [0, 1]$ on a finite set, the integral doesn't change
its still zero

I don't think I can understand this because for me the norm is some kind of energy and apart from the strict definition of the norm which is not satisfied as you say I think of defining the norm in the sense that the functions have finite energy on that interval
Do you think this is important for my intuition ?
my basic understanding minimal
of what is happening here?

You can still think of elements of $L^2$ as functions, but keep in mind they're not precisely functions and when you talk about this space as well as $L^p$ in general, you identify functions up to set of measure zero

@Jakobian okay
now in the integral $\int_{X}fdμ(x)$
which will become

When we want to talk about functions and not equivalence classes, even mathematicians usually forget this, but the usual notational distinction is $\mathcal{L}^p$ for functions and $L^p$ for equivalence classes of functions

now in the integral $\int_{[0,1]}fdt$

4:25 PM
If $X = [0, 1]$ and $\mu$ is the Lebesgue measure on $X$, sure
also $f$ has to be Lebesgue integrable

yes what is this t
?
is it a way to measure the interval?
is it a way to measure the interval?

or $f$ can be measurable and non-negative (if we allow for infinite values of integrals)
@giannisl9 integration?
oh $t$

yes $t$
what about if $t$ was $t^2$

Personally I'd never write $\int_{[0, 1]} fdt$, rather $\int_{[0, 1]} f(t)\mathrm{d}t$

that's how I thought about change of variables as measure
change

4:27 PM
$dt$ just means we are using Lebesgue measure
so its the standard integral

when we define as space

if you were to use $d(t^2)$, then I'd assume you're talking about Lebesgue-Stieltjes integrals
which can be defined as integral over the Lebesgue-Stieltjes measure

$(L^2 functions, [0,1], t)$?
this is where I get confused

One that would assing value $b^2-a^2$ to a subinterval $[a, b]\subseteq [0, 1]$

yes I get that
I mean we have a function $f$
defined on that interval
and $t$ is used to scan that interval no?

4:30 PM
@giannisl9 what do you mean

is that what we intuitively mean my measure?
I mean that in Rieman integral lets say
I think of the interval [0,1]
as some set
and then t
$f(t)$
wait

okay well, whats the goal here actually
what do you want to understand
concretely

what is $t$

a letter

thanks

4:32 PM
no problem

come on you can do better pls
I really appreciate it
in the integral
what is the relationship between what we call measure

you know, mathematics isn't about having this mush of things you throw at someone

and $t$

so you need to be precise about what you want

does the measure "measure" the interval?
I think I luck intuition about distance and measure
and the interval as a mathematical entity

4:34 PM
there are different measures
a measure is just a name for a certain concept
what you want is generalization of length
thats called Lebesgue measure

sorry to butt in here, but my knowledge in topology is limited, so I have a few beginner's questions about the following proof. 1) Why is the codomain of $F$ the set $\{0,1\}$? 2) why do we need to show that $F$ is constant? 3) Why is $f$ continuous?

you can measure intervals and you can measure various other subsets, not all of them, but a lot of them

@Jakobian so $t$ is vaguely here some kind of time length?

the Lebesgue measure of interval $[a, b]$ is $b-a$
there is no concept of time here

@Jakobian okay

4:36 PM
time is for physicists I don't care about time

@Jakobian okay
so what is $t$

most subsets of real line, the ones we call measurable, for we can measure them
or Lebesgue measurable
have assigned some value we call their Lebesgue measure

okay

For example, rationals numbers in $[0, 1]$ have Lebesgue measure zero
but irrationals in $[0, 1]$ have Lebesgue measure $1$

the thing for me here is that $[a,b]$ has a measure of $b-a$

4:38 PM
The $t$ in $\int f(t)\mathrm{d}t$ is basically a leftover from other definitions of integrals
one interpretation is the $t$ comes from how we measure the length of an interval

@Jakobian yeah that's what I want
that's what I thought
how we measure it

if it were $dt^2$ we could interpret it as Lebesgue-Stieltjes integral, and the measure of $[a, b]$, its "length" with respect to the corresponding measure would be $b^2-a^2$

as a function of $t$
yea that was my thought

and if we had some increasing function $g$ then $dg(t)$ would mean we integrate with respect to a measure that assigns length $g(b)-g(a)$ to $[a, b]$
For example if you look at Riemann integrals, we have Riemann sums $\sum f(t_i)(x_{i+1}-x_i)$

so here the $t$ plays the role of how we measure things ?

4:41 PM
but for Riemann-Stieltjes ones it would be $\sum f(t_i)(g(x_{i+1})-g(x_i))$

things = intervals

so the $dt$ here also indicates how we take the difference to be $x_{i+1}-x_i$ instead of the other, more exotic differences
of course this is for Riemann integrals which are simpler

@psie they are using the characterization that a space is connected if and only if all continuous functions from it to the discrete two-point space are constant

without measures
@giannisl9 you can interpret it this way, sure
$dt$ just means its integration with respect to Lebesgue measure

ah, I'll have to look that up

4:43 PM
but thats if this is a Lebesgue integral, there can be other integrals
which don't necessarily come from integration over some measure

so we could replace $0$ and $1$ with any other numbers, as long as it is a discrete two-point space I take it

@psie yeah
or any disconnected space

yeah, it doesn't matter what you call the two points
people just gravitate towards $0$ and $1$

So can we relate this to a Rieman sum where we change variables ?
lets say
$f:[0,1]->R$
and
we take the $\int_{0}^{1}f(t)dt$
and we make a change of variables
$u(t)=t^2$

change of variables is more or less glorified chain-rule

4:51 PM
I want to think of it analogously
now what happens is we change how we measure the interval $[0,1]$
so in essence there is a change in space

That depends on your point of view

I meant $t(u)=u^2$
so $f(u)$
is another function

I mean there'll always be going to change of interval but you can (usually) keep on integrating with respect to the Lebesgue measure

at another space
because the measure is different

You can write substitution as $\int f(x)\mathrm{d}x = \int (f\circ \Phi)(x) \mathrm{d}\Phi(t)$
but you can also just write it as $\int f(x)\mathrm{d}x = \int (f\circ \Phi)(x)\cdot \Phi'(x) \mathrm{d}x$

4:56 PM
My point of interest is if I am correct that the space changes
we have a different function
at a different space
that is related to the old function

what do you mean by space more precisely

at the old space
space in the sense of $L^2[0,1]$
analogous

analogous to what exactly

let me state it again

you mean you integrate over different measure?

4:58 PM
yes
because we have $t^2$
and also the interval changes somehow

well, like I said you can usually (and this is how we write it) keep on integrating with respect to the Lebesgue measure

this is what I don't understand

and the difference is accounted by the differential $\Phi'$

but we changed how we measure the interval
its not $t$ anymore
its $t^2$
okay we can write it in terms of $t$
because they are related

I feel lack of precision here
precise example

5:02 PM
$f(t) = t, t belongs [0,1]$

yeah sure but whats the question
is there a question?

take the integral $\int_{0}^{1}f(t)dt=\int_{0}^{1}tdt$
suppose a change of variable
$t(u)=u^2$
look at the integral
$\int_{0}^{1}f(u)dt(u)$

well it won't be the same integral
oh okay

okay now
let me tell you what I think

well like I said, you can look at it as $d(t(u))$ but you can also look at it as $\int_0^1 f(u)t'(u)\mathrm{d}u$ and thats usually how we do it
with the differential $t'(u) = 2u$ accounting for the change of variables

5:07 PM
wait
that's not my point of interest
sorry
:)
about the first integral: we have a function on [0,1] and we measure this by t
about the second integral: we have another function on [0,1] and we measure this by $t(u)$
and actually if the interval was $[0,4]$
it would become $[0,2]$
it means different function different interval different measure?
or different function same interval different measure
its like mapping a function from one space measured by t
to another function from another space measured by $t(u)$
I say different space because the same things is measured differently
Anyway there is my confusion as you can see
I want some clarification along those lines
Not definitions or completely irrelevant ideas
se can have some functions on $[0,1]$
integrate them like this and get a number
do a change of variables
get some new set of functions on some other interval (or the same measured differently)
and integrate some new function over this to get (the same?) number
when I say integrate them I mean integrate one of them

5:25 PM
I'd have to repeat myself again