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location | ||
age | ||
visits | member for | 5 years, 5 months |
seen | Aug 14 '10 at 15:55 | |
stats | profile views | 450 |
I like math.
Sep
24 |
awarded | Autobiographer |
Jul
7 |
awarded | Popular Question |
Mar
30 |
awarded | Nice Question |
Mar
21 |
awarded | Yearling |
May
20 |
comment |
Why is Lebesgue integration taught using positive and negative parts of functions?
Completion of what? I understand how the extended reals can be seen as an order completion of ${\mathbb R}.$ But what does metric completion have to do with vector valued integrals? What are we metrically completing? ${\mathbb R}$ is already complete. The connection between order and monotone convergence makes sense. But the examples of the "metric" approach seem off topic. Are you saying convergence in measure and completeness of $L_p$ are more natural if you avoid the extended reals? I don't get it. |
May
18 |
answered | Why is Lebesgue integration taught using positive and negative parts of functions? |
May
18 |
answered | a.e. convergence of the powers of an operator built from rotations |
May
10 |
comment |
How can one characterise compactness-by-experiment?
This was already covered in his question when he mentioned homotopy theory. |
May
10 |
comment |
Examples of common false beliefs in mathematics
Remember being confused by this too. It became much clearer when I formally was taught about short exact sequences. Then you can see exactly the obstruction to such a decomposition. |
May
2 |
comment |
Two commuting mappings in the disk
A related tangent: applying this other Kakutani Fixed-point theorem (en.wikipedia.org/wiki/Kakutani_fixed_point_theorem) to the multi-valued function $g^{-1} \circ f$ looks like it solves the problem... unfortunately the sets $f^{-1}( g(x))$ aren't convex (which is a pretty significant assumption!) |
Apr
14 |
comment |
2, 3, and 4 (a possible fixed point result ?)
@Ady: Since your question is whether or not $FP(2,3,4)$ holds, I'm confused. You seem to be claiming that BGK resolves your question. |
Apr
6 |
comment |
2, 3, and 4 (a possible fixed point result ?)
For fun, I started considering variants of this question but made no real progress on them either. Let ${\bf FP}(p,q,r)$ be the same statement with $2,3,4$ replaced by $p,q,r$. The BGK fixed point theorem gives us ${\bf FP}(p,p,p)$ for $1 < p < \infty.$ For what other values of $p,q,r$ can you prove or disprove this statement? Do you know of a counterexample to $FP(1,\infty, \infty)$? |
Mar
30 |
revised |
Topologizing free abelian groups
added 26 characters in body |
Mar
30 |
revised |
Topologizing free abelian groups
fixed the topology |
Mar
30 |
awarded | Commentator |
Mar
30 |
comment |
Topologizing free abelian groups
I've edited the uniqueness part to make it more convincing. |
Mar
30 |
revised |
Topologizing free abelian groups
clarified uniqueness |
Mar
30 |
comment |
Topologizing free abelian groups
Chris Schomer-Preis' comment made me think a little more. I was under the impression that my construction worked at least when $S$ was something nice like compact metric. If it goes wrong with a stranger space $S$ (as suggested by Dominguez below) then it would be really interesting to see some examples. |
Mar
30 |
revised |
Topologizing free abelian groups
Added existence portion of argument |
Mar
30 |
answered | Topologizing free abelian groups |