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Hey
What consequences has a function that satisfied the Lipschitz condition?
 
In mathematics, specifically real analysis and functional analysis, the Kirszbraun theorem states that if U is a subset of some Hilbert space H1, and H2 is another Hilbert space, and f : U → H2is a Lipschitz-continuous map, then there is a Lipschitz-continuous map F: H1 → H2that extends f and has the same Lipschitz constant as f. Note that this result in particular applies to Euclidean spaces En and Em, and it was in this form that Kirszbraun originally formulated and proved the theorem. The version for Hilbert spaces can for example be found in (Schwartz 1969, p. 21). If H1 is a separable...
 
Is the following true if $V$ is not finite dimensional?
Suppose $T \in \mathcal{L}(V)$ and $U$ is a subspace of $V$. Prove that $U$ is invariant under $T$ if and only if $U^\perp$ is invariant under $T^*$.
Forward direction is trivial and doesn't use finite dimensionality of $V$, but for the backward direction I assumed $(U^\perp)^\perp = U$.
And I'm wondering if there is anyway around this.
 
@user193319 what is $V$?
 
Well, I assumed $V$ was a finite dimensional vector space.
 
12:15 AM
what is $\mathcal L(V)$?
 
Linear operators acting on $V$
 
I interpreted your question very much differently...
ok
what is $U^\perp$?
 
orthogonal complement: all $v \in V$ such that $\langle v, u \rangle = 0$ for all $u \in U$.
 
what is $\langle,\rangle$?
 
Some inner product on $V$.
 
12:18 AM
you didn't assume that.
 
I'm not sure I follow.
 
you didn't assume that you have an inner product on $V$
 
Well, in the problem statement it isn't mentioned (it's assumed), but I definitely used it in my proof of the forward direction.
 
your statement is true in Hilbert spaces, of which any finite dimensional space can be one
 
Okay, so the backward implication is true for infinite dimensional vector spaces which aren't complete with respect to the norm induced by the inner-product?
 
12:23 AM
I don't know about inner product spaces in general
 
 
3 hours later…
3:14 AM
Hey someone is there?
 
Mog
4:04 AM
1
Q: Lebesgue Decomposition of Lebesgue-Stieltjes Measure with $F(x)=\lvert x\rvert\lfloor x\rfloor$

MogI have a similar question as here. Lesbesgue-Stieltjes measure $\mu_F$ with respect to the function $F=\lvert x\rvert\lfloor x\rfloor$. i.e. $\mu_F([a,b))=F(b)-F(a)$. I'm trying to find the Lebesgue-Radon-Nikodym decomposition of $\mu_F$ with respect to the Lebesgue measure $m$. Now, I know thi...

 
 
2 hours later…
6:07 AM
0
Q: What does the function $f(x,y)$ reduce to?

The Great DuckWhat does the function $f(x,y)$ reduce to? $$f(x,y) = \lim_{n \to \infty} \sum_{i = 0}^{n} (((x \bmod 2^{i-1})-(x \bmod 2^i)) \cdot ((y \bmod 2^{i-1})-(y \bmod 2^i)) \cdot \frac {1}{2^i})$$

 
6:59 AM
I have got 500 reputation what is the speciality of it
 
7:18 AM
@AkashRoy wdym?
 
7:56 AM
I mean , do I have any privilege if i earn 500 reputation on this site
 
8:20 AM
Is the number of ways to put 12 people into at most 4 groups = 1^12+2^12+3^12+4^12?
 
 
2 hours later…
Mog
10:18 AM
1
Q: Questions on differentiability of $F(x)=\int_{0}^x f(t)dt$, where $f$ is locally integrable

MogI'm going through Folland's Real Analysis, and has come to an impasse trying to come up with a couple examples in the topic of differentiability. On $(\mathbb R, \mathcal B_{\mathbb R}, m)$, for locally integrable $f$, define $F(x)=\int_{0}^x f(t)dt$, $A_rf(x)=\frac{1}{2r}\int_{x-r}^{x+r}f(t)...

could use some help!
 
 
3 hours later…
user131753
1:44 PM
-1
Q: Reference request for bounds of $n$-th composite

user 170039Motivation I will briefly elaborate here my motivations for asking the question. If you are not interested in it then please go to the questions. Recently during trying to understand and prove the Second Hardy-Littlewood Conjecture, I came to notice that the following slightly weak version of ...

 
Do anyone tell me why we are considering the ring $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$ in Golden ratio base number system instead of field?
Don't we need closure under division?
 
 
2 hours later…
3:26 PM
@AlessandroCodenotti hi
 
hi
 
@Astyx hi
@AlessandroCodenotti i showed that the map $t\to (t^2,t^3)$ is homeomorphism. i am asked to explain why it is not isomorphism $\Bbb A ^1 \to Y$ where $Y = Z(y^2-x^3)$
(3.2 in Hartshorne )
 
Look at the coordinate rings
 
i did. but what's $k[Y]$ ?
 
3:34 PM
Hey all, I need to prove:

Show that a ring is commutative if it has the property that $ab = ca$ implies $b = c$ when a is not equal to 0.
 
i need to show that $k[x] $ is not isomorphic to $k[Y]$ but im not sure what that is @AlessandroCodenotti
 
I did it by doing $b=c\implies ab=ac$. Since it is given that $ab=ca$, I got $ac=ca$ thus commutative
is it incorrect?
 
@Liad you should know
 
its $k[x,y]/I(Z(..)) $ @AlessandroCodenotti
 
Yup, so what's $I(V(y^2-x^3))$?
 
3:36 PM
i want to say $(y^2-x^3)$
its the radical of that ideal
 
Is that ideal prime?
 
that what i was going to say ^^
its prime because we are in an integral domain? @AlessandroCodenotti
 
@GaurangTandon It is incorrect
$a, b$ and $c$ are not specific numbers here
 
@Liad how are you using that we are in an integral domain?
 
$k[x,y] / (y^2-x^3)$ is a sub ring of $k[x,y]$
 
3:43 PM
Is it?
 
Ah? it is isomorphic to a subring of $k[x,y]$
 
@Astyx yah but we're given that if ab=ca then b=c. so doesn't it hold for any a,b,c? sorry but i do not understand where in my proof i have used specific a,b,c
 
Hmm I'm not seeing that and I still don't see how you're saying that this ideal is prime
 
Mog
1
Q: Questions on differentiability of $F(x)=\int_{0}^x f(t)dt$, where $f$ is locally integrable

MogI'm going through Folland's Real Analysis, and has come to an impasse trying to come up with a couple examples in the topic of differentiability. On $(\mathbb R, \mathcal B_{\mathbb R}, m)$, for locally integrable $f$, define $F(x)=\int_{0}^x f(t)dt$, $A_rf(x)=\frac{1}{2r}\int_{x-r}^{x+r}f(t)...

 
ok, what approach you see to show it is prime? @AlessandroCodenotti
 
Mog
3:54 PM
Could use some help!
 
Yes if ab-ca then a=b for any a, b and c
But it not given that ab = ca
 
What do you know about the polynomial $y^2-x^3$?
 
thought of proving its irreducible if that is where you going
 
@Astyx sorry but i'm not sure how i'm reading the questino...it does seem to be given that ab=ca. what am i missing?
 
No you have a ring
For which, if you find a, b and c such that ab = ca, then you can conclude b=c
And you need to prove for any a and b, that ab=ba
 
4:04 PM
i think my approach is fine.. @AlessandroCodenotti it is a subring of a domain
 
Why does your argument fail with (xy)?
 
$k[x,y] \to k[z]$ taking $x\to z^2$ and $y\to z^3$..
the kernal is $(y^2-x^3)$
and $k[z]$ is a domain..
@AlessandroCodenotti i must dont see something if you dont agree
ah i mixed that 2 and the 3
cant edit now
 
$k[x,y]/(y^2-x^3)$ is isomorphic to $k[z^3,z^2]$, I agree
Ok so know you know both coordinate rings
 
if phi is homomorphism from ring R to R' then phi(1) = 1' if R' is an integral domain .How to prove this ?
 
@Astyx oh thanks so much i understood the fault!
 
4:15 PM
if phi is homomorphism from ring R to R' then phi(1) = 1' if R' is an integral domain .How to prove this ?
for proving this we probably need to prove that phi is onto
 
@AlessandroCodenotti ok so if we assume it is isomorphic we get $k[x] $ isomorphic to $k[x,y]/(y^2-x^3)$
 
4:26 PM
Hi! If the formula is not true, then there's a seq $s$ in $\Sigma$ st $s$ satisfies $(\forall x_i)\mathscr{B}$ and $s$ doesn't satisfy $(\exists x_i)\mathscr{B}$ Here I want to ask if $(\exists x_i)\mathscr{B}$ and $\neg(\forall x_i)\neg\mathscr{B}$ are equivalent, since the lack of parantheses makes it hard to see this for me.
 
@GaurangTandon Glad I could help
 
How to prove that $Ф(1) = 1'$ if $R'$ is an integral domain? Since $R'$ is an integral domain , $Ф(b) = Ф(b.1) = Ф(b).Ф(1) = Ф(1).Ф(b)$ .But i can prove this only for those $r'∈R'$ for which there $∃r∈R$ such that $Ф(r)=r'$.
 
@neraj this is not true, the map $\phi(r)=0$ for all $r \in R$ is a ring homomorphism
 
@MatheinBoulomenos mean phi maps all r to 0 ?
 
yes
that's a possibility for a ring homomorphism
 
4:34 PM
but not always true
did you got my question
 
no, it's not always true
 
According to herstein my question is right
so how do i prove it ?
 
but if it's the case then $\phi(1)=1'$ is wrong
because $0\neq 1'$
 
Does that means herstein is wrong ?
 
that's possible
if you assume that $\phi$ does not map all $r$ to $0$, then you can prove it. In that case, you get $\phi(1)=\phi(1 \cdot 1)= \phi(1)^2$, so $\phi(1)$ is idempotent
the only idempotents in an integral domain are $1$ and $0$
 
@Liad $k[x^2,x^3], not $k[x]$
 
@neraj yeah, there is small mistake in Herstein
 
what's that ?
the example you gave ?
 
Ok .Thanks
@MatheinBoulomenos
 
4:49 PM
np
 
Hi @Mathei
 
Hi @Alessandro
 
What exactly does "a scheme over $k$" mean? I guess a scheme $X$ such that $\mathcal O_X(X)$ is a $k$-algebra?
 
it means a scheme $X$ together with a map of schemes $X \to \mathrm{Spec}(k)$
 
It's nice that you can algebraically kind of completely untangle the varying degrees of content in the statement "picking an element fiberwise in a projection map": If $\mathcal{F}$ is a presheaf on $X$, $\widehat{\mathcal{F}}$ be the sheafification and $\prod \mathcal{F}$ be the sheaf which takes value in product of the stalks above a given open set.
Then a section of $\mathcal{F}$ is kind of the most rigid thing, a section of $\widehat{\mathcal{F}}$ is inverse of the local homeomorphism that's the projection from the leafspace of $\mathcal{F}$, a section of $\prod \mathcal{F}$ is choosing a point from each stalk possibly "discontinuously"
There is a sequence of inclusions $\mathcal{F} \to \widehat{\mathcal{F}} \to \prod \mathcal{F}$, of course.
 
Pig
4:56 PM
hi all
 
Nevermind I was going in the wrong direction
 
hi @Mathein, a @Balarka, @Piggy, demonic @Alessandro
 
Hi @Balarka @Pig @Ted
 
Hi all
 
Pig
4:58 PM
Hi @TedShifrin and @MatheinBoulomenos
 
Hi @TedShifrin
 
Pig
also @AlessandroCodenotti your statement is strictly weaker (even if you fix the direction) i think
 
hi @neraj
 
Hello chat
 
@Alessandro it's the same as giving a $k$-algebra structure to $\mathcal{O}_X(U)$ for all $U\subset X$ open, such that all restrictions are $k$-linear
 
4:59 PM
I just solved an exercise asking to prove that it's equivalent (with the directions fixed) so I sure hope it is @Pig!
 
@Mathein: According to my definition, the zero map is not a ring homomorphism. Following Artin, I say it must map identity to identity. I don't remember what Herstein does.
salut @Astyx
 
Salut
 
My rings are commutative unital and my ring homomorphisms preserve identity
 
Pig
@AlessandroCodenotti what Mathein said is correct; not entirely sure if you can bootstrap stalk level information back up to open sets
 
@BalarkaSen Is this a riddle ?
 
5:00 PM
@TedShifrin Herstein doesn't require that the identity gets sent to the identity (if we do, the question is trivial)
 
Once again, a @Balarka is on the side of sanity.
 
@TedShifrin see the accepted answer of math.stackexchange.com/questions/3010551/…
 
I require unital ring homomorphisms as well. As for commutative, depends on the context
 
Right. It seems to me that problem is right if Herstein rules out the 0 map. Go back to the definition. Neraj, what's the definition of a ring homomorphism?
Word for word.
 
@MatheinBoulomenos I'm confused. On the Spec level there's only one map $f:\operatorname{Spec}(X)\to\operatorname{Spec}K$ so we only need a morphism of sheaves $\mathcal O_X\to f_\ast\mathcal O_K$, right?
 
5:03 PM
@TedShifrin second part of lemma probably rules out zero mapping ?
 
No, the zero mapping certainly satisfies that.
Please find the page where he defines ring homomorphism.
 
@AlessandroCodenotti $k[x]$ if we assume by contradiction that $\Bbb A ^1$ is isomorphic to $Z(y^2-x^3)$
 
Pig
$O_K \to f_* O_X$ @AlessandroCodenotti
 
Oh, sure @Liad
 
Pig
which is equivalent to $K \to O_X(X)$
 
5:05 PM
Yeah I just realized I wrote it the wrong way around again
@Pig Which is just another way of saying that we have a $k$-algebra structure, got it
 
oh yeah what I wrote is still correct, but to complicated @Alessandro
 
Pig
yea, but on the global section
what you mentioned earlier is equipping each stalk (well to be more precise, each residue field) a $K$-structure
lifting that back up feels uncertain :/
 
@Liad: I'm confused. That curve is rational.
 
you get a $K$-algebra structure on every section as the composition $K \to \mathcal{O}_X(X) \to \mathcal{O}_X(U)$
 
Sure, that's free :D
@Pig Hmmm I'll double check
 
5:07 PM
A mapping¢ from the ring R into the ring R' is said to be a
homomorphism if
I. ¢(a + b) = ¢(a) + ¢(b),
2. ¢(ab) = ¢(a)¢(b),
for all a, b belonging to R. @TedShifrin which point contradicts zero mapping ?
Exact same copy from herstein
 
I find schemes to be extremely counterintuitive objects to think about so far
 
@MatheinBoulomenos Do you find any contradiction that we cannot map to zero in a ring homomorphism
 
Nope. If that's the definition, then the zero map is a homomorphism and the exercise is in fact wrong. This is a different Herstein book than I used as a college student, but I don't remember anything wrong back then.
 
Sheafification is universal, right? As in, $A \to B$ is a morphism of presheaves over $X$ with $B$ a sheaf, it factors through $A \to \hat{A}$
 
@TedShifrin which exercise is wrong ?
 
5:10 PM
@BalarkaSen yeah, uniquely
 
Never mind. What Mathein says is correct if the zero map can be a ring homomorphism.
I really don't like that. That means the image of a ring homomorphism needn't be a ring (probably without identity). So this is stooopid.
 
Herstein's conventions are generally dumb
Remember he writes function composition backwards?
 
I'm not a big fan. Yeah.
 
Pig
i never know why we care about rng
 
I've seen rings without identity come up naturally in representation theory
 
5:12 PM
@BalarkaSen lol
 
Pig
but that's presumably why herstein formulates things that way
 
Actually, note that lemma says homomorphism into $R'$. What does "into" mean there?
 
or even measure theory, think about $L^1(G)$ where $G$ is locally compact, but not compact
 
Pig
@MatheinBoulomenos any examples?
oh that's a fair one
 
To me, "into" means injection. Otherwise I say "to."
 
5:13 PM
@BalarkaSen Sheafification is left adjoint to the forgetful functor Sheaves->Presheaves as @Mathei thaught me
 
Pig
but there's still some "approximate" identity anyway
 
tbh i think most of the time any algebraic constructions you see will "arise naturally in representation theory" lmao
 
@Ted I hate that book, actually. His exercises are hard but uninsightful
it feels like trick math
 
Neraj is using a different Herstein book, @Balarka. Yeah, it's too much about trickery, whereas Artin shows you mathematics.
 
oh ok different herstein.
 
5:15 PM
Topics in Algebra herstein
name of my book
 
@AlessandroCodenotti I think that's equivalent to what I asked, just parsed in fancy language
 
Oh, that is the book ... the type face looks like his newer, easier book.
 
I.N herstein
 
Yeah yeah
 
@BalarkaSen It is
 
5:15 PM
Maybe I just have forgotten after 45 years.
 
In my book also he does crazy things .writes function composition backwards
 
Yeah, Balarka just complained about that.
I think there are far better choices of algebra textbook.
 
I had a professor who wrote linear maps backward
 
@AlessandroCodenotti I think a scheme in general is so general that without more adjectives it's natural for them to be not too intuitive - after all even if you only cared about affine schemes - that's already the opposite category of commutative rings - and you can imagine one can cook up some pretty crazy examples of rings
 
see you all soon .bye
 
5:16 PM
As in $x\mapsto xA$ with row vectors urgh
 
@Alessandro: It always gets confusing with permutation cycle notation, too.
There are times I use the right action and row vectors. Certain things in differential geometry. Chern did it, and I blindly followed him instead of doing what other people do.
 
@loch I don't think we're going to care about non affine schemes a lot
 
Pig
and also galois theory
i remember the notation of herstein for galois theory is extremely nonstandard by modern conventions
 
But the definition of a morphism of schemes seems pretty involved and artificial
 
Pig
i would actually argue that's pretty natural
 
5:19 PM
@Pig yeah what I was thinking of is similar to the $L^1(G)$ example. If $G$ is a locally profinite group, then the Hecke algebra $\mathcal{H}(G)$ are locally constant compactly supported functions $G \to \Bbb{C}$ (also denoted $C^{\infty}_c(G)$), then this is a algebra without identity under convolution.
It turns out that smooth representations of $G$ are equivalent to certain modules over $\mathcal{H}(G)$ (the condition is that $\mathcal{H}(G)M=M$ (which isn't automatic due to the lacking identity)
 
I know that the whole point is to get the equivalence of categories with commutative rings with 1 but that doesn't make it more natural to me
 
Pig
a big theme of scheme is you can characterize the "geometry" of a space by specifying the functions (sheaf)
 
@Pig Can you elaborate on why you find it natural? Maybe it'll help me build intuition as well
 
Pig
if you want to charaterize maps between two objects
 
Good morning, chat.
 
5:20 PM
if $G=\mathrm{GL}_2(\Bbb{Q}_p)$ then this is related to the usual Hecke algebra from modular forms through adelization of modular forms
 
Pig
by this philosophy you want to specify the relationship between the space of functions on these separate objects
induced by the map between objects
 
morning @Fargle
 
Pig
the definition of morphism between schemes (or really, just ringed spaces in general) does exactly that
the first thing you face is you can't directly compare functions on different spaces
 
@AlessandroCodenotti probably it's easier to understand locally ringed spaces first, after all morphisms of schemes = morphisms of locally ringed spaces
 
Pig
which is why pushforward has to be involved
with that $O_x \to f_* O_Y$ should be thought of as "composition with f"
which is pretty muc hthe only thing you can do
 
5:21 PM
I mean you pushforward so that the "domain" of the sheaves are the same
 
@Pig Sure, I agree with that, it's the same as with prevarieties
 
Pig
Ah @MatheinBoulomenos I see
 
The most important condition is that it restricts to a map of local rings on the stalks
It's very natural, given that those are the three conditions a smooth map of smooth manifolds satisfy as well :p
 
Heya @Fargle
 
How is everyone?
 
Pig
5:24 PM
@MatheinBoulomenos I feel like this isn't a "true" example of lacking identity, in that we can probably enlarge $H(G)$ (p-adic or archimedean) to include delta functions
so it isn't "real" per se, but i can imagine in other cases the corresponding group algebra wouldn't have identity
 
My naive attempt would have been to try to define a morphism of schemes as a morphism of ringed spaces, but we saw in class that this cannot work, basically because the induced map on stalks can be a mess, right?
 
Pig
@AlessandroCodenotti i think morphism of schemes = morphism as ringed spaces
 
morphism as locally ringed spaces
 
Pig
surprised that it's not true - as long as the spaces involved are locally ringed space, induced map on stalk should come for free
even as ringed space should be fine
descent to stalk comes for free i believe
 
is it possible that a ring does not have identity element with respect to multiplication i.e dot ( . ) ?
 
5:27 PM
Yes but it doesn't necessarily respect the maximal ideals
Wait, let me find the example I have in my notes from a couple of lectures ago
 
Pig
oh
you are right
mb
 
Oh, yeah, the issue is that not every morphism $\operatorname{Spec}(R)\to\operatorname{Spec}(S)$ as ringed spaces is induced by a morphism of rings $S\to R$
While it holds for morphisms as locally ringed spaces
 
@Pig it's not a true example in another way (in the p-adic case at least), since $\mathcal{H}(G)$ is the union of subrings with an identity
 
Pig
right
anyway, gotta run - see you all :)
 
@Pig see you
 
5:31 PM
So anyway I guess the counterintuive part for me is how pullbacks are additional data one needs to specify in a morphism of schemes, while a map of the underlying spaces isn't enough
Which didn't happen with (pre)varieties
Bye @Pig!
 
My understanding is that schemes have smoothness inbuilt in them as the stalk, which is a local ring, and the tangent space is m/m^2 where m is the maximal ideal of the stalk. You want the map to preserve the maximal ideals stalk-wise
 
morally your stalk at p remembers the functions defined around p, and your maximal ideal consists of functions vanishing at p

and so asking for the induced map on stalks means that "you had a function on Y vanishing at f(p) and you want the function obtained via your map of sheaves O_Y -> f_*O_X to also vanish at p

I guess the point is that morphism of ringed spaces is too general to force the above (very natural!) condition
 
Do diffeomorphisms allow self-intersections? For example, is a line segment in $\mathbb{R}^2$ diffeomorphic to some smooth curve with a self-intersection?
 
what does that even mean
a smooth curve with self intersection is not a manifold
 
Yes, I know. I'm getting confused by this picture in a book.
 
5:35 PM
@loch Put like that it does sound like a natural condition to require
 
Looks like page 3 of a certain book to me
 
Oh. I had the same confusion semi-recently, @berrygreen. Think of that as a 3D picture.
 
I think that was very helpful to build some intuition, thanks @Pig @loch @Balarka @Mathei!
 
A helix, not a self-int.
 
I'm sure I've seen that picture before but I don't remember which book is it from
 
5:36 PM
Guillemin-Pollack
 
yeah a helix would make sense but couldn't he have marked the over and under passes? He did it immediate after in the same figure
 
I didn't pay attention to that illustration in Guillemin-Pollack when I read it. Fair point!
 
I think it's meant to be seen as a curve $t\mapsto f(t)$
 
@Astyx But then what does diffeomorphism mean?
They are supposed to be honest subsets of $\Bbb R^n$
 
I think possibly it was just a missed detail by the authors. There are quite a number of them.
 
5:38 PM
Well you'd have the graphs that are in $\Bbb R^3$ diffeomorphic I guess
 
ya
@Astyx That's too complicated
 
Yeah true
 
They were trying to illustrate shapes which are diffeomorphic in that sequence of images. This one seems to be a (minor) mistake
 
Actually if you take the bigger picture, you find that the middle illustrations are diffeomorphic to the left one (ie circle and sphere) and the rightmost are examples that are not diffeomorphic to the rightmost one
 
@Ted I believe has a list of errata for G-P somewhere. (This detail isn't in it IIRC, but there are a lot of other things that would otherwise make problems difficult and so on.)
 
5:43 PM
@Astyx Lol
 
Gotta eat, I'll be back
At some point in life
 
6:08 PM
@Balarka @Fargle: I don't see anything wrong. Those submanifolds are both diffeomorphic to a circle. There's no isotopy notion here ...
 
Wrong image
One of these is not even a submanifold
 
Oh ... It's meant to be a coil — the graphic is wrong.
 
Yeah
 
Agreed.
 
They should have done over-crossing.
I suppose I should add that to my errata. None of my students ever had an issue with that, but yeah. ....
 
6:10 PM
I never actually looked at that big image of examples of (non)diffeomorphic shapes
 
I remember being really confused by that and asking Mike about it when I first delved into the book.
 
There's another one that's wrong unless you visualize overcrossing in $\Bbb R^3$.
Funny that I never paid attention because no student ever asked. I shall add an erratum.
 
They give an example of non-diffeomorphic things with a V and a loopy curve. I don't like that too much, my diffeomorphisms always have domain/range manifolds.
They of course define it for subsets of R^n
 
They're doing submanifolds of $\Bbb R^3$, clearly.
 
I don't know about "clearly". Notice that they define diffeomorphisms for subsets of R^n, as restrictions of diffeomorphisms from open subsets containing them.
Not exclusively for submanifolds of R^n
 
6:15 PM
Oh, I didn't mean to say submanifolds, sorry. I meant everything sits in $\Bbb R^3$.
Plus some have boundary.
Hmm, I can't get to my UGA website to update ... I wonder if they killed me off.
 
For 3 independent events A, B and C
Is $P(A \cap B \cap C) = P(A)P(B)P(C)$?
Just like for two independent events $P(A\cap B) = P(A)P(B)$
 
@Fargle Shite Haken released a new album a month ago
 
6:42 PM
I'm back
 
@BalarkaSen wait what
How was I unaware of that?! I know what to listen to while studying tonight
 
Yup, "Vector".
I'm on it right now
 
Oh by the way @Balarka a question you might find interesting: I was trying to construct a connected topological space $X$ with at least two points such that there exist an $x\in X$ such that $x$ has a nbhd basis $(B_i)_{i\in I}$ and $B_i$ is homeomorphic to $X$ for every $i$
Kinda like an hawaiian earring but a little worse
But then I realized $\Bbb R$ works so I'm not sure what's the right question to ask to get interesting spaces :/
 
Hawaiian earring should be fixable if you add some sticks at the earring point
Or something
 
I don't think that works
Nvm, it does with countably many sticks
 
6:56 PM
yup
you have to wedge them, so it's not embeddable in R^2 anymore
@AlessandroCodenotti The Cantor set works as well, if you want slightly more interesting spaces
 
Nice one! It's not very connected though
 
Ah hrm
 
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