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12:00 AM
Invertible sheafs are boring if you solely look at the locally: the're just $O_X$ and the group is trivial
 
to me it's the Euler characteristic of the total space
 
The same way line bundles over a small enough ball are all trivial
 
Go find a correct definition?
 
Anyway, the Picard group can make its appearance in other contexts
 
Something like #zeroes - #poles of a generic meromorphic section? (i.e., intersection of section with zero section)
 
12:02 AM
@Astyx OK, I get it. I was just wondering how to write that pairing is an isomorphism; I suppose you'd tell me that its an isomorphism for sections over a specified open and so globally an isomorphism.
Remember I am not thinking so you should convince me why something is true. I agree that this Picard group is indeed a group now.
 
ah ok, so this is a different type of Euler characteristic
 
Right, it's a morphism of sheaves, so the behaviour locally is what you'd expect and you can glue things so you get an iso everywhere
 
Got it
 
You seem to ignore half of what I type today.
Yours is the definition of the Euler char of the tangent bundle, connection to homology/simplicial stuff being Poincaré-Hopf.
 
Back to Picard: for instance when looking at divisors. There are two ways (that I know of) to define divisors: Weil and Cartier divisors
 
12:04 AM
Ok tell me. I actually dont know the defns properly
 
I don't ignore it, I just don't understand it, though I guess it looks the same way to you
 
Yup.
 
So Cartier divisors are defined as follow on integral schemes: A Cartier divisor is an open cover with rational functions on each open set $(U_i, f_i)$, such that on intersections, the transition function $f_i/f_j$ are invertible regular maps
 
Natural
 
You quotient out by refining covers to only keep the non formal interesting part
 
12:08 AM
What's happened to this room?! I used to be able to understand most of what was said here :-o
 
So this is an element of the sheaf cohomology group $H^1(X, O_X)$
 
Then you define principal divisors, which are (X,f) (ie a global rational section)
@BalarkaSen I think so? I should know this
It turns out that in nice cases, the Picard group is also the quotient of cartier divisors by principal divisors
 
@robjohn Are you whining again? :D
 
@TedShifrin I'm always whining... I'm a crotchety old man.
 
what about in the bad cases?
 
12:11 AM
You're not nearly so old as I am.
 
@TedShifrin but I'm more crotchety ;-p
 
There apparently are ways to define Cartier divisors in the non integral case (I think you can drop out reduced or something) but things become ugly and you have to define a sheaf of rational functions etc
 
My friends might dispute that, @robjohn.
 
Anyway, the other kind is Weil divisors
 
@Astyx Hold on, let's figure out the cohomological interpretation?
 
12:13 AM
Sure
I might not be of much help for that though
 
My brain is mush so I am a little annoyed by not being able to do it on the spot. What is sheaf cohomology? If you have $X$ a space, $F$ a sheaf, a covering $\{U_i\}$ of $X$ by open sets, then define $n$-cochains to be collections $(U_{i_0, \cdots, i_n}, f_{i_0, \cdots, i_n})$ of sections of $F$ over $n+1$-fold intersections of these open sets of the cover.
The boundary map does a messy alternate sum by omitting a single, varying, index. The resulting chain complex has cohomology groups $H^n(X, F)$. Technically, this is Cech cohomology for the cover $\{U_i\}$, by refining over all covers you get sheaf cohomology $H^n(X, F)$, the absolute group.
What is your thing in this interpretation?
 
So just do $C^1$ and $H^1$ for goodness sake.
 
A Cartier divisor is $(U_i, f_i)$ where $f_i \in M_X^\times(U_i)$, such that $f_i/f_j \in O_X^\times(U_{ij})$.
 
I think the Picard group is meant to be $H^1(X, O_X^*)$ but I'm not 100% sure
 
You are correct, but we should see if everything checks out
 
12:18 AM
Yes, that's right, @Astyx.
 
Ah I see. $(U_i, f_i)$ can be treated as an element of $H^0(X, M_X^\times/O_X^\times)$.
Because if you mod out by the sheaf of regular functions, $f_i$'s patch over all open sets
 
What's $M_X^\times$ ?
 
Sheaf of rational functions over $X$ under multiplication, sorry.
$M$ stands for meromorphic in the complex category
 
Note that your line of reasoning only holds when X is assumed noetherian separated, otherwise Cech and sheaf cohom do not coincide
 
You are correct.
 
12:21 AM
And IIRC the statement itself doesn't hold
Anyway, should I move on to Weil divisors?
 
One moment, do you agree with my calculations?
 
I do
 
Now note that there is an exact sequence of sheaves $0 \to O_X^\times \to M_X^\times \to M_X^\times/O_X^\times \to 0$ which gives a long exact sequence in cohomology
 
for a measurable matrix-valued function, $(\int_0^tA_s \, \mathrm{d}s)x = \int_0^tA_sx \, \mathrm{d}s $ right
 
$0 \to H^0(O_X^\times) \to H^0(M_X^\times) \to H^0(M_X^\times/O_X^\times) \to H^1(O_X^\times) \to H^1(M_X^\times)$
 
12:24 AM
@user2103480 integration is linear
 
Ok so everything checks out if we prove $H^1(M_X^\times)=0$ right?
 
yeah; actually that's because sheaf of rational functions is gonna be flabby
 
Indeed
And flabby => acyclic
 
Sorta like me.
 
Cool!
 
12:28 AM
Are you acyclic ?
 
He's aspherical
 
No, very flabby.
 
OK, Weil it is
Weil weil weil
What do we have here
 
(btw is flask another term for flabby? or is that not a thing?)
 
flasque
 
12:29 AM
yeah haha i never remember this
 
Oh ok
thanks
 
flask is for drinking
 
And brewing
So Weil divisors are defined as formal sums with integer coefficients on the set of closed irreducible subschemes of codimension 1 of X.
 
@TedShifrin maybe if you drink from the flasque too much you become flabby
 
it's almost my martini-time.
 
12:32 AM
Principal Weil divisors are divisors of the form $\sum_Z ord_Z(f)[Z]$ for some rational function $f$
 
What is $ord_Z(f)$
 
@Thorgott I'm trying to figure the same out for bochner-integrals. I.e. for $s \mapsto T_s \in L(U,V)$, $(\int_0^t T_s \, \mathrm{d}s)(u) = \int_0^t T_s(u) \, \mathrm{d}s$ but I don't see directly how linearity solves this. Linearity means two things for me here: $f \mapsto \int f$ is linear and $A(\int f) = \int Af$
 
In dimension 1, that's the multiplicity of zeroes/poles
 
I mean, I can show the "linearity" for matrices just by some calculation, but here I need a more high level approach
 
Yeah I understand that but what about in higher D?
 
12:35 AM
@user2103480 I don't really know Bochner integrals. It's true for simple functions and just take a limit maybe?
 
@Thorgott smart yeah this might do it
 
In higher dimension you define it by localizing at the generic point of each codim 1 irreducible subscheme
 
what if you don't
 
nice
glad to know integration theory still works the same way as I remember it
 
simple functions for bochner are just functions that are constant on elements of a finite measurable partition and there I can exchange integration+
 
12:37 AM
Then look at $l(O_{X,x}/fO_{X,x})$ I think, let me check
 
When you perform such a localization you get a discrete valuation ring, and its valuation is ord()
 
Right
That's it
 
Why would you get a DVR? It's not 1-dimensional.
 
I should stick to doing math that can be solved by using the same trick over and over again instead of requiring sophisticated insights
 
Divisors and line bundles aren't "sophisticated insights." You just haven't learned any geometry. :D
 
12:38 AM
measure theory brain
 
I can't remember what the precise valuation is, something like the power required for it to be in the ideal of functions vanishing on the subscheme, or the opposite... something like that
 
@Thorgott I have to admit I feel the same.
The problem is everything that could be done by spamming the same trick again and again has been done
 
You guys stop feeling sorry for yourselves!
 
just invent a new trick
easy
 
@user2103480 this is the sophisticated insight part
 
12:40 AM
where's my fields
 
@TedShifrin Are motives considered sophisticated these days?
 
I know nothing about 'em.
 
geometry is too sophisticated
who even understands pictures
 
come to probability
 
Algebraic geometry isn't that kind of pictures, Thor.
 
12:41 AM
floppiest of the pictures is where it's at
@Astyx Dude define $ord_Z(f)$ for me I cannot guess what it is
Do I look like a scheme theorist
 
probability would kill me
 
It is the valuation IIRC
$O_{X, \eta_Y}$ is a DVR
 
But what is the DVR? A DVR happens when you have something 1-dimensional
 
@BalarkaSen If $f$ is in $\mathcal O^*_{X,\eta}$ then $\mathrm{ord}(f)=0$
 
(where $\eta_Y$ is the generic point of $Y$)
 
12:45 AM
If $f$ is in $\mathcal O_{X,\eta}$ then the valuation is the length of $\mathcal O_{X,\eta}/(f)$
 
This is all regular functions on $X$ which vanish along $Z$?
AKA the transverse direction to $Z$
 
to be honest, I think I'd understand more of my Riemann surfaces course if it was actually phrased in terms of bundles, but sadly it isn't
 
There's a closed immersion $Y\to X$ where $Y$ has its reduced structure
This is the map that is used to project $f\in M_X$ into $M_Y$
Then you compute the valuation by localizing
 
When I taught Riemann surfaces, I taught it to people at the end of a complex analysis course, so I minimized the bundle language. But in the grad diff geo and complex geo courses, of course I did plenty of bundles.
It's not hard to create a line bundle from a divisor by thinking of a Cech cocycle, as Balarka and Astyx have been discussing. Just give the divisor in the open sets of an open cover.
 
This is a very confusing phrasing. You're looking at germs of regular functions near $Z$ which vanish on $Z$, which is the ring $O_{X, Z}$ if I understand correctly. This is the DVR because $\codim_Z(X) = 1$, $Z$ is closed irreducible, so you can compute what the valuation of $f$ is along this
 
12:49 AM
I've just told you what it is
 
You wrote $O_{X, \eta}$ but never explained what it means.
 
Do you know what the generic point is?
 
Or if you did I didn't understand it
Nope.
 
Ah right
 
It took me a while to really have what a generic point is click
 
12:50 AM
So when you're dealing with schemes, you work with the Zariski topology
 
But do I need to understand it? $O_{X, Z}$ is what I described, isn't it? Germs of all reg. functions near $Z$ which vanish on $Z$.
 
If I understand you correctly yes
 
OK :)
 
But I think that's in the "nice" case
Because closed irreducible subschemes aren't always given by a single equation
 
$Z$ you took to be a closed irreducible subscheme anyway
OK fair enough
 
12:53 AM
Speaking of which, you always have a morphism from Cartier divisors to Weil divisors
 
You can take $Z(f_i)$
 
Which is not always surjective for that very reason
 
I think it would be instructive to do the calculation for the one-dimensional case, to see how the notion of order of a pole and zero naturally emerge
 
I know the calculation for one dimensions, which is why I am asking in higher dimensions :)
I know Riemann surface theory, but not this.
 
Well, you don't otherwise you'd have been able to relate the order of the pole to the length of the relevant module
 
12:54 AM
There's no difference between Weil and Cartier in 1D
 
The opposite goes for me
 
So that's my point, you should use the more abstract formalism to relate it to what you already know
 
So this morphism between Cartier and Weil is surjective when X is regular
 
@JamalS I think it's not helpful to be condescending if you're teaching someone anything :)
 
And in this case they define the same group
 
12:55 AM
I don't know algebraic geometry, I am a humble topologist.
 
super weird I thought you knew everything
 
(because then local rings are factorial (which I don't know how to prove) and everything is super nice)
 
@BalarkaSen Okay let me do an example.
 
I agree, Balarka should use more abstract formalism
 
lmao
 
12:57 AM
@Astyx That's Auslander-Buchsbaum or whatever
Regular local rings are always UFD's.
 
I'll take your word for it
 
@Thorgott The point I wanted to make was you use the more abstract formalism on a simple example to relate it to what you already know.
 
Just blackbox it. Why does UFD imply Cartier -> Weil is surjective?
 
Because in a factorial ring, every prime ideal of height 1 is principal
 
Ahh right.
 
1:00 AM
@Thorgott immense boredom? :D
 
So every codimension 1 variety is in fact cut out by a single equation. Neat.
 
@user2103480 brutal slaughter
my innards would get plastered on the walls
 
I think you'd survive
just a flesh wound
 
@Thorgott wtf suddenly death metal
 
@Astyx eyyy I knew this one at least
 
1:02 AM
@BalarkaSen Suppose you have $f = (z-1)^2(z-2)/(z-4i).$ You know the order of the pole is $2$, but let's do it more abstractly. We want to look at $z=1$ so we localise $C[z]$ at $(z-1)$, the ideal. Then the quotient module I mentioned before in this case is $C[z]_{(z-1)}/((z-4i)(z-1)^2) \cong C[z]_{(z-1)}/((z-1)^2).$
 
I wish I understood everything I've been talking about, but I don't really
 
algebraic geometry looks pretty neat later on
oh you kind of know the words to the story huh
I get that. that's good too- more than most you know
 
@JamalS Order of the zero you mean
 
@BalarkaSen Now we can form the maximal chain $(0) \subset C[z]_{(z-1)}/((z-1)) \subset C[z]_{(z-1)^2}$ which has length 2 (length is number of inclusions not number of components of the chain). So the order of the zero is 2.
@BalarkaSen Yes, sorry, zero.
 
Length is dimension of the quotient vector space...
 
1:04 AM
Yeah got it. This is obvious though because the ambient is 1-dimensional, you have a coordinate element $z$, if you localize everything is $z^n$ times a unit
 
the funny thing is that I can unironically imagine one of our probability professors to just continue a lecture if a student were to explode in the middle of it
 
This wasn't my issue, my issue was finding what the 1D ring was in higher dimensions
 
@Thorgott "Alright class, let's compute the odds of that happening"
 
As I understand it you extract something out of the transverse direction to the codimension 1 subvariety; for a principal subvariety I get it.
 
What should I read to get more comfortable with Riemann surfaces etc?
 
1:06 AM
Forster is a good book.
@Astyx And now I am convinced this is false
@Thorgott link
 
Well, I've read the material sufficiently many times to recite it to you. Doesn't mean I understand what it truly means in practice
 
Nobody really understands algebraic geometry
Or algebra for that matter
 
Thanks for the ref
 
@Thorgott the show must go on
 
ye lol
 
1:09 AM
Do you want to know what the generic point is?
 
Sure tell me
 
@Astyx $\mathrm{Spec}\mathbb Z$ might be a good example to use to explain it.
 
@Astyx since youre looking at it from an AG perspective, try Miranda perhaps
 
Why Spec $\Bbb Z$?
So as I was saying in AG we usually work with the Zariski topology, so we're not really considering points but rather prime ideals. In the Zariski topology of an integral ring there is a specific prime that is close to every other prime: the nilradical. This is the generic point. In a more general irreducible scheme, it turns out that you can also define such a point. The interesting property is that localization at this point (read: at this prime) yields precisely the ring of rational functions
For instance with $\mathrm{Spec}\mathbb Z$ (which is integral affine), the localization at $\sqrt{(0)}$ gives $\Bbb Q$
 
Chuck all zero divisors, then why is this useful? You're just localizing at (0)
 
1:21 AM
In the affine plane $\Bbb A^2 = \mathrm{Spec} k[X,Y]$, you get $k(X,Y)$
So when you deal with an irreducible subscheme $Z$ of codim 1, this lets you define precisely $ord_Z(f)$
 
Oh, fine, $O_{X, Z}$ should just be localization of $O_X(X)$ at the prime ideal corresponding to $Z$?
Is that what this means
 
Because you're localizing everything except the codim 1 part
Right
 
Why in some sense? On every affine open it is that, right?
$O_{X, Z}(U) = O_X(U)_{[Z]}$ where $[Z]$ denotes the prime ideal corresponding to $Z \cap U \subset U$.
 
:)
 
OK, got it.
 
1:26 AM
Anyway I need to sleep
'night
 
Thanks! Bye
 
@Thorgott Thanks
 
night astyx
 
Night!
 
@JamalS alas, I cannot answer your question
you can get $T_m f = a_2(f) T_m f$ for odd m just by looking at the relevant formula right?
idk how that helps
 
 
1 hour later…
3:03 AM
Can anyone check if my part 2 proof is valid?
like what i missed and what are redundant
 
3:40 AM
you wrote $T\in\mathbb{R}$ multiple times where it should say $T\subseteq\mathbb{R}$
 
3:53 AM
thanks
new to latex, is there anything else, hopefully that's logic related?
and typos
 
 
3 hours later…
123
6:32 AM
Hi All..
 
7:06 AM
Can anyone tell me a reference or any proof of this line: Any geodesic on any surface must intersect itself transversely if it is not simple or a covering of a simple closed geodesic.
 
If not then the geodesic must intersect itself somewhere tangentially
Do you know why that's a contradiction
 
No, I donot know, but what about "covering of a simple closed geodesic"
 
Let there exist a multivariable function $f(\mathbf x)$, where it's domain is $D\subseteq \mathbb R^n$. Let there be an open set $S \subset D$. It is given that $\partial f/\partial x_i$ is defined at every point $P \in S$, for all $i$ from $1$ to $n$, where $x_i$ are the $n$ orthogonal coordinates forming up the $\mathbb R^n$ space. From the above given information, can we conclude that $f$ is continuous at every point $P\in S$? I think yes, we can.
 
@User873110 Given a point $p$ and a tangent vector $v$ through $p$, how many geodesics pass through $p$ with initial vector $v$?
 
Ohh uniqueness of Geodesic
 
7:13 AM
Yeah. If you scale $v$, then you just scale the parametrization; the image of the curve remains the same.
This is why you can rule out the covering of a simple closed geodesic case.
(That just means going around a single closed geodesic 100 times)
@FakeMod This is not true. Consider $f(x, y) = (xy)/(x^2 + y^2)$ for $(x, y) \neq (0, 0)$ and $f(0, 0) = 0$. Around any neighborhood of the origin, both the partial derivatives exist.
This is nevertheless discontinuous at $(0, 0)$.
 
@BalarkaSen But it says transverse self-intersection not tangentially
 
I told you why tangential intersection cannot happen, unless it goes around itself multiple times.
Proof by contradictions; if not it contradicts uniqueness of solution to the geodesic eqn
 
ok
now got.
 
Cool
Sorry if I was being confusing
 
@BalarkaSen Oh, I see. If I apply one extra constraint that the partial derivatives must be bounded, would it then allow me to deduce the continuity of $f$.
 
7:29 AM
It's still not true; maybe you can cook up an example. Try $(x^2 - y^2)/(x^2 + y^2)$ or something of this nature.
 
I just found these notes which state this (albeit in a slightly different way). Image below.
 
Ah ok yeah maybe it is true then. If the partials are continuous it is definitely true but I didn't remember this.
Bounded is probably enough
 
@BalarkaSen Yeah, in the continuous partial case, it is true. I was just curious about tis case. Thank you for the help.
Can anyone please explain why the converse of the above theorem is not true?
 
They are only continuous at $X_0$ and not on a neighborhood.
That is not enough
 
@BalarkaSen So is the given theorem also incorrect?
 
7:42 AM
Oh I misread. I mean differentiable doesn't imply derivatives will be continuous
 
@BalarkaSen Yeah, could you please provide an example of such a case?
 
No need to go to 3 dimensions, just take 1D examples of differentiable but not continuously differentiable functions
Well this is single variable analysis so you should come up with examples
That's a prereq for multicalc
 
oh, probably a stupid question but does 1D differentiable implies derivative over a closed interval is bounded?
 
@EdwardH no.
 
ok
 
7:49 AM
it is a standard counter ex which i am busy looking up right now :-)
 
@BalarkaSen But in case of single variable calculus, the point where the derivative of the function is not continuous, that either has to be a point of non-differentiability or a point which doesn't belong to the domain. I am not able to think of any case, where a single variable function is differentiable at a point but it's derivative isn't continuous.
 
@FakeMod Huh? $f(x) = x^2$ for $x \geq 0$, $f(x) = -x^2$ for $x \leq 0$.
 
yeah, somehow the word "no" just reminded me of a counter-example, so no worries
 
It's probably a good idea to be aware of calculus before calculus in n dimensions.
 
7:50 AM
@BalarkaSen The derivative of that function is continuous, isn't it?
 
@FakeMod look at the cex i posted.
 
the analyst's sine curve is so over powered
 
@BalarkaSen Of course the derivative is not differentiable, but it is indeed continuous.
@copper.hat Yes, I remember seeing that example. I had forgotten about it.
 
Oh ok I misunderstood your question again. You want $f'$ to be not continuous, but you can cook up such examples.
 
@BalarkaSen Yes, like the sine bounded by a parabola one.
 
7:53 AM
$f'$ cannot be too wild. worth keeping Darboux's theorem in mind.
 
$x^2 \sin(1/x)$ doesn't quite work I think, you want $x^a \sin(1/x)$
$a < 2$
 
why doesn't
 
@FakeMod you are looking for a function whose partials exist but the function is not continuous?
 
@EdwardH Ah yeah $2$ already works, more than $2$ doesn't. Something like this
I am admittedly not computing
 
probably works as long as power $>1$, $2$ is the safe choice
 
7:58 AM
No $x^3 \sin(1/x)$ is $C^1$ what do you mean
$2$ is the critical exponent
 
oh true
I was just thinking the sine part can be as oscillatory as one wants
 
yeah something
 
@copper.hat I am looking for a function which is differentiable but its partials are discontinuous.
 
Surely the product of two one variable ones will suffice?
Let $f$ be the one in the counterexample above (link) then let $(x,y) \mapsto f(x)f(y)$.
actually i need to check if that is differentiable at 0
silly me, of course it is, it is the product of two differentiable functions.
 
8:15 AM
just take $(x, y) \mapsto f(x)$
 
better still :-)
 
@copper.hat Ah, okay. I see. This seems like a satisfactory counter example.
@BalarkaSen :)
@copper.hat Are there any other such (possibly common or typical) examples of functions which can be written using elementary functions?
 
there is an old but still useful book that is worthwhile having on a friend's shelf: Counterexamples in Analysisby Gelbaum& Olmsted
 
Oh, okay. Thanks. I will surely have a look.
 
@FakeMod i can only recall similar ones. they have $x^2$ to get differentability at zero, and then some bounded function of ${1 \over x}$ to get the wild derivative.
 
8:21 AM
@copper.hat I see. I am also not able to think of any other way to achieve non continuous derivatives along with differentiability right now.
Thank you @BalarkaSen and @copper.hat for the help.
 
you're welcome!
 
glad to be of no help
 
i was in darjeeling many decades ago.
 
9:01 AM
The defintion of rational equivalence on a variety $X$ is as follows, for two cycles $A_0$ and $A_1$, I say both of them are rationally equivalent there exists a cycle in $\Bbb{P}^1 \times X$ such that the fiber at $t_0$ is $A_0$ and the fiber at $t_1$ is $A_1$. Very naively this suspiciously looks a lot like two things, one a "rational" homotopy and also like a bordism between two manifolds. One even has a transversality type result here.
Where you say that two cycles are transversal upto a rational equivalence
So intersections make sense in the Chow ring. But I was asked to be extremely vary in thinking of this equivalence as homotopy or bordism, why is that? Is there an issue in thinking of this as a $\Bbb{P}^1$ homotopy?
 
Why is it a homotopy? Homotopy of what map? It's a family of cycles indexed over $\Bbb P^1$
Rationally equivalent cycles are "cobordant" as a special case because you can take a path joining $t_0$ and $t_1$ in $\Bbb P^1$ and take preimage
But even this is fidgety because these are singular manifolds
At least that proves that $CH_* \to H_{2*}$ is a well-defined map for varieties over $\Bbb C$, because you can triangulate the cycles and then rational equivalence gives a simplicial cobordism or whatever between the cycles; $\pi_X : \Bbb P^1 \times X \supset \pi_{\Bbb P^1}^{-1}(\gamma) \to X$ where $\gamma$ is the path.
 
Right okay I get this, but why do you call cobordant a fidgety idea for singular manifolds?
 
9:28 AM
Well first of all it is a lot more than a cobordism; you have a subvariety of $X \times \Bbb P^1$ which is a flat family over $\Bbb P^1$
By restricting to paths over the base you get certain singular cobordisms, which is fine.
 
 
3 hours later…
12:10 PM
@copper.hat @BalarkaSen From our older discussion, I wonder whether the following function $$f(x, y) = x^2\sin\left(\frac{1}{x}\right) + y^2\sin\left(\frac{1}{y}\right)$$ differentiable at $(0,0)$? I am not sure about this because both $\partial f/\partial x$ and $\partial f/\partial y$ are discontinuous at $(0,0)$, but the functions $$g(x, y) = x^2\sin\left(\frac{1}{x}\right) \quad \text{and}\quad h(x,y) = y^2\sin\left(\frac{1}{y}\right)$$ are differentiable...
so their sum ($g(x,y) + h(x,y) = f(x,y)$) must also be differentiable.
 
 
1 hour later…
1:22 PM
3
Q: What are some books or state of the art papers about the development of a strong-AI?

pasaba por aquiI am looking for books or to state of the art papers about current the development trends for a strong-AI. Please, do not include opinions about the books, just refer the book with a brief description. To emphasize, I am not looking for books on applied AI (e.g. neural networks or the book by No...

 
 
1 hour later…
2:38 PM
Neat little fact I'm surprised I never saw before: If $X$ is compact metric, $f\colon X\to Y$ is continuous and $f(X)$ is Hausdorff, then $f(X)$ is metrizable
 
@AlessandroCodenotti doesn't that just follow from compact + hausdorff => metrizable
 
It would if that implication were true, which it isn't
 
oh, you need second countable as well
 
(counterexamples are $\omega_1+1$, $\beta\omega$, $[0,1]^{[0,1]}$ just to name a few)
 
> a compact Hausdorff space is metrizable if and only if it is second-countable
 
2:43 PM
yeah that's Urysohn's metrization theorem
to see the hard direction I mean, metrizable+compact hausdorff implies second-countable is easy
 
and is it true that image of metric space is second-countable?
 
It's false in general, I'm not sure if you also want the image to be Hausdorff (R with standard top into R with cofinite top)
Ah of course not, pick the identity on an (infinite dim) Banach space with the norm topology on the domain and weak topology on the codomain
 
Hi
i have confusion in mod of x . Pls help me
It says that always mod of 3x = 9 , then 3x =9 and 3x=-9.Is it right
if we say mod 3x + 4 = 9. Then why don’t we say 3x-4 =9 but 3x+4 = 9 and -9
 
If |(some expression)| = (something), then (some expression) = (something), or (some expression) = -(something).
(That is, it should be "or" instead of "and".)
 
2:56 PM
It isn't "and" because it's not both at the same time.
 
Ok,I got it
 
Right.
 
You could also state it as, in your last example, 3x+4 = 9 or -(3x+4) = 9. The minus can go on either side of the equation---but it has to apply to the whole side, so 3x - 4 = 9 isn't true if |3x+4| = 9 is true.
You would need instead to say that 3x+4 = 9 or -3x - 4 = 9, which is the same as saying 3x+4 = 9 or 3x+4 = -9.
 
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