@TobiasKildetoft Thanks, maybe let me try it on this way. For the real version of Stone-weierstrass theorem, we claim that the set of all polynomials a subalgebra of $C([0,T],\mathbb R)$. Do we mean the set of convergent polynomials?
@TobiasKildetoft I don’t really know about these topics , am more of a geometry guy but my point isn’t that we should not use it , am sure it comes in handy , but not to treat it as the holy grail like some fanatics today do , just talking about functions all day without any relation to concrete things like manifolds or rings
Does anyone here know about the rate of convergence of continued fractions? In particular, is it guaranteed to be linear (for quadratic irrationals, at least)?
@Zee Can you give an example say with (smooth?) manifolds where category theory is used instead of more elementary techniques? (Just asking because I'm interested and haven't come across it being used that way)
is there measure that collects all the function spaces that are in L^p space and makes the entire L^p space complete in that it is infinitely dense or something
@geocalc33 if the measure space is infinite, the equivalence is true only if both sides are false. If both sides were true, then $1\in L^p$ would imply the measure space is finite.
@Zee Showing that the quotient of a square by its boundary is homeomorphic to a sphere
I actually think I have a way of doing it, I just have to work out some hairy details and I got stuck on one of them
I posted a link to the question I asked on main earlier (just scroll up and you'll find it), where I tried to solve it and just needed to verify continuity for the function that was going to be the homeomorphism
So if we view $S^1 \subseteq \mathbb{C}$, then we can define $f : I / \partial I \to S^1$ by $f([t]) = e^{2\pi i t}$, then $f$ is bijective and continuous and by the closed map lemma a homeomorphism
@TedShifrin For the $S^2$ picture I thought about doing the same thing you said, but open sets around $\infty$ were hard to characterize in a nice way (if I recall correctly from my post)
That's why I'm suggesting you think about $\epsilon$ balls around $\infty$.
Maybe it would be easier to use $D^2$ rather than $I^2$. Then you'd have stuff outside a ball centered at the origin. You can do something corresponding in $I^2$.
@TedShifrin Here's another way to prove it. There's this theorem that states the following "If $A \subseteq \mathbb{R}^n$ is a compact convex subset with nonempty inteorior, then $A \cong D^n$" So using this we obtain $I^n \cong D^n$ and since $D^n /\partial D^n \cong S^n$ it follows that $I^n / \partial I^n \cong S^n$
That's written quite sloppily though so I hope it makes sense
Yeah, I wouldn't want to quote such a theorem. You can prove it directly for the open square. [Do 1/4 of it.] But most people won't even bother by the time you get to algebraic topology exercises.
Actually, a stronger theorem is true, @Perturb — star-shaped will do it — and in fact you get diffeomorphic rather than homeomorphic. But that's hard!! I just want to make sure you see how to do it for the square and disk. But you should, in general, just say these things are homeomorphic, as you proceed in mathematics, rather than checking it every time.
Yippee, Demonark. Did you learn a word or two?
Have they made the math part more advanced for science types, as they threatened to?
There was some talk a few years ago about actually going up through calculus in the general math exam for science types taking it; I hoped they'd ease up on the verbal a bit when they did that.
There was one problem having to do with geometry which took me embarrassingly long because I forgot the business about sum of interior angles of a polygon
That cost a few points because I ran out of time, and the verbal had a few words which threw me off but on the whole I had a good enough score on each by the looks of it that no retake is needed
@geocalc I currently live in Uppsala, Sweden. I was born in Västervik, which is in the south of Sweden, but raised in Skellefteå, which is in the north of Sweden.
Does Lagrange's theorem say anything about the number of subgroups of order $n$?
Say Lagrange's tells you that any subgroup of your group must have orders $p$ and $q$. Does it say or is there a theorem that tells me the number of possible such subgroups?
@Nebulae That's more along the lines of a converse to Lagrange's Theorem. In this direction, one finds the Sylow Theorems, which inform you about the existence and to an extent the number of subgroups of certain orders.
