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Feb
18
comment If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$?
@A.S. could you add some details on how to construct the transitions while ensuring $g\geq f$?
Feb
18
comment If $f$ is bounded non-negative $L^1$, is $f\leq g$ a.e. for some continuous integrable $g$?
Wlog $0< f <1$, in which case $E_1=\mathbb{R}$ has infinite measure. I think indeed something like Lusin's theorem may be needed, but I like your idea of approximating by open sets.
Nov
16
comment Prove or Disprove: if $\lim\limits_{n \to \infty} (a_{2n}-a_n)=0,$ then $\lim\limits_{n \to \infty} a_n=0.$
What about the constant sequence $a_n=1$?
Nov
8
comment First-order definition of “$f$ is continuous at $x$” using just $<$
@avid19 I see, is that the problem, that there are no such symbols in the language? But how is $\Bbb R$ to be understood in this context?
Nov
8
comment First-order definition of “$f$ is continuous at $x$” using just $<$
Wouldn't something like $$\varphi(x)=\forall\epsilon >0~\exists\delta>0~\forall y~ (x-\delta<y<x+\delta\rightarrow f(x)-\epsilon<f(y)<f(x)+\epsilon)$$ do the trick?
Sep
27
comment Is the homotopy category cartesian closed?
If you consider unbased paths, then there is a homotopy equivalence $X\simeq X^I$.
Sep
27
comment Is the homotopy category cartesian closed?
@MikeMiller ok, but why then does the OP say that the path space is isomorphic to the discrete set of path components? This doesn't seem to gel with a basepoint approach.
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?
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.
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
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
10
comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
Actually, I was probably wrong, I think I can see a construction.
Jun
10
comment Does every bijection of $\mathbb{Z}^2$ extend to a homeomorphism of $\mathbb{R}^2$?
The bijection you present seems unrealizable, I would be shocked if the answer was yes in this case. I vote no.