Mariano Suárez-Alvarez
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8h
comment Why $[0,1)$, i.e. “closed/open” in defining Borel sigma algebra?
Since a sigma algebra is closed under intersections, complements and countable unions, it does not matter if you start with $[a,b)$s or with $[a,b]$s or with $(a,b)$s: you get the same sigma algebra in all cases.
9h
comment What does path-connectedness of $I$ have to do with this at all?
Moreover, you «what elation does it have with $X$'s path-connectedness?» is very weird. The very definittion of of path-connectedness is that there be a path from $0$ to $1$ in $X$!
9h
comment What does path-connectedness of $I$ have to do with this at all?
If you can only imagine that there is no path $w:[0,1]\to\{0,1\}$ from $0$ to $1$ but not proving it, I suggest you stop looking at the question and think about this instead.
14h
comment Chromatic Polynomial of a graph
The chromatic polynomial of a graph always has degree equal to the number of vertices of the graph.
16h
comment $Tf = xf(x)$ is not compact in $L^2([0,1])$
Restrict your $T$ to $L^2([1/2,1])$. Notice that $f/2\leq Tf\leq f$ for nonnegative $f$ in that subspace.
1d
comment Almost-invariant polynomials under dihedral group action
You have an action of group $G$ and a character $\chi:G\to\mathbb C^\times$, and what you are looking for is the so called semi-invariants. Indeed, this set is a module over the invariants, and in many cases we know how to describe them. Google a bit for examples.
1d
comment Almost-invariant polynomials under dihedral group action
This is at odds with the claim made by the OP that his almost invariants do not form a ring.
2d
comment Why is $\frac{1}{1-x} = 1 + \Theta(x)$ for $x \in (0,1)$?
Heh :-| ${}{}{}{}{}$
2d
comment Is the inverse limit of exact sequences exact?
Exactness at $M^i$ depends on a very small part of the complex, so the fact that the latter is unbounded is irrelevant.
2d
comment How do you pronounce $\preceq$?
Just "Less than or equal". if you have both $\leq$ and the curly one, you invent some convenction with the person you are talking with to tell them apart, like calling "curly less than or equal". Sometimes $\preceq$ is called "preceeds", but that only makes sense in some contexts.
2d
comment Existence of Certain Lie Groups
Oh, if your Lie algebras are infinite dimensional then the statement is not true. One has to specify exactly what is meant by an infinite dimensional Lie group, what exactly you mean by the Lie algebra of such a group, but for most sensible choices of these meanings the result is not true (for example, countable-dimensional Lie algebras cannot be Banach of Frechet spaces, and the Lie algebra of an infinite dimensional Lie group is usually one)
2d
comment Existence of Certain Lie Groups
This is called the fundamental theorem of Lie theory, Lie's theorem or the Lie-Cartan theorem and, in particular, is true.
2d
comment What is the motivation of studying $P[A]$ in operator K-theory?
Every finitel generated projective module is isomorphic to the image of an idempotent map $p:A^n\to A^n$ for some $n$.
2d
comment What is the motivation of studying $P[A]$ in operator K-theory?
Projections correspond to vector bundles. You should at least browse a book like Rosenberg's "Introduction to K-Theory" or Max Karoubi's book on the same subject.
May
3
comment Show that exist a “good” cover $\{U_{\alpha}\}$ of the Differential Manifold
If you search in MathOverflow you'll find one of my questions which had a fee answers explaining a couple of approaches to do this.
May
2
comment Complex manifold with subvarieties but no submanifolds
It is much better to move questions to MO than to duplicate them.ç
May
2
comment Constructing representation of $G$
In genral this is the only thing you can do. For every $n$ there is a group $G$ whose irreducible complex representations are, apart from the trivial one, of dimension larger than $n$ (for example, the alternating group of degree $n$ has minimal nontrivial degree qual to $n-1$)
May
2
awarded  ideals
May
2
awarded  Nice Answer
May
1
comment If I and J are isomorphic ideals of a ring R, does it follow that $R/I \simeq R/J$?
Why would we need the direct summands to be indecomposable?