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12:56 AM
@XanderHenderson That is quite SOP, no?
 
Bob
1:15 AM
Hello Ted
 
@TedShifrin Not here, no.
The college business office is very nervous about having anyone spend their own money and then ask for reimbursement (though, in my opinion, this is the way that things should be done).
Oh, wait.
Maybe? I've never had this problem in the past.
E.g. when I went to Phoenix last year, they ran the company card, despite the fact that I did not physically possess it.
As that was the card used to make the hotel reservation in the first place.
 
Can 4-dimensional object (e.g. Klein bottle) be visualized using 3d video? Considering time as 1-dimension.
 
@onepotatotwopotato Yes?
 
the klein bottle is 2-dimensional
 
True... I don't know what to say about that maybe embedding?
 
1:57 AM
No, potato.
@Xander I always had to produce the credit card used.
 
2:27 AM
@TedShifrin Why not? there is a parametric equation of klein bottle into $\Bbb R^4$ math.stackexchange.com/q/759264/668308
 
2:57 AM
Of course. But the Klein bottle is 2-dimensional. Learn what dimension of — say — a manifold is. Nothing to do with where you might or might not embed it.
 
Yes that '4-dimensional' is misleading. Klein bottle is 2-dimensional manifold but cannot be globally embedded in $\Bbb R^3$ so can't be visualized (only as a immersion) but it can be embedded in $\Bbb R^4$. So I wondered if one can visualize the globally embedded Klein bottle using video or animation considering time as additional 1 dimension.
 
Yes, or color. That is discussed in many books.
 
3:41 AM
Oh I didn't know it's in wikipedia en.wikipedia.org/wiki/…
 
4:34 AM
I got $3$ different answers by using $3$ different methods while solving, $$\int \frac{1}{\sin^2x \cos^2x}\ dx$$
Just want to know if there is any problem in any of the method.

Method 1:
$$\int \frac{1}{\sin^2x \cos^2x}\ dx\\ =\int \frac{\sin^2x + \cos^2x}{\sin^2x \cos^2x}\ dx\\ = \int \sec^2x + \csc^2x\ dx \\= \tan x - \cot x + C $$

Method 2:
$$\int \dfrac{1}{\sin^2x \cos^2x}\ dx\\=\int \dfrac{\sec^4x}{\tan^2x}\ dx \\ = \int \dfrac{(\tan^2x + 1)\sec^2x}{\tan^2x}\ dx\\ =\int \dfrac{(t^2 + 1)}{t^2}\ dt\quad {\text {where }}t = \tan(x)\\= \tan(x) - \frac{1}{\tan(x)} + C $$
I wanted to post it as a question but am not allowed to post more than 1 question in a week.
1st and 2nd are just identical... wanted to know about 3rd.
 
So figure out why the third is the same as the first two.
With integrals you can have very different-looking results and you have to use your brains to understand what's going on.
Here only half a brain is required.
 
lucky for me
 
@TedShifrin Well well well.. double angle formula i guess?
 
Indeed @leslie. Indeed @HelpMe
 
@TedShifrin Thanks a lot.
 
4:45 AM
But be prepared that often integration by parts and integration by substitution will give wildly different results, and you have to figure out that they differ by a constant.
 
@TedShifrin Yes. Sometimes it makes me doubtful if I'm going wrong way.
 
Be careful to check your steps and don't be doubtful. And remember that with antidifferentiation, you can always check your answer(s) by differentiating and doing — as needed — some algebra.
 
trig identities are spooky to people well before they get to calculus because they are examples of different looking expressions nevertheless being the same. 'the same, plus a constant' is even trickier. particularly when at least some famous trig identities involve a constant looking like something like sin^2 x + cos^2 x. super helpful to check by differentiation.
people always forget they can do that. it's like a law of the universe that people need to be reminded that they can do that.
 
Thanks to both of you for your guidance. I appreciate your time.
 
Do $\int \frac y{\sqrt{1-y}}dy$ or $\int t^3\sqrt{1-t^2}dt$ in two different ways (parts and substitution) and you'll see what I mean.
 
5:01 AM
indefinite integration seems a recipe for mistakes
 
You prefer infinite regression?
 
definite regression
 
 
4 hours later…
9:15 AM
@TedShifrin Ohh yes, I tried first one. Using integration by parts gave $-2y\sqrt{1-y} - \dfrac{4(1-y)^{3/2}}{3} +C $ while integration by substitution gave $-2\sqrt{1-y} + \dfrac{2(1-y)^{3/2}}{3} + C$. That seems like a magic xD Plotting these antiderivatives reveals that both the functions are exactly same. (Don't even differ by any constant)
I would like to simplify them and prove that they are equal to each other but that's really a hard nut to crack.
 
 
2 hours later…
11:31 AM
dense set intersect every closed uncountable set. Is it true? I believe yes. Attempt: Let $F\not \subset D$.Since $F$ is closed uncountable set in a lindelof space $F$ , $\exists \alpha\in F$ limit point of $F$ such that $\alpha\notin D$ . From here I have to show that $\alpha\notin \overline{D}$ to conclude that $D$ is not dense.
 
11:47 AM
The complement of the Cantor Set is dense, is it not?
 
11:59 AM
@robjohn is the place where you owe your 2 bananas the same as the place where I bought my 3 bananas?
 
Why should it matter? I carry my -2 bananas in a paper bag.
 
And I carry mine in a paper bag also.
 
What is with you people and bananas? Bananas are gross. I'll take those $-2$ bananas off of you any day of the week, @robjohn.
 
Negative bananas have a certain appeal.
Or perhaps, they are only a peel.
 
@robjohn Yes. As The Cantor set is n.w.dense , it's exterior (same as compliment) is dense. Now suppose $D$ is saturated non measurable set. Does $D$ intersect every closed uncountable set?
@XanderHenderson What are some interesting examples of subset of $\Bbb{R}$ which intersect every closed uncountable set ( except Bernstein set)?
 
12:14 PM
wait, what is a saturated set? Nothing was mentioned about non-measurability before. I was just replying to the statement "dense set intersect every closed uncountable set"
 
@SouravGhosh Why are you asking me?
I am not a point-set topologist (do such animals even exist any more?).
 
@robjohn Yes. You are right. This is my second question.
 
as I have no idea what a saturated set is, I am not the right person to ask.
 
@robjohn Thanks.
@XanderHenderson Sorry.
 
Nothing to be sorry about---I just do know why I would have any special insight.
 
12:37 PM
if $p_1\subseteq p_2$, where both prime ideals in $R$, then $R_{p_1}\subseteq R_{p_2}$?
where $R_{p_i}$ is localization of $R$ at $p_i$
 
12:48 PM
@XanderHenderson would that be the rug in the Hilbert Hotel?
 
@robjohn Yeah, but this is real problematic, as the Hotel can only hold a countable number of things.
So the rug ends up all lumpy, trying to cover up the reals.
 
Yeah, I keep tripping when I go there.
 
1:01 PM
Oct 25, 2011 at 9:46, by robjohn
You can check out, but you can never leave.
 
May I know websites youtube channels and books to learn number theory!
even any website providing short notes for number theory would also help
 
in integral domains, a prime ideal $p$ in $R$ has height 0 iff $p=0$?
 
1:38 PM
@robjohn usually it means something consisting of equivalence classes
So if $q$ is a quotient map, $A$ is saturated iff $A = q^{-1}q(A) = \bigcup_{a\in A} [a]$
@XanderHenderson kind of? People usually specialize in something but have really good knowledge of point-set topology. I'd say Alessandro Codenotti or Henno Brandsma on this site (I hope I didn't butcher anyones surname sorry if I did), are examples of such people.
Not sure what saturated means in this context though
@SouravGhosh I don't think there are any other interesting examples? Those depend on axiom of choice in the first place.
 
@Jakobian That is my point. There are folk doing, for example, low-dimensional topology, or knot theory, or whatnot who are really good at point-set topology, but I feel like most of the material that falls under the umbrella of "point-set topology" is pretty resolved, and that there isn't a lot of active research in that area any longer.
M'kay... I am off to Convocation. After more than two years at this institution, I will be meeting some of my coworkers in person for the very first time today. :/
 
2:07 PM
@monoidaltransform yes, height 0 = minimal prime ideal and in an integral domain, there is only one minimal prime ideal (which is minimal amongst all ideals, even) and that is 0
@monoidaltransform no, first think about localizations at arbitrary multiplicative subsets
that is, if $S\subseteq T$ are multiplicative subsets of $R$, there is only one reasonable map between $S^{-1}R$ and $T^{-1}R$, in which direction does it go?
 
@Jakobian here is the full question compiled with all details math.stackexchange.com/q/4512540/977780
0
Q: Does a saturated non measurable dense subset of the Real line intersect every closed uncountable sets?

Sourav GhoshLet $D\subset \Bbb{R}$ be a dense set. Question: Does $D$ intersect every closed uncountable sets? The answer is negative. As the Cantor set $\mathcal{C}$ is n.w.dense , it's exterior $\Bbb{R}\setminus \overline{\mathcal{C}}=\Bbb{R}\setminus {\mathcal{C}}$ is dense in $\Bbb{R}$ but it doesn't i...

@Jakobian Saturated means inner measure of the set and inner measure of it's complement both are 0.For an example Bernstein set (one of my favorite toy)
 
2:40 PM
@Jakobian I have seen many answers of Henno and found them to be very helpful and out of the box. Their answers have helped me a lot. :)
 
2:53 PM
Henno helped me a lot too (both answers to other people questions and mine own), but not always. Sometimes they've treated my question as something obvious and not worth elaborating on
I had different opinion in those kind of moments
 
3:13 PM
@XanderHenderson most of research in point-set topology nowadays falls under set theoretic topology or continuum theory, but there's still plenty happening in the field. In the last 10 years or so we've had the classification of homogeneous planar continua, the classification of hereditarily equivalent planar continua and the first ZFC construction of a countably compact group with no convergent subsequences to cite a few very big results
 
3:30 PM
Re point-set expertise: There's Brian M. Scott, whose research was in point-set.
 
@TedShifrin oh yes!
His answers have also helped me a lot.
I remember the way he proved that every open set in R can be written as a countable disjoint union of open intervals, using equivalence relation :).
That is so out of the box.
 
For some reason, I don't remember how to get to the page listing everyone's rep, but I think he may be at the top of that list. Not saying that means much ...
 
Ted: click users
 
Where??!!
This is ironic. I know how to get the page on my iPad, but I'm not finding it from my desktop!
 
Ted: on the page where you see questions.
On the left column if you're on desktop.
 
3:41 PM
Nope. Not there.
OK, I found it. Had to click on the triple bar at the top.
OK, I was right. Brian M. Scott does have the highest rep.
 
continuum theory is cool when you do it up to homotopy
 
rolls $6^\pi + e^e$ eyes
 
one-dimensional Peano continua are determined up to homotopy by their fundamental group in a way that's almost a categorical equivalence (except for some basepoint business that's slightly less trivial than you'd want it to be)
 
OK, time for me to go eat breakfast. Thanks, Koro.
 
:)
 
3:56 PM
I was using Geogebra and I produced 4 3D graphs and I wanted to know how to get their intersection?
 
I think that for every order topology on X, the following statement is true: cl(a, b) =[a, b] for any distinct a and b in X.
By definition, cl (a, b)= (a, b) U (a, b)'. Suppose that x is not in (a,b) but is a limit point of (a, b), then it can be shown that x is either a or b.
So cl(a, b) is a subset of [a, b].
And the reverse inclusion is obvious.
Is this correct?
 
it's obvious that cl(a,b) is a subset of [a,b], because [a,b] is closed and contains (a,b). what you first talk about is the other inclusion, but of course "it can be shown" is hiding the core steps of the argument.
 
4:33 PM
"It can be shown" is just as bad as "Obviously, ..."
 
Let $t\in [a,b]$. If $t=a$, then t is a limit point of (a,b). Because any open set that contains t also contains a basis which looks like one of the following: 1) [t, 'something in (a,b)], 2) (p, 'something in (a,b)) for some p<a, 3) (p, something in (a,b)] for some p<a
which brings the question what if (a,b) has only finite elements in it.
 
4:53 PM
What if we have $\{0\}\cup (1,2]\subset\Bbb R$ in the subspace topology?
What is the order topology?
 
you mean in your example? Or definition of the order topology?
 
My example.
Interesting? $(0,2)=(1,2)$ ….
 
It will be generated by [0, t), (t', 2], (a,b), where a,b, t,t' are in the set.
 
But 1/2 has no meaning.
 
yeah.
@TedShifrin 1 also has no meaning?
 
5:02 PM
Right.
So what space is this homeomorphic to?
 
I don't yet know much about homeomorphism(s).
My question is from section 17 of Munkres.
homeomorphism is in the next section.
I just know the definition and some examples of homeomorphism.
 
What space does it feel like we’re talking about, based on the comments?
Compare $/{0\}\cup (1,2]$ and $\{0\}\cup [1,2]$.
 
@TedShifrin I can find a bijection between the two. But not sure about continuity of the bijection.
 
Between which two?
 
between (1,2] and [1,2].
I also observe that {0} is an open set in subspace topology but in the order topology {0} is not open.
because any basis element containing 0 must contain an element from (1,2].
 
5:17 PM
Since we’re talking about the order topology, you probably want your bijection to be increasing or decreasing. So My question was what our space $\{0\}\cup (1,2]$ might be.
Correct.
 
5:28 PM
I'll think about that.
 
6:25 PM
@Koro In order topology on $X$ the basic open sets are of the form $(a, b), [a_0, b), (a, b_0]$ where $a_0, b_0$ are least and greatest elements of $X$ respt (if exists) . For (a, b) two ends points are also limit points as any basic open set containing a, b must contain a point of (a, b) other than a, b respectively.
 
6:42 PM
What's a good way to make a bijection between $\Bbb{R}^1$ and $\Bbb{R}^2$?
Correction: $\Bbb{N}^1$ to $\Bbb{N}^2$
 
For the following example, I'm a little confused about Axler's solution for b)
How are all the eigenvectors of T in U as well? I understand that if I can show this, then U and W will not be direct sums and thus there is a contradiction. I also can see how 0 is the only eigenvalue
 
The only way I've thought of thus far works fine by simply subdividing $\Bbb{N}^1$ using $1 = \sum_{n=1}^{\infty} 2^{-n}$, however, when we apply this to a finite subset, the obvious problem arises that less and less space remains available proportional to a curve $O(-2^x)$.
 
7:00 PM
@igameonamac Since you know that eigenvalue of T is only 0, you can find eigenvectors (x,y)$\ne (0,0)$ by solving T(x,y)=0(x,y)=(0,0).
This gives: (y,0)=(0,0).
@SouravGhosh I know the definition. I used the definition earlier in the comments here.
 
7:43 PM
sometimes I wonder how much power tripping mods have cost the company in money
 
8:06 PM
In cofinite topology on R, the sequence (1/n) converges to every real number.
 
@Koro Fun fact: if you quotient the rationals by the "same denominator (in simplest form)" relation, you get $\Bbb N$ with the cofinite topology
 
 
2 hours later…
9:59 PM
@AMDG $(i,j)\mapsto \sum_{k=1}^{i+j-1}k + j$
 
 
2 hours later…
11:34 PM
@Astyx Does this hold as i or j increase or decrease?
 
For the proof above, how does Axler get the last inequality?
Is it from the fact that every subspace of V is a part of a direct sum of V?
 
Oh wait, I think I'm just being slow.
He gave me a bijection between N and N^2, so it doesn't matter the size of i or j.
Am retard
 
if U is a subspace of V, then dim(U)<=dim(V)
 
ty LOL i cant believe i missed that
the sum of subspaces was still a subspace
thank you!
 

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