i have to run.

Bye

Oh there is an engineering stack exchange, maybe we can start moving all our junk over there...

@DiscipleofBarney can you help me with the question thus is the only question I am stuck at in vector spaces

What is the question? @Rememberme

chat.stackexchange.com/users/138497 This one

That takes me to your account

Oh I gave you the wrong link

chat.stackexchange.com/rooms/13473/linear-abstract-algebra This one

You can go here

And scroll a little back

You will find me asking the question to incurrence

What are you having trouble with?

How to go in showing linear independence do I have to write them in a matrix@DiscipleofBarney

Do you? Do you know what it means to be linearly independent?

I have to show that for any scalars a, b, c, if af_1 + bf_2 + cf_3 = 0, then a = b = c = 0.@DiscipleofBarney

@DiscipleofBarney can I multiply a row like this (0,y_2) by anything

Okay so what are you actually having trouble with, what have you tried?

Multiply a row?, what are you talking you can multiply anything you want. I think you have to be more specific

Let me say if a have a row can I multiply the row by something@DiscipleofBarney

For eg let me have a row (2,3) can multiply it with let's say 4

You can multiply anything by anything you want it is not a crime. So you will have to be more specific about what you are talking about and what the context is, what you are worried about, what you don't want to change, etc

The context is that I have $\alpha = (x_1,x_2) \beta = (y_1,y_2)$ I have two show its a basis and given that x1y1+x2y2=0 and x1^2+x2^2 = 1 and the same with y @DiscipleofBarney

Now I did stuff and got a matrix like (these are rows of the matrix) {1,0}{0,y_2} @DiscipleofBarney

Now I have to get a 1 in place of y_2 so in order to do that I have two multiply the second row by something and I am asking can I do that

Well the matrix changes. You have the same solutions to the homogeous equation equation if you multiply by a non-zero scalar. @Rememberme

So am i right ?@DiscipleofBarney

Right about what?

The multiplying thingy

Explain why it would be okay. You shouldn't need my approval to be able to do something. It does not sound like you know what you are doing or why you are doing it @Rememberme

Anyways according to the question geometrically it means that $\alpha $ and $\beta$ each is of unit length and perpendicular to each other

Ok it is okay because if I multiply the row with some scalar then the solutions will still be the same because later in can do row reduction and bring it back to the earlier row

Ahh......linear algebra is so damn tiring

Yah, that is sort of the whole point of the elementary row operations, they don't change solutions to the homogeneous equations

@DiscipleofBarney So my argument about multiplying is right ?

Its the right idea @Rememberme

okhay

hi @iwriteonbananas

wazzup

Hello

good how are you?@iwriteonbananas

im quite good

solving problem sheets all day

I am really tired after doing linear algebra it has so many computations

very less proofs

Sorry

someone needed linear algebra help?

@Rememberme was it you?

@Rememberme the real mathematics is all about computations, not fancy-looking words. :)

hey! @BalarkaSen

heya

so, you're studying algebraic topology?

I'm doing a course titled "Topology and Groups"

It has an intro to algebriac top

and probably some geometric group theory

cool, that sounds nice.

do you know whether simplicial complexes are locally path-connected? finite ones are

and i'd say probably in general they are

i just answered that above.

oh sorry!

to repeat : yes, they are locally-path connected.

yes

nice

in general, CW complexes are locally path connected

is there a proof in Hatcher?

or is it obvious

oh

yeah I think i see

yes, there is a proof in the appendix. either way, it's not hard.

thanks

if $x$ is not a vertex we can just take the interior/inside of the simplex it sits in

and then that's homeomorphic to some open disc

if it's a vertex we can consider the star or something

the point is that CW complexes are all locally contractible. this is obvious if the point is inside the interior of a cell. so you just need to check the nbhds doesn't do anything crazy when it is in the boundary, which you identify with another cell via an attaching map.

@JC574 right, it's even simpler for $\Delta$-complexes. you need to look at nbhds of points on the corners.

you should work out the whole detail though.

@BalarkaSen cool

yeah I can't use the word contractible, this isn't for my course, it came up in an exercise a guy in the year below me was doing, taking an introductory point-set topology course. He asked for help

I can see how the argument goes and it's not vital i work through it now

ah, i see. well, save it as a mental-note. it's a good fact.

yeah the exercise was to show the "comb space" cannot be triangulated

@JC574 ok, if you wish. personally, i don't care for the point-set topology of CW-complexes :p

@JC574 ah.

haha yeah i know right

not sure why their course focuses so much on it

i have taken a course on alg top and the professor only spent two or three days on the point-set topology of simplicial/CW complexes.

i mean, yeah, better know the space you are going to work with (like whether it's even hausorff or not, how can one define simplicial structure on product spaces, etc) but there is more to life that this.

@BalarkaSen Has the existence of life beyond point-set topology of simplicial/CW complexes been conclusively shown?

of course : that's what algebraic topology is about :p

Greetings

@r9m hope you're in a better mood now. :-)

@robjohn did you try that question I showed you yesterday? I'm thinking to add it to my book too.

Hey @Chris'ssis

@Nickolas Hey

@Chris'ssis hows your book going

Hi @Nickolas

my textbook suggests there exists a set where $\overline{E_1\cap E_2}\not\supset\overline{E_1}\cap\overline{E_2}$ where the overline denotes the closure of the set

what kind of examples should I be thinking about?

I haven't found anything that seems to apply here

are any such examples sets unusually complicated?

(here the closure is the points that are either members of the sets or accumulation points of the set)

aw I think I finally found something

something like $E_1=(0,1)$ and $E_2=\{0,1\}$

$\varnothing\not\supset\{0,1\}$

Which is the simplest figure in 3 dimensions?

I feel its a sphere

@TruthSerum

You can make everything in this world into a sphere

Inflate it

But how do I prove it

I have not touched statistics I like geometry and abstract stuff more...@TruthSerum

Then what is the simplest figure

@GBeau not particularly. there are infinitely many examples with open sets in $\Bbb C$ which intersects at the boundary but are disjoint otherwise. the one you wrote works. in general, $A, A^c$ works where $A$ is an open set in $\Bbb C$ with the standard topology.

Hey @Balarka If i start crushing all the things in this world one by one which is the simplest mathematical figure that i will obtain and how will i prove that that is the simplest mathematical figure i will obtain

Sorry if this is nonsense

SO simplexes are the simplest mathematical figure in 3 dimensions?

it took me a second to realize how $[0,1]$ could be the intersection of open sets

under the same section the book asks the same of a union of open sets (to construct [0,1])

that's...not possible, right?

oh I think the book has a theorem that would apply

"Any union of an arbitrary collection of open sets is open."

it's not a theorem. it's one of the axioms of a topology :)

@BalarkaSen depends on the viewpoint

fair enough. i guess @GBeau is working with metric spaces.

hi, btw

hi@BalarkaSen

it's an introductory analysis course, this was the last chapter we're doing before finals

chat.stackexchange.com/transcript/message/21364599#21364599 @BalarkaSen what do you think about it

i think that it is nonsense.

why?

thus, i am not bothering to answer.

If we have the interval set $[0,1]$ with the induced topology from $\Bbb R$, can there be a continuous surjection $f: [0,1] \to \Bbb R$ since every element in $[0,1]$ can be found in an open set?

oh....okay

@Rememberme lots of undefined terms. "crushing", "simplest", etc.

Oh okay undefined terms .....

i will try to bring it you with definitions

@ᴇʏᴇs There is not only a continuous surjection, but also a continuous bijection.

Even a continuous bijection with continuous inverse :p

Try to find it.

oh, wait, $[0, 1]$

@BalarkaSen you said maths is computations but isnt math totally a proof subject :)

ummm

(0,1) works, right?

yeah, (0, 1) is homeomorphic to R.

@Bal I'm not sure about the agreed-on name of the topology but it's $[0, 1]$ with this topology from $\Bbb R$: en.wikipedia.org/wiki/Subspace_topology

So that would make $[0, 1)$ an open set with this topology

try to see if there is a continuous surjection from [0, 1] to (0, 1) first.

is that not the same problem?

yes, but it's easier if one shrinks the domain.

Are you familiar with compact sets?

one doesn't need compactness to do this, but ok..

@BalarkaSen sure, but that makes for a very easy argument

We haven't learned compactness, so I don't think we're allowed to use that argument although $\Bbb R$ not being compact would make the argument easier

I know the usual $[0, 1]$ isn't surjective onto $\Bbb R$, but with the subspace topology on $[0, 1]$, sets like $[0, 1)$ are considered open so it seems like it might be possible

@Nickolas All is fine so far. Thanks.

More important than the book itself is to reach the level of Ramanujan in terms of integrals, series and limits.

(Ramanujan was also great in this area because of the creation of some tools with which he solved a lot of impossible stuff - like mathworld.wolfram.com/RamanujansMasterTheorem.html)

wait

it just occurred to me that theorem isn't sufficient

since closed doesn't mean not open

(and visa versa)

I think the only sets both open and closed are $\varnothing$ and $\mathbb{R}$ but I think my knowledge of that is out of the scope of the question

(can a union of open sets be used to construct $[0,1]$)

oh I just have to note [0,1] is not open :<

question: if I have a function f:R^n -> R^m

and the partials of f are defined everywhere

and the partials are bounded

then this implies f is continuous

does this imply that f is differentiable everywhere?

I just need to know whether this is true or false

I think one potential counterexample would be f(x,y) = \sqrt{|x| + |y|}

so it shouldn't imply that f is differentiable everywhere, should it?

Good morning, everyone!

Hi @teadawg1337

@teadawg1337!!!!!!!!!!

check my answer once math.stackexchange.com/questions/1257411/…

@Rememberme You seem to be doing several things that are questionable since this is a sequence of complex numbers.

@TobiasKildetoft i dont find any problem with my proof

@Rememberme Actually, it has a very basic problem from the outset: You need to pick branches a lot of times to even make sense of the sequence.

but @Tobias does my proof have any faults

@Rememberme manipulating a not well-defined sequence seems like a pretty big problem to me.

@Tobias i found this problem a lot same to this one math.stackexchange.com/questions/1170303/…

isnt it the same

@Rememberme ohh, I didn't read it properly it seems

so my proof is right?

I missed that you would only need odd roots

@Rememberme Yes, but it need a lot of arguments that are not there for why you are allowed to do those manipulations

@TruthSerum it does not even make sense to ask that, unless you define what the expected value of a number is (expected value is defined for stochastic variables)

oh I had another question

that I was independently curious of

@TruthSerum that suffers the same problem

if $A\subseteq\mathbb{R}$ is a closed interval and $f:A\rightarrow\mathbb{R}$ is a bounded continuous function, can we say with certainty that $f$ has a maximum?

I was wondering in class when I could say some function had a max, and it occurred to me that it wasn't true for $A$ being an open interval and it wasn't true if $f$ wasn't continuous, but once applying the above restrictions, is the above true?

@TruthSerum ahh, so it is the usual misuse of notation, using $E[X]$ to refer both to the value and to the constant stochastic variable

in qhich case, it is trivial that expected value is idempotent

@TruthSerum You mean that the expected value of a constant stochastic variable equals the value it takes everywhere?

@GBeau Well, assuming you mean that the interval is bounded

in which case, the function is automatically bounded

and will indeed obtain its maximum

@TruthSerum right, that is the same statement

@TobiasKildetoft I do not mean to imply the interval is bounded

perhaps just "A is closed"

I mean the range is bounded

gah, i am missing something obvious. i don't get why TQFT of the bordism between the empty manifold and two points with opposite orientation has to be the evaluation map $Z(pt) \otimes \overline{Z(pt)} \to k$.

@GBeau well, the only closed interval that is not bounded is the entire real line

@GBeau image of a closed interval in R is compact and bounded, thus closed.

@BalarkaSen of a bounded and closed interval

right, right :p

mumbles something about being nitpicky (i kid)

@TobiasKildetoft yes I mean to include this

in my head I'm thinking "R and things like [1,2]"

@GBeau those behave a lot differently

I understand the question is probably out of my scope rigorously

I was just curious

clearly, we cannot be sure to have a maximum for a function defined on all of the reals, even if it is bounded (just take $1-\frac{1}{x}$ and continue however below $0$).

(I am missing any good definition for continuous)

@TobiasKildetoft which is why I applied continuous and closed

which seems to preclude such examples

@GBeau yeah, you need to tweak it slightly for that to work

@GBeau closed will not help when you do not disallow the entire real line

what is the example on the entire real line that is continuous?

@TruthSerum to see what I said above, just apply the definition of the expected value

@TobiasKildetoft I don't see how the construction works for continuous?

@TruthSerum whatever definition you have been given

@GBeau Take $1-\frac{1}{x}$ for $x > 1$ and $0$ elsewhere

@TruthSerum that is not a definition

@TruthSerum do you see how to make the expected value of $X$ into a stochastic variable itself?

otherwise it does not even make sense to speak of the expected value of the expected value

@TruthSerum What about it does not make sense?

If $X$ is a constant stochastic variable with $X(x) = a$ for all $x$ then it should be easy to see that $E[X] = a$.

@TruthSerum indeed, it is not in general multiplicative

@TobiasKildetoft I see! so, if we have a strict subset then $A\subset\mathbb{R}$

@GBeau Then we are in the case of a closed and bounded interval

which may still not have a max

can anyone look over my proof about lipschitz functions?

a finite supremum, certainly

@GBeau No, that will definitely have a max

@TobiasKildetoft with my full list of requirements, or only "closed and bounded" ?

@GBeau any continuous function on a closed and bounded set will have that property

ok thank you

$f(x)=1-x$ when $x\neq 0$ and $f(x)=0$ when $x=0$ for interval $[-1,1]$ was my original counter-example that made me suggest continuous

@GBeau right, you need continuous

@GBeau A fun exercise: Let $f: [0,1]\to [0,1]$ be continuous. Show that $f$ has a fixed point.

@TobiasKildetoft The same example shoved against an open interval made me suggest closed

(since it's now continuous)

How do you take cross product of cross products ? I'm having trouble simplifying (ba)*(bc) .

wait I mean (b cross a)cross(b cross c)

@AGoogler The cross product is associative and anticommutative

so is it like b cross (b cross c) + (a cross b cross c) ?

Woops, sorry, it is not associative

So how can I simplify it?

@TobiasKildetoft oh [0,\infty) is closed

so it must be a bounded strict subset

@GBeau Right

@TobiasKildetoft can you help me with a real analysis problem?

hi @DanielFischer

@JohnDoe I don't think he's ere

Okay no prob

Well, now I am.

@DanielFischer Now that you are here, do you think you could help me out with a real analysis problem?

It goes like this:

Let $f:\R^n \to R^m$ be a differentiable map. Prove that $f$ is $C^1$ iff $Df : \R^n \to (L(\R^n,R^m), || \cdot ||)$ is continuous where we view the target as a metric space using the operator norm $||\cdot||$.

@Newb For me, that's the definition of $C^1$. How is $C^1$ defined for you?

if I recall correctly, $f$ is $C^1$ if $f$ is differentiable (at least) once, and its derivative is a continuous function.

But what do you mean by "its derivative is continuous" except the above?

Hi @Mike.

that's what had me confused -- it seemed kind of tautological

@Newb Then you should check how the book/course is defining $C^1$. It might be that they define it so that all partial derivatives are continuous.

@DanielFischer I'm think the definition I gave was the instructor's...

Then you'd need that a) continuous partial derivatives imply differentiability, and b) the matrix of the derivative (in the standard bases) is given by the partial derivatives. And c), all norms on a finite-dimensional space are equivalent.

@Newb Well, for the definition "$f$ is differentiable and the derivative is continuous", it is tautological.

But while I'm asking -- what really had me confused was the statement $Df : \R^n \to L(\R^n \to R^m), ||\cdot ||$

is that a usual way of interpreting the multivariate derivative?

(I'm new to analysis in R^n)

hmm

@Newb The derivative of a differentiable function $f\colon \mathbb{R}^n \to \mathbb{R}^m$ at a point $x\in \mathbb{R}^n$ is a linear map $\mathbb{R}^n\to \mathbb{R}^m$.

I guess it makes sense because when we take the derivative of some multivariate function

we get a matrix that maps R^n \to R^m

I'll check my instructor's notes for his definition of C^1

Hi @DanielF.

@DanielFischer Aha! You were right about the definition

I was using the definition of C^k for $R \to R$

Not the one for R^n \to R^m

morning

@DanielFischer For the Lebnitz rule as indicated in the wiki post it states that $f$ and $f_{x}$ are required to both be continuous, but in the measure theortic formulation further down it only requires that $f$ be Lebesgue integrable, $f_{x}$ exists and that $f$ can be dominated by some integrable function. Do you know why the conditions on $f$ are relaxed in some sense for the measure theoretic formulation of the Leibnitz rule?

Hi@KarimMansour

Hi @Rememberme

@StanShunpike I remembered you talking about the $\Bbb R^n_{++}$ notation, and saw this question that had a similar notation for strictly upper triangular matrices

@JohnDoe We have stronger convergence theorems for the Lebesgue integral. I can't say anything detailed off the top of my head. If you try to prove it, you should be able to identify points where a lack of regularity would pose problems. How these problems can be overcome - how much regularity is needed - depends on the theory you're working with.

@DanielFischer Okay I see. Have you done any quantum mechanics?

@JohnDoe No, my physics is basically all $19^{\text{th}}$ century and earlier.

@DanielFischer Is it easily seen why $f$ and $f_{x}$ have to be continuous for the non measure theoretic Leibnitz formulation?

@TruthSerum linearity of expectations

@JohnDoe It isn't required. Having both continuous is just an easy sufficient condition. If you go beyond that, it makes sense to go far beyond that, since proving it in the measure-theoretic form isn't much harder than proving versions more general than the one demanding continuity but still less general than the measure-theoretic version for the Riemann integral. And for the Riemann integral, the complexity of the proof grows much faster than the generality of the result you prove.

why didn't you do quantum mechanics @DanielFischer

@DanielFischer Okay I see. What properties of a function $f$ satisfying the condition $\int_{- \infty}^{\infty}|f(x,t)|^{2}dx = 1$ are apparent? It seems clear that $f$ is at least bounded. Can you see anything else?

@JohnDoe Need not be bounded. Except that $x\mapsto \lvert f(x,t)\rvert$ is in $L^2$ with norm $1$, you can't say anything.

@DanielFischer I thought about your remark earlier on how to prove the claim, but I'm a little bit stuck

Do you think you could flesh out the directions a little bit?

@Newb At what point are you stuck?

@DanielFischer Is it clear to you that $f$ goes to zero as $|x| \rightarrow \infty$?

@DanielFischer for one, I'm not sure how I'm meant to use the fact that on a finite-dim vector space, all norms are equivalent

@JohnDoe It need not.

how do I use that in the proof?

@Newb It's pretty immediate that the components of the matrix of $Df$ are continuous if you have continuous partial derivatives. From that, you can directly get the continuity if you endow $L(\mathbb{R}^n,\mathbb{R}^m)$ with the maximum norm with respect to the entries of the matrix wrt the standard bases. Then, by equivalence of norms, it follows that $Df$ is continuous wrt the operator norm.

how does putting the max norm on L(...) yield continuity?

@Newb Continuity in the maximum norm is continuity of all matrix entries. That is easy to see. It's less easy to see for the operator norm.

Dinner time, bbl.

@DanielFischer Okay, thanks. I'll try to put together a proof.

@DanielFischer In the book I am studying it states that $f$ must go to zero faster that $\frac{1}{\sqrt{|x|}}$ as $|x| \rightarrow \infty$.