Olivier Bégassat
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 Sep 27 comment Is the homotopy category cartesian closed? "the path space of a path-connected space is contractible" I very much doubt that's true. Am I wrong? Sep 12 answered show that $\lambda(x+y+z)^2+2c(x^2+y^2+z^2)>0$ for any $x,y,z$ if and only if $c>0$ and $\lambda+\frac{2c}{3}>0$? Sep 12 asked Existence of a measurable $\theta$ such that $\frac{f(y)-f(x)}{y-x}=f'(\theta_{x,y})$? Aug 21 awarded general-topology Aug 12 awarded Enlightened Aug 12 awarded Nice Answer Aug 7 comment Describe a group $G$ that acts on a set $X$ of 4 elements such that the action of $G$ has 2 orbits. An explicit example comes from geometry: $\mathrm{GL}_2(\Bbb F_2)$ acts on $\Bbb F_2^2$, a four element set, and has two orbits. Similarly, yet less interestingly, the multiplicative group of units $\Bbb F_4^\times\simeq \Bbb Z/3\Bbb Z$ acts on the four element set $\Bbb F_4$ by homothecies and has two orbits. Jul 30 revised Homeomorphisms of X form a topological group added 92 characters in body Jul 9 awarded Nice Answer Jun 28 comment Do we have a “short five lemma” for any two of the isomorphisms? Yes, $A$ and $A'$ are the kernels of $B\to C$ and $B'\to C'$, and the vertical map $B\to B'$ maps the kernel of $B\to C$ isomorphically to the kernel of $B'\to C'$. Jun 24 comment Applications of Complex Analysis. Laplace transforms are used in engineering, they produce the transfer functions of a mechanical system, whose poles and zeros and phase have physical meaning, but I doubt techniques from complex analysis is being used regularly. Jun 18 comment Looking For a Neat Proof of the Fact that the Grassmannian Manifold is Hausdorff @caffeinemachine yes indeed. Jun 12 comment Existence of diffeomorphism through convergence in Hausdorff distance It's strange indeed, as the Hausdorff distance isn't a metric on open sets, you can have open sets that are distinct and have zero Hausdorff distance... Jun 12 comment Is the sum of the following series a finite number or not? Explain. $\sum_{k=1}^\infty \frac{5\sin^2k}{k!}$ Where are you stuck, and what have you tried? Jun 12 revised Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$? added 368 characters in body Jun 11 comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$? Wow, this is very neat! +1 Jun 11 comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$? @JimBelk You are right, and in any case the process I propose doens't do what is expected, it's a little more complicated than I originally thought: the banana shaped regions should extend along the complete positive and negative orbit of a point. I think there might be a problem if $f$ isn't nice enough, in particular, it should work if all of $f$'s orbits are finite. Jun 11 comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$? Actually, one could forego the infinite composition altogether by just integrating $\sum_{n=1}^\infty X_n$. Jun 11 comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$? There is a slight problem with the final composite (it may not be finite on every compact set), but it can be fixed. You just need a collection of disjoint, banana-shaped open sets $B_n$ that contain $\phi(n)$ and $f(\phi(n))$ for all $n$ and no other lattice points, and have the $X_n$ have their supports in there. Actually, one could forego the infinite composition altogether. Jun 11 revised Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$? added 376 characters in body