@BalarkaSen I'm pretty sure an alternating k-tensor is an element of $\wedge^{k}(V)$, not a linear transformation defined on it.

@Eric Thanks, yes. I was thinking of an alternating k-form.

Another way to look at is that $\frac{d}{dt}\ln x(t)=x'/x$, so you've got $\frac{d}{dt}\ln x(t)=c$.

Which is an element of $\wedge^k V^*$

Integrating both sides then gives $\ln x(t)=ct+C$, which you can rearrange to $x(t)=ae^{ct}$ (identifying $a=e^{C}$)

Though I seem to have gotten the labels backwards. Oh well.

@Semiclassical are you talking to me?

Yeah.

I know this way works. I just don't know why I can not take the derivative of $x=ce^{at}$ to show it is the general solution

You mean, is it enough to verify that that's a solution?

yes

I suspect it depends on the rigor of the course.

At a formal level, it's certainly enough.

But I forget if there's some things you need to check for that to be rigorous.

I don't like plugging in stuff in differential equations to verify they are solutions.

ok

There is something to be said for concluding that "this is what the solution must be" rather than guessing and checking, yes.

Hence why I do like the above presentation.

On the other hand, I don't mind a well-chosen ansatz.

If I have a system $X'=\begin{pmatrix}0 & 1 \\ -9 & 0\end{pmatrix}X$ and need to change the coordinate, like $Y'=(T^{-1}AT)Y$, what is $T$

Depends what you're looking for.

(Should it be X for the second equation as well?)

not sure, that is what I got from the question

Hmmm.

I think they probably mean this, actually.

You've got a linear system $X'=AX$.

yes

Suppose you want to change coordinates from $X=(x_1,x_2)$ to $Y=(y_1,y_2)$. That should correspond to some transformation matrix $T$ such that $Y=TX$.

I'm assuming $T$ is just a coefficient matrix, so therefore you'd also have $Y'=TX'$.

Hence you'd have $Y'=TX'=TAX=TAT^{-1}Y$. I guess this is backwards, though, so I should've said $X=TY$.

If I solve $Y'$ , then the solution for $X$ is $TY$

Right.

This is clear from your original equation if you multiply both sides by $T$, since then you've got $TY'=A(TY)$ which is the same system as $X$ satisfies.

So $T$ is whatever transformation matrix is used to transform $Y$ into $X$.

How do I find it

Depends what you're looking for. Typically, you want to have $\Lambda = T^{-1}A T$ a diagonal matrix so that $Y'=\Lambda Y$.

In that case, you'ove got $y_1'=\lambda_1 y_1$ and $y_2'=\lambda_2 y_2$ which are both of the form you had above.

the eigenvalues are $\pm 3i$

should $T$ just a matrix of those two correspond eigenvectors

by eigenvalue decomposition

Well, suppose that's true. Then you'd have $T=(v_1\,v_2)$ where $v_1,v_2$ are the (column) eigenvectors of $A$.

Hence $AT=A(v_1\,v_2)=(Av_1\,Av_2)=(\lambda_1 v_1\,\lambda_2 v_2)=(v_1\,v_2)\text{diag}(\lambda_1,\lambda_2)=T\Lambda$.

So yeah, you want the columns of $T$ to be right eigenvectors of $A$.

so my $T$ is $\begin{pmatrix}1/3i & 1/3i\\ -1 & 1\end{pmatrix}$

@Semiclassical could you verify my statements above?

from MatLab, $T^{-1}AT=\begin{pmatrix}3i & 0 \\ 0 & -3i\end{pmatrix}$

[1..2] is supposed to be what? Not familiar with that notation.

@Semiclassical $\{x \in \Bbb R | 1 \le x \le 2\}$

Ah. I typically see that written as [1,2].

$\overline{A}$ is supposed to be the complement of $A$?

@Semiclassical no, the closure of $A$

Oh.

Well, every irrational number in [1,2] can be found as the limit of a sequence of rationals in [1,2]. So yeah, the closure should be the entire interval.

what's A^\circ intended as?

the interior of $A$

Hmm.

It's been too long since I did real analysis, as you can tell.

So I think I'll have to decline to say more on the grounds that I'm likely to say nonsense.

I like to hear nonsense

@Vrouvrou oh, and $\partial(\partial A) = \{1,2\}$

Eh, I like interesting nonsense.

I don't think anything I could say re: real analysis would count for that.

what on earth is this topology

Is it ok to write a system of ode in the form $X'=\begin{pmatrix} 3i & 0 \\ 0 & -3i\end{pmatrix}X$

@Simple sure

that means $a' = 3ia$ and $b' = -3ib$

where $X = \left[\begin{matrix}a\\b\end{matrix}\right]$

just to be clear, in the problem above you'd have $Y',Y$ in place of $X',X$ there.

yes,

So you'd have to transform back by using $T^{-1}$ to get the solution in $X$.

I get this

sorry - I left to eat food. The reason I prefer the term basis independent is because coordinates don't make sense to me without defining a basis first because with a vector space like (R^+, *), it doesn't always make sense to me to define a coordinate system first if that makes sense to say?

Given the reals in the usual topology (thus all open intervals are open sets). Suppose we use the Euclidean metric, then 2 and 6 are closer than 2 and 8 because d(2,6) < d(2,8). We can also see that 6 and 8 are closer to the aforementioned pairs since d(6,8) < d(2,6) < d(2,8). But if I don't have any metric and just the topology, then a open set (1,9) contains 2 and 8 but another open set (1,4) contains 2 but not 8. Therefore I had a weird scenario where 2 and 8 are close to each other and far

away at the same time in this topology?

yeah, it's not a perfect intuition but more or less, just kind of let your knowledge of metrics fade away

they're close because they're in an open set toghetr

timothy gowers explained it better

let me find it

nvm I misattributed the source

here we go

it's a little different than what I took from it

to be honest that's a better intuition than I had

thanks

@Secret Could you check my statements above?

@DHMO Then what i say is correct

Hello, @MartinSleziak

@DHMO A recently posted answer here gives a link to a paper mentioning that: it was striking that one could not even decide whether or not $(3/2)^n \pmod 1$ has infinitely many limit points in $[0, 1/2)$ or in $[1/2, 1)$.

@Vrouvrou the last sentence of what you said is incorrect

@MartinSleziak thanks

Which also implies that your question about $(3/2)^n$ is an open problem. (Or was in 1995.)

@MartinSleziak can we know if (3/2)^n mod 1 has infinitely many points in [0,1/2) or [1/2,1)?

@DHMO ok

can you please help me here math.stackexchange.com/questions/2160430/…

what is $\varphi(u)$?

$\varphi:(0,+\infty) \rightarrow \mathbb{R}$

je veux remplacer $u$ par $u^-$

@Vrouvrou sorry I'm busy

okkkkk

@DHMO A longer quote from the paper linked in that answer: "One of the few positive results known for (non-integer) rational $\theta = p/q$ is that of Vijayaraghavan (1940) who showed that the set $(p/q)^n \pmod 1$ has infinitely many limit points. V Vijayaraghavan later remarked that it was striking that one could not even decide whether or not $(3/2)^n \pmod 1$ has infinitely many limit points in $[0, 1/2)$ or in $[1/2, 1)$."

So $a^n$ for $a$ rational has infinitely many limit points in $[0,1)$. If I understood your chat message correctly, you asked this for $a=3/2$.

@MartinSleziak I asked for points

follow the definition of probability generating function, I could have $G_X(s)=\sum_{k=0}^{\infty}s^kP(X>n)$

I notices that $E\left(\frac{1-s^X}{1-s}\right)=(1-G(s))/(1-s)$

@DHMO Isn't the sequence $(3/2)^n\bmod 1$ injective? (I have a feeling that we discussed this.) If yes, than it is immediately clear that it attains infinitely many values.

But maybe I have misunderstood what you are asking here in chat.

@MartinSleziak I asked for infinitely many values in [0,1/2) or [1/2,1)

Yes we did here.

@MartinSleziak I am asking for infinitely many values in [0,1/2)

or infinitely many values in [1/2,1)

If we know that a set infinitely many values in $[0,1)=[0,1/2)\cup [1/2,1)$, then it is immediately clear that it has inifnitely many values in one of the two intervals.

I know that it must have infinitely many values in at least one of the intervals I mentioned

I'm interested in whether we can prove it for either of the two intervals

I understand now.

I hope I make my intentions as clear as I can to you.

It was caused by the phrasing - In understood the question: "Doe it have infinitely many values in [0,1/2) or [1/2,1)?" as "Is this true: (infinitely many values in [0,1/2)) $\lor$ (infintely many values in [1/2,1))

So you are asking whether it has infinitely many values in $[0,1/2)$, And a separate question where you are asking the same about $[1/2,1)$.

@MartinSleziak I just followed this phrasing

@MartinSleziak yes

In the other words: We know that this is true for one of the intervals. Can we say which one? Is it true for both of them?

Yes

@DHMO that looks ok. As for your topology question, I am not sure given a few hours ago I have been mixing up limit points and boundary points and many other concepts

@Secret alright

How can i approach a proof without epsilon delta and balls that any map f from (a,b) to [a,b] must be discontinuous unless f is the constant map, using reals in the usual topology?

@Secret how can you prove without epsilon-delta or balls, if "continuous" is defiend using either of them?

by the way, $f(x) = a+b-x$ is continuous

@DHMO I am trying to make a proof using the topology of the reals alone, so I can use this case to get more intuition on how to approach continuous maps in general topological spaces. While Yash and a MO answer have provided me the intuition on open sets, I want to see what contradiction will arise if the preimage of a open set is closed in order to further justify the definition of continuous map on how it recovers the real analysis notion of continuous

But so far, i struggle on getting a contradiction because it seems no matter what I tried to do, somehow I am forced to introduce a jump in the function near the boundary points a and b

Have you considered my counter-example to your claim?

@Secret Are you missing some conditions there? The map $f(x)=x$ is clearly a continuous map $(a,b)\to[a,b]$.

I do not see an easy answer to the DHMO's question about values $(3/2)^n \bmod 1$ in the interval $[0,1/2)$.

@DHMO o sorry I initially miscalculated and get a+b as one of the points in the image. Yes that will work and it is not constant

@MartinSleziak let's add "bijective" to the claim

@MartinSleziak in many topology books , continuity of a map is defined to be the preimage of any open set is open. This definition makes me wonder, what's wrong with the preimage of an open set is closed. As your and DHMO's counterexample showed, those two functions are clearly continuous in the epsilon delta sense, thus I don't see what's the problem

@Secret I am not sure what is actually the question here, but it sounds a bit as if you thought that negation of open is closed.

@Secret different meanings of "preimage"

"preimage" means "inverse image" here

A function $f:X\to Y$ is continuous if for every open set $A \subseteq Y$, $f^{-1}(A)$ is also open.

Of course, continuous in epsilon-delta sense and continuous in premiage of open is open sense are equivalent. But you can only define continuity in epsilon-delta way if you have metric space. The definition with open sets is more general.

For metric spaces (in particular for functions $\mathbb R\to\mathbb R$) those two definitions are equivalent.

And, of course, there are many other equivalent ways how to express that a function is continuous.

what I don't understand is the following: let g be the preimage of f. Continuity is commonly defined to be g(open)=open g(closed)=closed. But why we cannot avoid "jumps" for a g(open)=x g(closed)=x where x is either neither, open (resp closed) or clopen?

clopen is both closed and open

Yeah, but my question still stands. Basically why there must be jumps when going from open to closed (or vise versa, or going into a neither set)?

(Cause if there are no jumps, then there will be no reason to rule out e.g g(closed)=open from the definition of continuity

@Secret be careful that sets are not doors

(I still can't think of an explanation)

I know, the cases g(open)=neither and related are also ruled out from the definition of continuity, so that means somehow such f defined on it must have at least one jump

But I have no idea how to prove that. It is also complicated now by that you and Martin have shown counterexample that suggests g(closed)=open can be continuous and not a constant map

@Secret it must be surjective

refer to my statement

$f^{-1}(A)$ here is called the "preimage"

But if that is the case, the bijection between (a,b) and [a,b) (they can biject into each other as both are uncountable sets with cardinality continuum) cannot be continuous?

indeed

then how to prove a jump must exist and where it is?

would you be happy if I consider $f:[a,b)\to(a,b)$ instead?

Continue

Let $f(a)=c$

I think that would be the critical point

but as I have said earlier, I have no idea how to prove.

what is "g(open)=open" supposed to mean?

@arctictern If $A$ is open, then $g(A)$ is open

where $g$ is a function

g is a function? I thought g is the preimage of f?

preimage means inverse image here

although "preimate of f" itself doesn't make sense

treat $g$ as $f^{-1}$

that's the inverse function, not the inverse image

and $f^{-1}$ only exists if $f$ is bijective first of all, which needs to be a stated condition if we're going to use it

...

I don't want to get too rigorous here

Let $f:X\to Y$

Let $B \subset Y$

The preimage of $B$ under $f$ is $\{x|f(x) \in B\}$

Alright?

yes

g is the preimage, that is the subset A of the domain that maps to a subset B of the codons in such that f(A)=B

Autocorrect error: codon=codomain

$f^{-1}(B)$ is the preimage of $B$ under $f$

yes

if you say "$g$ is the preimage" then you're declaring the letter $g$ refers to a set, in which case I have to ask what you mean by "g(open)=open"

In that case I think I should be writing g(open) is the preimage, not g itself

just write $f^{-1}(B)$

or f^-1(open)=open

Ok

Now my question is why the following must contain a jump thus excluded from the definition of continuity:

f^-1(open)=closed f^-1(open)=neither f^-1(open)=clopen f^-1(closed)=open f^-1(closed)=neither f^-1(closed)=clopen f-1(neither)=open f^-1(neither)=closed f^-1(neither)=clopen f^-1(neither)=neither

what do you mean by contain a jump?

i am guessing, a jump means that there are two points in a neighbourhood being mapped by f so that they end up located at disjoint neighbourhoods, one for each point?

Analogous to in the reals, the graph of the function has a break somewhere

That is a property that even continuous functions have.

For example: given any continuous monotonic increasing function on R, like f(x)=x, given any nbhd U of two points p,q there exist disjoint nbhds V,W of f(p),f(q).

Ok, why limit points must map to limit points and boundary points must map to boundary points? Why we cannot have boundary points mapped to non boundary points if f is continuous?

You can have boundary points go to non-boundary points. For example the inclusion map [0,1]->R.

Limit points map to limit points because we want continuity to preserve limits.

So if I define a map $f : (0,1) \to [0,1]$ where I take e.g. (0,0.3) to 0, (0.6,1) to 1 and (0.3,0.6) to (0.3,0.6), then f is not continuous because f does not preserve limits?

the graph of that obviously has one of those jump discontinuities you were talking about. yes it fails to preserve limits.

(although you have not finished defining the function. you said the doain of f was (0,1) but didn't say where to send 0.3 or 0.6)

the "preimages of open sets are open" definition is a generalization of preserving limits (with epsilon and delta balls being replaced by open sets)

I am typing that now...

where for the (0.3,0.6) -> (0.3,0.6) segment, for any x arbitrarily close to 0.6 from below, f(x)=1 Similarly for any y arbitarily close to 0.3 from above f(y)=0. Anything else k in the middle is defined as f(k)=k. So I should end up with a unbroken sigmoid like graph?

hence continuous?

huh?

are you confusing the words "below" and "above" with "left" and "right"?

or, I guess we do use the words below and above for numbers on the number line

just not a good idea to use them in the context of x-values when we also have a y-axis that goes up and down

how can you have a function (0.3,0.6)->(0.3,0.6) for which there are x values close to 0.6 from the left such that f(x)=1?

that makes no sense

(0.3,0.6)->(0.3,0.6) means that for all 0.3<x<0.6 we have 0.3<f(x)<0.6

so f(x) cannot equal 1 for any x in (0.3,0.6)

O sorry I mean [0.3,0.6] -> [0,1] where f(0.3)=0 and f(0.6)=1. Then f as a whole maps the open set (0,1) to the closed set [0,1] yet it has an unbroken graph hence continuous

"has an unbroken graph hence continuous" - well, a function can have a connected graph and still be continuous (standard example: the topologist's sine curve). but yes, if you want a function that is identically 0 on [0,0.3], identically 1 on [0.6,1], and restricts to (say) a linear map [0.3,0.6]->[0,1], then f will be continuous and the image of (0,1) will be [0,1].

woops, mean to say a function can have a connected graph and be not continuous

What is the topologist's sine curve?

google

wait what?? The graph will look something like this _/$\overline$. I don't see any jumps in the graph nor regions of wild oscillation like those in the topological sine?

@Secret what's your point?

Like I said, this piecewise linear function f is a continuous map (0,1)->[0,1].

@BalarkaSen I should have guessed

what does connected graph mean?

Ok sorry, I think I should be typing: Our f in the above case corresponds to f^-1(closed)=open. But the definition of continuity as covered by any textbooks is f^-1(open)=open or f^-1(closed)=closed. How do we rationalise that our case being continuous yet it is not covered by the definition of continuity?

it's a connected topological space, given the subspace topology from R^2

@Secret f^-1([0, 1]) is closed in (0, 1)

that means cannot be separated to disjoint closed sets?

@Secret f^-1(closed)=open does not contradict f^-1(open)=open

any topological space is closed as a subspace of itself.

@DHMO Right

@secret a riddle for you: separate [0,1] into two closed and one open set, all mutually disjoint

Nonempty closed sets anyway

@DHMO the graph is a subset of the cartesian product, we can put the product topology on the product and then the subspace topology on the graph and see if it's a connected space

@DHMO [0,c],(c,d), [d,1]

@Secret in other words, your $f$ does have the property that for all open subsets $U\subseteq[0,1]$, the preimage $f^{-1}(U)\subseteq(0,1)$ is open. it does satisfy the condition for being continuous.

@secret now in my question replace "two" bye "one"

by*

@DHMO empty set and [0,1]. The empty set is clopen while [0,1] is closed

Hello @DHMO

i got a combinatoric question

@secret actually you can just union the two closed sets

@KasmirKhaan sorry ask others

O yes, finite union of closed sets are closed

Okay thanks anyway :)

@KasmirKhaan draw a venn diagram, put variables in each spot, set up a system of equations

@secret find |{|x|:x is an interval}|

@DHMO How to find the sum of this series $$1.2.3.4 + 2.3.4.5 + 3.4.5.6+ ....$$ upto n terms ? Any idea?

@DHMO its countable if the intervals are pairwise disjoint

@Secret you misunderstood my question

@anonymous it must be a quintic polynomial

@DHMO |{|x|:x is an interval}| is the cardinality of the set of all lengths of intervals?

@DHMO That is too huge a method and I don't remember the summation of 4 degree terms..

There should be some easy way out

@arctic yes but I am asking (testing) Secret

@DHMO just seems like a weird question to ask

@arctictern Looks promising. How do you sum them up ?

@anonymous hockey-stick identity

@arctictern Oh oh ! Great..thanks!

mmhmm

smart

didn't know there was a name for that

yup

I found another method probably $\frac{1}{5}(1.2.3.4(5-0) + 2.3.4.5(6-1) + .....)$....seems okay?

The terms except the last and the first all cancel out

nice

essentially equivalent to the telescoping proof of hockey stick

@DHMO continuumly many. The only two free parameters of any interval x of any length is the end points or the bounds (infrima and suprema). For each of these there are continuum many choices therefore it is the same problem as the cardinality of the set $\mathbb{R}^2$ which has cardinality continuum

wrong

you are still misunderstanding

But aren't all intervals are of the form (a,b) [a,b) (a,b] [a,b] regardless of what length they are, and they will be uniquely specified by a and b?

yes

@arctictern oh, not the length

the cardinality

and each of these will have cardinality continuum since they all biject with the whole real line?

not all

?

that is what i am asking you

There are 355 students in a college who have taken a course in calculus, 212 who have take a course in discrete mathematics, and 121 who have taken a course in both calculus and discrete mathematics. How many students at this college have taken a course in either calculus or discrete mathematics?

What do they mean by either or last statemnent

@KasmirKhaan xor

xor ?

I still dont understand

(A and not B) or (not A and B)

hmm

Can give me a simple example ?

ahh

you are either male or female

they want the number of student who took calculus ( just calculus)

and just discrete math

but not both

Right?

yes

thanks ! :)

355 +212 - 121

no

@DHMO How, if they don't biject with the reals, that means there are intervals in the reals that are countable or larger than the reals (but that will make the reals themselves have cardinality > continuum). A singleton is not an interval

we include those who took calc and those who took discrete

then we delete those who took both right?

because they got counted twice

@kasmir subtract twice

@secret why is a singleton not an interval?

hmm

@DHMO according to Wikipedia, an interval is a set containing elements between two numbers x and y. Also, even if we do count singletons as intervals, there are still only continuumly many of them and the cardinality of the set of all intervals will be $|\mathbb{R}^3|=\mathfrak{c}$

if |x| is the cardinality of an interval x, then {|x|:x is an interval} is not the set of all intervals

the above

O I was screwed. That thing is a proper class of cardinals up to continuum, not a set. In that case it has countable number of members (the cardinal numbers) right?

what...

wrong

what can the cardinality of an interval be, secret?

list the possibilities

(I suppose it depends on if you count the empty set (a,a) in this)

0 (empty), 1 (singleton [a,a]), continuum (The rest). So... cardinality is 3...?

bingo

@DHMO I want to test my knowledge on finding limit points, closed set and other basic topology notions. Can you give me a topology that is "very nonintuitve"?

My answer is 64*2+1

is that correct?

@secret Look at the topology on $\Bbb R$ with basis $\{[a,b),a,b\in\Bbb R\}$. Is it comparable to the euclidean? Is $[0,1]$ compact? Does the sequence 1/n converge? Does the sequence 1-1/n converge?

Consider the product topology of this space with itself, what's the subspace topology induced on the diagonal of the product?

Anti-diagonal* sorry

$\{(x,-x):x\in\Bbb R\}$ that one

@Alessandro Any more interesting algebraic/differential topology in a while?

I'm going to have another AT lecture on Monday, I looked at the professor's notes, we should have some more stuff about CW-complexes and then start talking about fundamental groups

@secret why is "x is open in this topology iff x is closed in the usual topology" not a topology on R?

@AlessandroCodenotti is {x|x<3} open?

in which topology?

in tu topologia

yes, you can write it as a union of basis elements

cuales?

$\bigcup\limits_{n<3} [n,n+1)$

oh, countable union

and I thought you would have to declare it separately

that raises a question

can any basis generate a topology?

the conditions in the definition of basis are there to ensure that it does generate a topology

what are the conditions?

If you have a topology a basis is just a collection of open sets such that all open sets of the topology can be written as union of those in the basis. If you go the other way around you can pick a set $X$, a family of subsets of $X$ such that 1)they cover $X$ and 2)if $B_1$ and $B_2$ are in the basis then for every $x\inB_1\cap B_2$ there is a $B_3$ containing $x$ and contained in the intersection

If a family of subsets satisfies 1 and 2 then it generates a unique topology (and is a basis for it)

@AlessandroCodenotti consider the basis generated by the singletons...

I mean, the topology generated by the singletons

That's the discrete

is [0,1] open...

(1/n, 1], take union

Or rather [0, 1-1/n)

that isnt our bases

aren't we working on half-open intervals as base elements

i'm not following the convo

is uncountable union allowed?

arbitrary union of open sets is open