« first day (2165 days earlier)      last day (2783 days later) » 

6:00 AM
If you consider expansion of a solid then the molecules in it just vibrate, and thermal expansion is due to increased amplitude of vibration. So the molecules on average stay in the same place.
 
user228700
@JohnRennie On average...OK..?
 
Yes. But in diffusion molecules don't on average stay in the same place. Diffusion involves movement of molecules on length scales that are large compared to molecular sizes.
 
user228700
Hm, OK...
 
In gases I guess the distinction is less clear since in gases the molecules do move large distances ...
 
user228700
Yeah, that's what.
 
user228700
6:06 AM
But alright.
 
Well in gases every property is ultimately described in terms of molecular motion.
 
user228700
Hm, yes.
 
That doesn't make diffusion and expansion in gases the same, just because both are ultimately explained in terms of molecular motion.
 
user228700
OK, I guess I understand it for the most part now.
 
user228700
OK, next question? This is turning out to be like a rapid-fire round, sorry :-P
 
user228700
6:12 AM
..? Shall I ask u..?
 
Yes, go ahead
Sorry, I was waiting for the next question :-)
 
user228700
Oh :-P Okay.
 
user116211
@Kaumudi Smart school.
 
user228700
@MAFIA36790 I have so many arguments against that but I haven't got the time, so OK.
 
user116211
._.
 
user228700
6:16 AM
(@JohnRennie: Checking to see if questions are worth asking...)
 
user228700
OK, so out of ALL this:
 
user116211
turned my head... pains...
 
user228700
 
user228700
 
user228700
 
user228700
6:24 AM
This is what I gathered, because I didn't understand most of what they are going on about:
 
user116211
@Kaumudi Which one did bother you?
 
user228700
 
user228700
 
user116211
pink...
 
user228700
Are my notes correct..? :/ (I hate my textbook, urgh
 
user116211
6:27 AM
What do you mean by size of the orbital?
 
user228700
@MAFIA36790 ..?
 
user116211
@Kaumudi I hardly knew such ink exists.
 
user228700
@MAFIA36790 Uh, I mean exactly what size of orbital sounds like..?
 
user228700
@MAFIA36790 I have a set of 10 differently colored pens, not a rainbow pen BTW.
 
user116211
@Kaumudi WTH...
 
user116211
6:29 AM
A wavefunction is decomposed into three parts in spherical coordinates; radial is one of them.
 
user228700
@MAFIA36790 It makes the book like livelier and also, makes me feel like revising these notes from time to time.
 
user116211
@Kaumudi Your book looks really dull though.
 
user228700
@MAFIA36790 Even this:
 
@Kaumudi yes, the notes look fine.
 
user228700
 
user228700
6:31 AM
? (@MAFIA36790 :-P)
 
user228700
@JohnRennie OMG really?! :O
 
@Kaumudi were you worried some bits were wrong? If so, which bits?
@Kaumudi: look what someone has just asked:
0
Q: Is this the correct way to think about why static friction is directed radially during a turn on a level surface?

rb612So after much pondering of the fact that the net static friction force points in the center, perpendicular the tangential motion, I thought of this explanation. If we look at a car travelling around a circle, the wheels are always turned. There is a static friction force causing the tires to ro...

 
user228700
@JohnRennie Honestly, I was worried that everything was messed up, 'cause I understood nothing of what my textbook has given.
 
@Kaumudi I recall you saying you didn't understand polar coordinates - at least I think so.
The descriptions of the atomic orbitals are using polar coordinates so that would make them confusing.
 
user228700
@JohnRennie Oh, wow. OK, I'm gonna check that out (properly) as soon as these questions are over.
 
user228700
6:35 AM
@JohnRennie Yeah, I haven't learned about it yet.
 
Polar coordinates are really very simple. I suspect once you look into them you'll wonder why they ever seemed hard.
 
user228700
@JohnRennie No sir, it's not that they seem hard or anything. I just haven't had the necessity/time to learn it yet.
 
Atoms are spherically symmetric, and for any spherically symmetric system polar coordinates are the obvious choice.
 
user228700
@JohnRennie Hm, OK...
 
user116211
@Kaumudi They really make Schrödinger's Equation easy to solve for hydrogenic atoms.
 
user228700
6:40 AM
@MAFIA36790 Don't need to solve it :-P
 
user228700
@JohnRennie: Do u have to start work now?
 
I probably should start work now, but I'm happy to postpone it if you have an interesting question. After all, it is a Saturday :-)
 
user116211
The whole thing is separation of variables.
 
user116211
We generally start with Coulombic potential.
 
user228700
:-) All of my questions are very basic sir, so I dunno if they will be of interest to you but my next question is about relative angular velocity and I am just not at all understanding what the heck my book is telling me. So I was gonna post a pic...
 
user116211
6:47 AM
And spherical coordinate system is quite handy when you are working with Coulomic potential.
 
@Kaumudi carry on ...
 
user116211
If yu want, @Kaumudi, check this related post:
 
user116211
6
A: Trying to understand the necessity of orbital approximation or the independent electron model

WildcatSetting the stage Once the Born-Oppenheimer approximation is introduced, we have (first) to solve the so-called electronic Schrödinger equation that could be written as follows, \begin{equation} \newcommand{\el}{_{\mathrm{e}}} \newcommand{\elel}{_{\mathrm{ee}}} \newcommand{\elnuc}{_{...

 
user228700
@MAFIA36790 OK, thank you :-)
 
user116211
@0celo7, go back to bed. Have a sleep.
 
user228700
6:52 AM
Stupid phone, logging me out every now and then. Sorry.
 
user228700
 
user228700
@JohnRennie: Do u remember how I asked u about the definition of a reference line? I'd asked to make better sense of this only.
 
Ok, so now my question is on hold. and I allready edited once. In my point of view this should be within these limits; meta.physics.stackexchange.co… "For example, a question that proposes a new concept or paradigm, but asks for evaluation of that concept within the framework of current (mainstream) physics is OK."
So should I edit it (how?) or delete it?
 
user228700
@MAFIA36790 Didn't he say that his phone keeps logging him back in?
 
6:58 AM
@Kaumudi OK. Incidentally the first statement on that page angular velocity is also always relative is not true.
 
user116211
@Kaumudi Maybe.
 
user228700
@JohnRennie Oh..?
 
user228700
"There is no such thing as absolute angular velocity"
 
There is no way to measure absolute velocity. However any rotation creates pseudoforces, so it is always possible to measure absolute rotation by measuring the associated pseudoforces.
 
user116211
@Kaumudi Check Newton's Bucket Experiment.
 
7:01 AM
But I suspect this is a distraction.
It is certainly possible to measure relative angular velocities.
 
user228700
@JohnRennie What is? The thing about pseudo forces?
 
I mean that if you're asking about relative angular velocity we probably don't need to worry about whether angular velocity can be absolutely defined.
 
user228700
@JohnRennie That doesn't make sense. Why not? If a quantity is absolute, there is no talking about relative, no?
 
Mass is absolute, but I can still say mass A is 1kg lighter than mass B.
 
user228700
Right. Alright then, let's skip that discussion.
 
7:04 AM
Probably wise
 
user228700
But I wasn't able to understand how they arrived at those formulae... :/
 
Probably because that's a really unhelpful way of describing it.
 
user228700
Also, is it OK to start/end sentences with the word "but"? I remember my English teacher in primary school told me it was incorrect but I've read this usage in books and all and I've always wondered if my teacher was correct that day, lo those many years ago :-P
 
@Kaumudi no, that's generally regarded as poor grammar, but on a chat room no-one cares.
 
user228700
OK :-)
 
user228700
7:07 AM
This:
 
Are you happy you understand how angular momentum is defined?
 
user228700
"Angular velocity of a particle 'A' with respect to the other moving particle 'B' is the rate at which position vector of 'A' wrt 'B' rotates at that instant" I cannot understand this AT ALL!
 
@JohnRennie Hi, You must have been the first to put this on hold. physics.stackexchange.com/questions/284717/… I am not interested to start to edit it x-times to try to find out if it will be aloud here. Can you be more specific?
 
@Kaumudi well that's a rather wordy definition. A more precise definition is $$\mathbf v = \mathbf r \times \omega$$
 
user228700
But they're talking about relative angular displacement, no?
 
user116211
7:11 AM
@Kaumudi $\mathbf r$ is relative.
 
user228700
So that's all there is to it? The displacement vector is swapped for the relative displacement vector?
 
@Kaumudi: when we define vectors like $\mathbf r$ and $\mathbf v$ we do so using some coordinate system that we've chosen.
Typically this is an inertial frame with ourselves at the origin
 
@Obliv Not just definitions, but also important features such as ideals and congruences. I have not included homomorphisms because they are not very important unless I need to compare between two algebraic structures
 
But we can choose a rotating coordinate system. In that rotating coordinate system $\mathbf r$ and $\mathbf v$ will have different values so the angular velocity we calculate will be different.
 
NB I copied all of this from wikipedia, because I found I always confuse the various types of ring generslisations
 
user228700
7:17 AM
@JohnRennie But won't these values be continually changing..?
 
user228700
Sir, I need to have lunch. I'm starving. You please carry on with ur work :-) I'll ask the rest of my questions another time.
 
@Kaumudi yes, choosing some random frame will generally result in the angular velocity being time dependent.
But that's not anything unusual.
It just means you're working in non-inertial frame.
 
user228700
@JohnRennie OK...
 
At undergrad level I generally advise students to be very careful about using non-inertial frames because your intuition may fail you and it's eaasy to make mistakes.
 
user228700
@JohnRennie But it so happens that I need to work with them every now and then because I have to solve complicated problems :/
 
user228700
7:22 AM
You generally advise students..? Were u a teacher at any point?
 
user116211
@Kaumudi He is teaching you.
 
user228700
@MAFIA36790 Of course he is :-) I was just wondering if he was ever a teacher IRL.
 
No, I've never been a teacher. I was speaking figuratively.
 
6 mins ago, by John Rennie
At undergrad level I generally advise students to be very careful about using non-inertial frames because your intuition may fail you and it's eaasy to make mistakes.
For starters, tons of fictitious forces to worry about
 
user116211
8:00 AM
Hallo @loong!
 
@MAFIA36790 hi
 
0
Q: If a system is deterministic, will it still be deterministic if time is reversed?

DanIf you were to drop a ball, it would be easy to calculate when it will hit the ground and how much energy will be absorbed by the ground (let's assume there's no air resistance and the ball does not bounce). If you were to then reverse time, would energy from the ground gather rapidly toward the...

Well it has to be, otherwise time reversal symmetry will be broken
But even then it is not the main point, what will it mean for a system that evolve backwards to be nonrandom nondeterminstic and what happens when such is interacted with a system that is deterministic in forward evolution?
 
8:30 AM
Almost feel bad for this guy:
 
I think this should be ok now;
-5
Q: Can Planck's constant be defined through the speed of light

JokelaTurbineThis is the question; Could it be defined through the speed of light using only known mathematics?

 
@Danu The future may be already there, except there might be never any way to know...
 
@JohnRennie Hi, You vote the first close for my question. -Is it now ok?
@innisfree Ok?
@bosoneando Ok?
 
user228700
@JohnRennie Oh, OK :-)
 
9:38 AM
0
Q: Can I copy a whole section from a book as an answer?

Bruce GreethamWe have a question that has been unresolved for a few months. I have now found a section of a book (1-2 pages) which is excellently written and I wish to share as a possible answer. Am I allowed to just copy the whole section out and reference the source. I fear any attempt by me to paraphrase it...

 
10:02 AM
@Danu "Instead, one could assume a spaceshit traveling at 99,99% c." -spaceshit? :-) I thought the maximum speed of spaceshit was something like 99.97% c. but maybe the shit is different in n-dimensions, and my shit in 4-dimensional space just stinks.
 
10:43 AM
@JokelaTurbine lol
 
anyone currectly not afking can help me on a monoid concept question?
 
Hi nerds
 
user228700
11:07 AM
Hi, fellow nerd :-)
 
11:28 AM
physics.stackexchange.com/questions/277183/… what an impressive first post :O
 
user116211
@Secret What question?
 
user116211
@Sanya Thank you Sanya! Is it you who gave me a bronze badge? thank you for that as well in that case! BrouH's question is a very important one, and I've never seen it formulated as neatly and simply as he/she did. I wish BrouH would send it to pedagogical journals like American Journal of Physics or European Journal of Physics and see what answer it'd receive there... — pglpm 8 mins ago
 
@MAFIA36790 ?
 
user116211
@Sanya Thank you Sanya! Is it you who gave me a bronze badge? thank you for that as well in that case!
 
"A subset S of M is said to be a generator of M if M is the smallest set containing S that is closed under the monoid operation, or equivalently M is the result of applying the finitary closure operator to S"

@MAFIA36790 Is it saying that $S$ closed under the monoid operation of $M$, or only $M$ is and applying the monoid operations on $S$ gives $M$?
 
11:38 AM
@MAFIA36790 you mean I should reply to that or sth?
I'm unsure of what point you are making
 
user116211
@Sanya No; he looks happy ;)
 
well, that's something I'd be happy about; I was really happy reading the post
it came up in the late answers review; otherwise I'd probably have missed it
 
user116211
@Secret Monoid is defined to be a semi-group which itself is a magma and thus closed.
 
user116211
But I'm reading still the upper para...
 
@EmilioPisanty : Just logged in. Did any mod help you?
 
user116211
11:46 AM
Hmm, I've not come to the concept of generator but let me understand the part...
 
@Qmechanic I left a comment on the answer
 
I want to understand some trivial things about baryogenesis and statistical mechanics. Anyone know about Sakharov's 3rd law?
 
user116211
@Secret The operation $M$ must be closed for $S$ to be the generator of $M\,.$
 
But is S does not need to be closed under the operation, as otherwise, I am not able to see how one can make M by applying the operation on all elements of S?
That is, if the operation always maps any element in S back to S, then it cannot give elements in M-S?
 
user116211
Basically, One of the conditions for $S$ to be the generator of the algebraic structure $(M, \circ )$ is that $\forall x,y \in S: x\circ y \in M\,.$
 
12:03 PM
ok that makes sense, since based on what I understood, a generator should act like for monoids as basis sets to vector space
 
In Sakharov's 3rd, does thermal equilibrium <-> Gibbs' canonical ensemble?
 
Meanwhile, I am on the group article. I am currently trying to understand the interesting properties of each algebraic structure that are derived form their axioms
e.g. ideals in rings, set of idempotents in semigroups, symmetry in groups etc.
 
user116211
I have not still encountered ideal.
 
There's a notion of ideals in semigroups too. But ideals are most important in rings
An ideal $I\subset S$ is such that $\forall r\in I, \forall a \in S, ar\in I \text{ or } ra\in I$
 
user116211
@Secret okay.
 
12:10 PM
ACM said that ideals are generalisation of the notion of even and odd numbers
For me, its more like a generalisation of an absorber, because anything multiplied by an element in an ideal get sent into the ideal and thus there is no escape
 
1:06 PM
Delete & re-aske proposed for this question;
-6
Q: Can Planck's constant be defined through the speed of light

JokelaTurbineThis is the question; Could it be defined through the speed of light using only known mathematics?

in it's comments.
I would like to hear some hint's before doing this. 1. Pure question with no context from my own thoughts? or 2. Possibly detailled path-of-thoughts which produced this question?
 
@ACuriousMind What I wanted to be true, isn't.
 
2:01 PM
I just added my explanation "why double-slit experiment" on research-gate; researchgate.net/publication/… ...Dont want to spam this place with post that will be deleted anyhow.
 
2:52 PM
Got my Robinson book
Robinson Crusoe and the non-standard analysis
 
In abstract algebra, a normal subgroup is a subgroup which is invariant under conjugation by members of the group of which it is a part. In other words, a subgroup H of a group G is normal in G if and only if gH = Hg for all g in G; i.e., the sets of left and right cosets coincide. Normal subgroups (and only normal subgroups) can be used to construct quotient groups from a given group. Évariste Galois was the first to realize the importance of the existence of normal subgroups. == Definitions == A subgroup, N, of a group, G, is called a normal subgroup if it is invariant under conjugation; that...
I think group theory is so large that just that a book can be written on normal subgroups alone
 
user116211
People do PhD in concerned topics of group-theory.
 
user116211
3:08 PM
1
Q: Difference between $SU(2)_L $ and $SU(2)_R$

DanielLike my title, what is the difference between $SU(2)_L$ and $SU(2)_R$? I know they transform differently, but I don't know how to write down the transformation. I'm not familiar with group theory, so I would very much appreciate a detailed answer.

 
user116211
> I'm not familiar with group theory, so I would very much appreciate a detailed answer.
 
user116211
Weird.
 
In abstract algebra, a basic subgroup is a subgroup of an abelian group which is a direct sum of cyclic subgroups and satisfies further technical conditions. This notion was introduced by L. Ya. Kulikov (for p-groups) and by László Fuchs (in general) in an attempt to formulate classification theory of infinite abelian groups that goes beyond the Prüfer theorems. It helps to reduce the classification problem to classification of possible extensions between two well understood classes of abelian groups: direct sums of cyclic groups and divisible groups. == Definition and properties == A subgroup...
Slereah, I heard you like p adic stuff
 
3:27 PM
@MAFIA36790 I keep losing ;_;
 
user116211
@0celo7 Use Trump cards!! You can't loose. You are the boss!!
 
I can lose
 
I don't
p-adic numbers are shit
 
3:44 PM
Proof?
@MAFIA36790 I think I won
maybe
this is a fierce battle
 
user116211
This may be an illusion or a tactic played by the enemy; you know like the Greeks in Trojan War; people of Troy thought they won but....
 
user116211
So, be aware.
 
aha
I have figured out that really hard analysis problem
 
user116211
Meanwhile, I'm also studying Royden's Real Analysis...
 
that's a really advanced book
maybe learn topology first
 
user116211
3:51 PM
Reading Heine-Borel Theorem.
 
user116211
@0celo7 It's an undergrad book.
 
It's graduate level
 
user218912
@0celo7 question asks to find an expression for linear momentum in terms of creation and annihilation operators using a free scalar field and the integral of the energy momentum tensor.
 
user218912
is that hard?
 
user218912
@MAFIA36790 that has really bad reviews.
 
3:53 PM
Royden is definitely not an undergraduate book
 
Robinson has a lot of tedious first order logic
 
user116211
 
user116211
 
Should use Rudin
 
user116211
It's recommended in our course.
 
3:58 PM
Royden starts with measure theory, your list of topics is almost the contents of Rudin
 
user116211
Analysis really starts from 4th semester.
 
Rudin is literally the best book, it might take 3 years to make sense of it, but one day it'll dawn
 
user116211
But I'm studying Royden by myself. So far, it is quite clear although not lucid.
 
user116211
The exercise is bleh though.
 
Hey @yuggib
Robinson has arrived
 
4:02 PM
@bloo you follow the same idea you use when doing it for energy, plug a fourier transform in
Royden is really different to Rudin, just look at the contents, Royden is comparable to the other two Rudin's
 
@MAFIA36790 that's so weird
that's a single variable analysis course
Royden is calculus on Banach spaces
oh well I guess the second have is multivariable
 
No, Royden is measure spaces, then metric spaces, then Banach, Hilbert and Topological spaces, then more measure spaces, calculus on Banach spaces is in Dieudonne or Lang or Cartan
 
I might skip chapter 2 of Robinson
It's a bit tedious
And not too salient to the point
Oh I might need to check what internal sets are tho
Everyone keeps talking about them and I still don't know what those are
"General properties of enlargement"
heheheh
 
4:18 PM
holy crap this proof is getting out of hand
@ACuriousMind I think I have a 5-epsilon proof on my hands
 
Divide by 5
It will only be a single epsilon
 
In abstract algebra, a heap (sometimes also called a groud) is a mathematical generalization of a group. Informally speaking, a heap is obtained from a group by "forgetting" which element is the unit, in the same way that an affine space can be viewed as a vector space in which the 0 element has been "forgotten". A heap is essentially the same thing as a torsor, and the category of heaps is equivalent to the category of torsors, with morphisms given by transport of structure under group homomorphisms, but the theory of heaps emphasizes the intrinsic composition law, rather than global structures...
Big brother of torsors
 
Awesome
 
4:38 PM
@yuggib
 
@yuggib isn't around
 
Why has he forsaken me
 
@MAFIA36790
Let $(X,d)$ be a $\sigma$-compact metric space, $\mathscr F\subset C_E(X)$ a family of continuous functions with $E$ Banach. If $\mathscr F$ is equicontinuous at each point of $X$ and $\mathscr F(x)$ is precompact in $E$ for each $x\in X$, then any sequence of functions in $\mathscr F$ has a subsequence converging uniformly on compact sets.

*Proof.*
Since $X$ is $\sigma$-compact, it has a countable dense subset $\mathscr D$ (see problem 6). Let $(f_n)\subset\mathscr F$ be a sequence of functions. We enumerate $\mathscr D$ as $\{d_1,d_2,d_3,\dotsc\}$. The set $\{f_n(d_1)\mid n\in\Bbb N\}$ i
 
user116211
4:56 PM
I can only say I know what compact metric space is.
 
user116211
I don't why Kelley wrote supremum is unique in linear (total) ordering.
 
user116211
But it's true for partial ordering too.
 
user116211
The main thing is asymmetry of the ordering.
 
@Slereah hey
I'll be around only for ten minutes or so
 
user116211
@Slereah HURRY!!
 
5:07 PM
Oh well it can wait then
 
@Slereah internal sets are those defined by transfer if I recall correctly
 
That doesn't really help me
 
like $^*\mathbb{N}$ within $^*\mathbb{R}$
 
What's a non-internal set, by contrast?
$F$?
 
it's a set that cannot be written by transfer
I forgot the standard examples
 
5:10 PM
Does it mean that there's no subset of $A \subset R$ such that the set is the transfer of $A$
 
something like the reals are non-internal
exactly
 
Is it just the transfer of the elements themselves or the whole structure?
Since $^*N$ is one I'm guessing the whole structure
 
the whole structure yeah
you should really read the book by goldblatt
everything is carefully explained there
 
Well I got Robinson
Chapter 2 is about that but it's pretty dry
He's stingy on examples
 
@yuggib I need A-A help
bottom of page 3
I don't understand the $3\epsilon$ argument
 
5:15 PM
@Slereah yep, that's why I suggest to read Goldblatt
@0celo7 what do you not understand?
 
about to send this note to the prof:
but if you can save me some embarrassment, cool
 
$d_i$ is a point right?
 
Yes
We know that the series converges pointwise on $d_i$
and the family is equicontinuous on the whole space
 
I mean, if you choose $x_i=d_i$ isn't your hypothesis to get the $\varepsilon$ satisfied?
 
no, $x_i$ are chosen from a finite subcover
$d_i$ are chosen from a countable dense subset
they won't necessarily be equal
well, what you could do is cover by balls on each $d_i$, then take a finite subcover of that (but he didn't write this)
 
5:20 PM
I mean, it suffices that you choose $d_i$ as a point with distance $<2\delta_i$ to obtain the estimate
 
@yuggib did you read what I wrote
 
yes, and I find it mysterious unclear
 
the $2\delta_i$ estimate applies only to $||f(x)-f(x_i)||$
we're not comparing those two
we're comparing $f(x)$ and $f(d_i)$
either of those are $f(x_i)$ in general
 
I seriously suspect that the estimate applies to any $y$ such that $d(x,y)< 2\delta_i$
 
Why?
 
5:23 PM
(with $y$ that could either depend or not on $i$)
and it is roughly stated in the fifth line of the proof in the linked pdf
 
The equicontinuity estimate is that $(\forall x\in K)(\exists \delta>0):d(x,y)<2\delta\implies ||f_n(x)-f_n(y)||<\epsilon(\forall n)$
 
so why would not that apply to $y=d_i$?
 
@yuggib no, the center of the ball is kept fixed
@yuggib dude, the center of the ball is $x_i$
The estimate we have is from the center of the ball to some other point in the ball
But we WANT to estimate the distance between two different point in the ball, neither of which is the center
So we have to use the triangle inequality to go from one point to the center, then from the center to the second point.
 
ok now I see
unhappy choice of notations he used
the later $x$ plays the role of the former $y$ and the later $x_i$ plays the role of the former $x$
 
yes
more clearly written:
The estimate we have is from the center ($x_i$) of the ball to some other point in the ball. But we want to estimate the distance between two different points in the ball ($x$ and $d_i$), neither of which is the center. So we have to use the triangle inequality to go from one point to the center, then from the center to the second point.
 
5:29 PM
anyways, equicontinuity applies to any ball
 
@yuggib what does that mean?
 
in other words, $x\in B_{2\delta}(d_i)$
so you can apply the equicontinuity estimate
 
Yes, that's clear from the triangle
@yuggib why.
 
you apply it with former $x=d_i$ and former $y=x$
in what you wrote above
 
But the $\delta$ might not be the same...
 
5:32 PM
it is chosen to be the same
 
No, delta depends on $x_i$
it's $B_{2\delta_i}(x_i)$.
 
I rectify, it is $x\in B_{2\delta_i}(d_i)$ because $d(x,d_i)<2\delta_i$
 
@yuggib Yes, that's clear, but so what?
 
such $\delta_i$ is the same as above
 
The definition of $\delta_i$ is that $x\in B_{2\delta_i}(x_i)\implies ||f_n(x)-f_n(x_i)||<\epsilon$, do you agree?
 
5:33 PM
so you apply the equicontinuity estimate with the same delta $\delta_i$ but different centers
 
@yuggib Are you saying the following is true:
 
if you want to be extremely picky, choose $\delta_i$ to be the minimum between the delta for the ball centered in $d_i$ and the one centered in $x_i$
 
With the same $\delta_i$ as above, $x\in B_{2\delta_i}(d_i)\implies ||f_n(x)-f_n(d_i)||<\epsilon$?
 
@0celo7 yes, as I said if you want to be 100% sure take $\delta_i=\min\{\delta_i(x_i),\delta_i(d_i)\}$
where $\delta_i(x_i)$ is the one defined by :32799875
 
I'll email the prof
My 5epsilon proof is right
I don't really know what you're doing
 
5:37 PM
and $\delta_i(d_i)$ is the one defined by :32799907
@0celo7 yes, but also is the one of the prof
I gotta go, cya
 
no it's not
not without more work
 
@0celo7 It's not wrong, I just forgot a single symbol on the $K_j$ :P
 
@ACuriousMind No, it is wrong.
I have a counterexample.
What symbol did you forget?
 
$K_j^\circ$.
 
Interior could be empty
(as it is in my counterexample)
 
5:40 PM
Mh. Then I think you need to require $K_i\subset K_{i+1}^\circ$ for the statement to hold.
 
That's an exhausion by compact sets, you need local compactness for that to work
i.e. not gonna happen for Banach spaces
:)
@ACuriousMind Let $X=\{0\}\cup\{1,2,3,\dotsc\}\cup\{\frac{1}{2},\frac{1}{3},\dotsc\}$ with the subspace topology.
This is a separable, noncompact metric space
Write $K_n=\{0\}\cup\{1,2,\dotsc, n\}\cup\{\frac{1}{2},\frac{1}{2},\dotsc,\frac{1}{n}\}$.
Each of these are compact, they're nested
 
Ah, yes, I see
 
But they don't contain $\{0\}\cup\{\frac{1}{2},\frac{1}{3},\dotsc\}$, which is compact.
@ACuriousMind I found an alternative proof of what I wanted anyway, and I found an error in the lecture notes
@yuggib seems hell-bent on defending my prof, but it's certainly not correct as stated
 

« first day (2165 days earlier)      last day (2783 days later) »