Hi, I have a simple question that is probably not worth posting to the main site. So if I and J are ideals, with $I\subseteq J$, then clearly $rad(I)\subset\rad(J)$, where rad means radical. What about the converse? If $\rad(I)\subset\rad(J)$, would this imply that $I\subset J$? I am guessing no, but can't seem to find a counter-example.
@TedShifrin: Well, if $f\in I$, then $f\in rad(I)$, so $f\in \rad(J)$, so $f^{n}\in J$. So if we can manage that $n\geq 2$ (and $n$ no lower), then we get our counter-example.
@anon Well, since the curve is just a very curled-up line, we can "grab" the two extreme "ends" of the curve and straighten it out like a thread, is what I was saying. But it won't work, as your link says.
@MichaelAlbanese: OK, I've finally got a result I'm satisfied with. We have the nice form $\text{Diff}(S^n) \simeq O(n+1) \times \text{Diff}(D^n, \partial D^n)$. So we see all the complication comes from this second factor. It'd be nice to have a similar thing for Homeo. I think I've finally found it but I don't have access to all the papers I need right now, I'm writing them down so I can take a look when I get home.
Here's a taste. Cerf proved that $\pi_{i+m+1}(\text{Homeo}(S^m),O(m+1)) \simeq \pi_i(\text{Diff}(S^m),O(m+1))$. He also calculated the smaller homotopy groups of the former. I think (hope?) this really takes the form of something like $\text{Homeo}(S^m) \simeq O(m+1) \times \Omega^m \text{Diff}(D^n,\partial D^n)$. But in particular this tells us the homotopy of $(\text{Homeo}(S^m),O(m+1))$ is also wild for $m \geq 5$, and in the same way!
When I get a chance I'm going to try to take a look at the papers this comes from to get a more precise version, and a taste for why it's true. I don't know if I'll have time; hopefully it's not too wild.
@SohamChowdhury: I'm traveling right now. I meant when I get back to the university. I should be able to set up a VPN to get journal access even when I'm away but I forgot to.
@MichaelAlbanese: I have a sign wrong up there. I really want to write $O(m+1) \times \Omega^{-m} \text{Diff}(D^n, \partial D^n)$, where by that I mean some space whose $m$th loop space is the desired object. It'd also be nice if there were a concrete descriptor of this delooped space.
Consider $$ I = \int_0^{\omega}\int_0^{\alpha}\int_0^{\alpha}F(\beta){\tilde{F}(\gamma)}e^{i\beta t}e^{-i\gamma t}R(\alpha)d\beta d\gamma d\alpha$$
In this triple integral,I want to bring about, a change of order of integration, where in I take integration with respect to $\alpha$ inside most an...
@Semiclassical I dont understand exactly... you left, some time ago, a link for exactly I was searching for, yes. I leave a comment here but I dont know if you read it. Yes, it is very interesting, thank you very much.
@BalarkaSen Yeah. If there were, then I guess those two would be homeo, which they are not (because the unit interval minus pt is not connected, although I can't prove that yet).
@SohamChowdhury The Hilbert and Peano curves from $[0,1]\to[0,1]^2$ are Hölder-$\frac12$ continuous, and not rectifiable. This means you cannot straighten them out to unit length.
Path-connectedness is a intristic property of the topological space you are working with (i.e., two points can always be joined with a path). What has that to do with continuity (which is a property of a map)?
@robjohn he's there on my facebook friend list .. I doubt the name he put on fb (the same as the one that is currently his M.se alias) is not his real one too .. :|
it's a subspace of $\prod \Bbb Z/p^k\Bbb Z$, which is compact by Tychonoff. but you need more work than this, as it's not nessesary that subspaces of compact things are compact.
@Chris'ssistheartist: I don't know if you would be interested in my latest answer, but it does use some interesting sums. I need to shorten my derivation of the recursion I use.
@r9m I saw you above talking about identities referring to sos, and the funny thing is that many of the people I talk to they think I'm 50-60 years old (referring to Romanian). That's pretty funny, and many refuse to believe my real age. :-)))
I was talking last night with a cute person who told me that I might probably be 60-65 years old, but maybe also because of my poetic speech in Romanian. :-)
Actually after a certain point almost all our talk was exclusively about age, and less about other things that might have been far more interesting. :-)
Anyway.
@robjohn I can finish both identities nicely, in my style.
$$\sum\limits_{m=1}^{n}\frac{H_{m+n}}{m}=\sum\limits_{m=1}^{n}\frac{H_{m}+\frac{1}{m+1}+\frac{1}{m+2}+\cdots+\frac{1}{m+n}}{m}$$ and then split. You have the well-known sum $\sum\limits_{m=1}^{n}\frac{H_{m}}{m}$ and the rest part to be done by sumamtion by parts.
This makes the theorem so tautological . . . Let $X$ be a compact space and $A$ a closed subset. Take any open cover of $X$. By compactness, there is a finite subcover. This finite subcover also covers $A$. Done.
I started with open cover for A but wasn't able to go much far :)... I got stuck after defining an open cover and then anon gave me a hint of the subspace thingy.. and I chose the values according to it (X-A here) which gave me an open cover of X@Soham
@Soham I am really irritated today... All these dumb educational institutions are playing with my photos
I switched classes okay from one institution to another since I was getting free educations and some paid .... Now when I got some marks in my examination they are using my photos in their pamphlets ...... Unknown institutes are also coming from nowhere@Soham
And every other guy comes to my house saying ... Hey you are the topper right and all sorts of nonsense ... tell me how should I study?, what books did you refer.... Ahhhh..............
@robjohn I can't figure out how to show that $\sum\limits_{m=1}^{n}\frac{H_{m+n}}{m} - H_n^2 = \frac{1}{2}H_n^{(2)}$ .. (that is if at all what I did is correct)
@Huy Crank post about the Hodge conjecture. Total nonsense. Second post by the same guy on the subject. I think he has a question ban now, since he hasn't posted again.
@robjohn That would be amazing if it's posted somewhere. Well, it's possible but I created it independetly of any other work. Let me know if you find it somewhere or anything similar.
