Question: Am I right in thinking that $A := \{(\frac{1}{n},0) \mid n \in \Bbb{N} \}$ as a subset of $I_o^2$, the unit square in the dictionary order topology, has no limit point in $I_o^2$ and therefore $I_o^2$ is not metrizable?

@user193319: It certainly does have a limit point. What things do you know about compact metric spaces in terms of countability axioms?

I know that a compact metric space has a countable basis.

hi

@user193319: So, is that true of $I_o^2$?

heya @Meow

@user193319: BTW, you should find that limit point to make sure you understand the dictionary order topology.

@TedShifrin I don't know. I tried showing that it doesn't have a countable basis, which proved to be difficult; so then I tried to an infinite subset with no limit point.

Well, I'm telling you there is a limit point. But if you have a countable basis, how many disjoint open sets can you have?

@TedShifrin I am guessing an uncountable number, but I don't know how to show this.

Oh wait!

Uh huh?

Each vertical strip is an open set, which are uncountable in number, and each one would have to contain a basis element from an arbitrary basis. Since these vertical strips are disjoint, each basis element in each strip will be distinct from the others.

Hence, any basis will be uncountable.

Good!

Excellent! Thanks for the hint!

Now, to finish up, what's the limit point of your sequence of points?

It think it would be $(0,1)$, the upper left vertex, since any open interval around $(0,1)$ would "bump" up on the bottom edge of the square.

Yup.

OK, you have learned something today :P

Obviously that could be made more rigorous, but I think it works for now.

Yes I have! Thanks again.

Sure.

heya anon

Is the product between a matrix and a vector usually called "dot" product or can we also call it matrix-matrix multiplication?

NO, not dot product

what's wrong with matrix-vector multiplication? :)

@TedShifrin gasp

smacks Demonark

Nothing wrong, it's just my TA which is arguing that the product of a matrix and a vector is a dot product.

I used the notation $W \cdot x $, yes, but I clearly didn't mean the dot product...

Its entries are dot products of rows of the matrix with the vector, but I would NOT call it dot product, EVER.

I don't understand these TAs which write the assignment in an ambiguous away and then they take you points away because you didn't write exactly the solutions in they way they wanted. Ridiculous.

Like, write the solutions in "vectorized" form!

What the hell does it mean?

Well, if what you wrote is 100% correct, then I agree. But often students think things they've done are correct when they are far from correct.

Is your TA a native English speaker?

I think it means to use vector notation rather than components.

why would you not call the product between matrices and vectors just matrix multiplication lol

yes, I used vector notation, but a vector notation "in the way they didn't mean", so they took me points away...

I usually distinguish, Eric, but it's fine by me.

@nbro: I would need to see what you wrote to comment.

The usual mean squared error...

So no dot.

$\mathbf x_i$ are a bunch of vectors?

xi is just one vector

And $W$ is a row vector?

No, you have a subscript being summed over.

W is considered a matrix.

OK, I think we agree.

But it's a $1\times n$ matrix?

ok I would be confused as to how to interpret this notation if i were grading

Yeah, I'm still losted.

This looks totally wrong to me.

It looks like components whose squares are being added, but we haven't taken the $i$th component of $Wx$.

So something's wrong.

So I'm guessing it has to be an honest dot product because t_i seem like real numbers, so subtraction wouldn't make sense if Wx_i were a vector

Damnz, yes, I realized that in the way I wrote W, $W \cdot x_i$ ends up being a dot product, but in my head that was simply a matrix.

@nbro: This really is not right at all.

But then W really feels like something you'd denote as a matrix, like this is actually a bit iffy

What vector are you computing the length of? Forget the summation.

They are two column vectors.

Who are?

Demonark: It is, however, standard to "confuse" a $1\times 1$ matrix and a scalar. I'm fine with that abuse.

are the $\mathbf{x}_{i}$ components? is $W$ a matrix or a vector?

$W$ and $x_i$ are two column vectors, so that's really a dot product, so my TA is right.

I think we're looking at $\mathbf t - W\mathbf x$ as a vector.

My worry was more that if W was a non-vector, Wx_i is a vector, so t_i minus that makes no sense

I guess I don't really know least squares, even though I wrote two books with 'em in 'em.

ah ok

But if W is a vector as well, we're taking the dot product of two vectors and subtracting it from a scalar, that I can accept

how are you people?

I would write this as $W^{T}\mathbf{x}_{i}$ and make really damn clear that $W$ is a column vector

I don't think we really have $n$ different vectors $x$.

I think one needs to start at the beginning and really understand what's going on.

But the use of capital W as a column vector makes me quite sad

I also would never do this^

I realized my mistake while talking with you, thanks!

won't give a lecture on least squares solutions of $Ax=b$

I dunno if I've heard of least squares

@TedShifrin Yes, probably they meant something like that by "vectorized"

Hi handsome ppl

Souganidis didnt do it with you? He did it with us @Daminark

Yes, @nbro, I approve. And I think you really need to make sense of this from the beginning.

You can see my lectures on least squares on my YouTube videos :P

@Daminark @TedShifrin extra hello to you extra handsome folks

hi Kasmir

:)

it's like super useful if you like, do things that matter to people that aren't math people

It's way past your bedtime again, Kasmir.

Yes, my bad! Sometimes, I just feel more comfortable by thinking in terms of components.

I just woke up Ted haha

slpet 8 pm

It's really not good, @nbro ... you lose understanding and notation gets horrendous.

@Eric Maybe we did it by a different name but remember that Soug just gave us a bunch of problems from a book which has no applications each week and never touched LA in class

Sometimes matrix notation is not clear, dimensions are missing, etc. So, to clarify my thoughts, I think in terms of components

bad idea

my life has too many indices already

LOL

Ted ! what should be studied first

topology or algebraic top , or diff top

topology ... diff top requires serious multivariable analysis, which you have not had yet

proofs with the derivative as a linear map, inverse function theorem, etc. (That stuff is all in my lectures, but you didn't do them in your courses.)

Okay then thank god i applied for that :D

have you had a serious analysis course yet? proofs with metric spaces?

we do things differently here than in US

No ><

@nbro in math you really want to be eclectic, is the vibe I'm getting. If a situation is best framed in terms of coordinates, you should be able to do coordinates/matrices, and if a situation is best framed in terms of linear transformations, you should jump to that.

you should usually do analysis before topology ...

But we call that course real analysis

yes, we do too. You should do that before topology.

Oh :(

is there like no chance without real analysis?

Topology is a generalization of that stuff, up through continuity.

If you've never done $\delta$-$\epsilon$ proofs seriously, topology will be super abstract.

It's quite hard, and I'm absolute shit at that, but at least try, because the less religious you are about a specific mindset, the better when that becomes cumbersome

because I just took that course for curiosity only , my main Courses are numver theorey and rep theory

I did epsilon delta in detail

Woo number theory!

:D

I'll be back in a bit.

THere is a book written by ireland and rosen dami

very good i hear

@TedShifrin Okay Ted :D i still got some Q's on the material :D

I've got that one. And there's one that Mathein recommended which is also worth looking into

Algebraic number theory by Jarvis

@Ted catch you around!

I hate it when Ted leaves :(

Ehm okay ill check that book too thanks dami

@Daminark do you have the ISBN of it?

I can find you an Amazon link

Or just send over/find you a pdf

pfd would be great :D

but ill check if the librray have it or better yet if they sell it here

Okay, what's your email?

@Daminark got it? ><

Nope, not yet

Try again

there :D

tell me when u done

copying it

Okay

come on dami just copy it :D

hoo :)

I hope no one else was here on chat

I dont want to be hacked ._.

I doubt you'll be. And I sent the book

brb hacking

AAAAAAAAAAAAAAAAA

Rest in hax0rz: Kasmir

hello

@EricSilva Dont you ever try !

sup cantor

@Daminark ill check it now, thanks dami :D

Yo @io_cantor!

I'm going to read your pdf and see if it helps me

on sylow

Damn, a full 5 people will have read my paper then. Two were revising but still

Do I qualify as a world-class expositor?

yes 5 ppl is the precise requirement

u made it

congrats

\('-')/

haha

it was good dami

how do i set a thing on top of an arrow in tex

Trying to write a commuting diagram I see. One sec, lemme see if I've got something

I guess that or an exact sequence

You'll want to use \overset{i}{\longrightarrow}

Dami not sure what happned bu

did not get the mail: (

Check spam?

good stuff

i did

Nothing there new either

maybe the file was too big?

Strange. Email me at BALEETED @Balarka

okay

Done :D

Okay now check to see

@Daminark Yes I got it :D thanks alot ! :D

Woot woot

DEBEETED

I am really unsure how that particular Homestarrunner bit became so widespread

Spam! It's pink and it's oval! Spam! I bought it the Mobile! Spam!

It's made in Chernobyl! Spam!

@AkivaWeinberger Seems also relevant for reading papers from other natural sciences as well

It's not spam if it is not this:

i.huffpost.com/gen/1667145/thumbs/o-SPAM-570.jpg?6

I'm lolling irl

So ive been watching the visual group theory youtube videos for the past little while now

and man

it's been eye opening

It makes so much more intuitive sense. I finally "get" what it means that a group is meant to act on a set

I'd hope it is, doing visual group theory with your eyes closed is a bad idea

haha

I have a question that I'm having trouble with

Consider the rectangle with labeled vertices X = {1,2,3,4}. Consider the set of actions G= {e, h, v, t}. h is horizontal flip, v is vertical flip and t is 180 degree turn. Now this is obviously the klien 4 group. But I want my set G to act on X. I get that $g \cdot x = x \implies g = e$, but this doesn't have to hold true for general actions. Is there a geometric shape for which this implication fails?

I know such actions exist because actions are homomorphisms from G to Sym(X) and we can consider the trivial homomorphism so that all x are sent to themselves, but what does that mean visually?

it means the action is transitive for one

visually?

stab(x) need not have only the identity btw

I totally understand that lol

I'm not disputing that

What do you mean by visually then

if you let (12) act on the vertices on the square, [3] , [4], it dont move them

and (12) is not the identity

visually i guess you need to find a rigid motion that preserve somthing

exactly, let me draw it tout

out

vertices edges or something stuff but that is normally hard to find when shapes are not simple

that is not something you want to do for ever, maybe few examples of it to get the intuition

Hmph

So, I drew a square. I get 12 different states of the square.

I have 3 generators for my set of actions <h,v,r>, horizontal flip, vertical flip, and 90 degree rotation. Yet if I let my group act on this the shape, it still has tthe property that $g \cdot x =x \implies g = e$.

also commutative

Let $f$ be the diffeomorphism between the torus $\Bbb T^2 := \Bbb S^1 \times \Bbb S^1$ to itself by swapping the two coordinates. $f$ is not homeotopic to the identity map in the ambient space $\Bbb R^3$. Is there an ambient space $\Bbb R^n$ such that $f$ is null-homotopic?

@LeakyNun do you know basic group theory?

@io_cantor I know advanced group theory

Do you know of an example of a geometric object

that when acted upon

you have $g \cdot x = x \land g \neq e$

where are the quantifiers?

$\exists g \in G$ such that $\forall x \in X$ we have $g \cdot x = x \land g \neq e$

Where X is the set of vertices for our geometric object

i Ddotn know where you going with this but

(56) acts on a set the set {1,2,3,4}

@io_cantor I don't think those are the right quantifiers

(56) element in S_17

@LeakyNun why not

so S_17 acting on {1,2,3,4}

and I'm asking because different quantifiers can have profoundly different meanings

and all of those meanings are used

@Leaky it's pretty clear what he meant

Morning

@io_cantor so you want an element that is not the identity and fixes every object?

How's @LeakyNun

morning faust

long time no see

@io_cantor don't count on me for geometry in general

I like saying Faust :D

it's in the stabalizer

yes

I'm absolute shit at it and don't even like it much. But...

but I liked you more when you were Faust7

Been busy doing stuff and things so much math

Do you know about matrix groups?