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NKSSS attendee


Oct
4
awarded  Popular Question
Sep
24
awarded  Autobiographer
Aug
22
awarded  Yearling
May
14
comment What are the minimal conditions for the exactness of a 1-form on an open connected subset?
Yes - I mean they still need to be connected, but the paths from $\textbf{a}$ to $\textbf{x}$ are only parallel to the axes in some open set about $\textbf{x}$. If $n$ such paths exist and have the same integrals, one for each dimension, then do we have exactness?
May
14
accepted What are the minimal conditions for the exactness of a 1-form on an open connected subset?
May
14
comment Is there a way in matrix math notation to show the 'flip up-down', and 'flip left-right' of a matrix?
@Spacey no - nothing widely used.
May
14
comment What are the minimal conditions for the exactness of a 1-form on an open connected subset?
I see. Could you please clarify what the minimal condition for exactness is then? It seems that all that's needed is that for every $\textbf{a},\textbf{x}$, that there really only need to be $n$ paths with equivalent integrals, one for each dimension, where the paths don't even need to trace out $\partial\Delta$ fully, but just near $\textbf{x}$.
May
14
comment Is there a way in matrix math notation to show the 'flip up-down', and 'flip left-right' of a matrix?
For a matrix $M$, will $DM$ and $MD$ suffice for up-down and left-right flips, respectively, where $D$ is the unit anti-diagonal matrix?
May
14
asked What are the minimal conditions for the exactness of a 1-form on an open connected subset?
May
10
comment Why does $\exists x\,\ x = x$?
@RickyDemer So, basically, the difference between your formalization and the one in Wikipedia is that there's an additional existential qualifier $\exists\emptyset$ in the Axiom of Infinity?
May
7
comment Prove that an open interval and a closed interval are not homeomorphic
Right, it should be the induced topology. What I'm saying is that it isn't clear that on the induced topology that the whole unit open interval is not a compact set. On the induced topology it is closed after all.
May
7
comment Prove that an open interval and a closed interval are not homeomorphic
Why is $(0,1)$ not compact? Its compactness needs to be evaluated relative to its induced topology, not some unrelated superset like $\mathbb{R}$.
May
6
comment Prove that an open interval and a closed interval are not homeomorphic
Didn't you make assumptions about how the image looks like?
May
6
comment How come Stone-Weierstrass theorem does not imply that in a given interval every continuous function has a power series expansion?
I'd also note that Stone's extension to the Weierstrass Polynomial Approximation theorem generalizes its conclusions beyond the scope of polynomials anyway. Any point-separating real continuous function algebra on any compact $T_2$ space suffices for the approximation (complex algebras must also be self-adjoint).
May
6
answered Prove that an open interval and a closed interval are not homeomorphic
May
3
comment Show that the functions $m(x) = \inf_{a\leq \xi \leq x}{f(a)}$ and $M(x) = \sup_{a\leq \eta \leq x} {f(\eta)}$ are both continuous from left.
The continuity of the function $f$ is a big assumption which was not stated in the question.
May
3
comment Show that the functions $m(x) = \inf_{a\leq \xi \leq x}{f(a)}$ and $M(x) = \sup_{a\leq \eta \leq x} {f(\eta)}$ are both continuous from left.
If $M(x)$ is not continuous from the left, then $\exists \epsilon >0\forall\delta...$, which is different from what you said. Further, $M$ is monotone increasing, not decreasing.
May
3
comment Show that the functions $m(x) = \inf_{a\leq \xi \leq x}{f(a)}$ and $M(x) = \sup_{a\leq \eta \leq x} {f(\eta)}$ are both continuous from left.
How would you show that the Heaviside theta is continuous from the left at 1? It is not.
May
3
comment Show that the functions $m(x) = \inf_{a\leq \xi \leq x}{f(a)}$ and $M(x) = \sup_{a\leq \eta \leq x} {f(\eta)}$ are both continuous from left.
But do you agree that by your definitions for the Heaviside function $M(x)=0$ whenever $x<0$ and $M(1)=1$, so we do not have left-continuity, so the question is ill-formed?
May
3
comment Show that the functions $m(x) = \inf_{a\leq \xi \leq x}{f(a)}$ and $M(x) = \sup_{a\leq \eta \leq x} {f(\eta)}$ are both continuous from left.
Take $f$ to be the Heaviside Theta on $[-1,1]$, which is 0 for $x <0$ but $1$ for $x\ge 0$, both defined and bounded. It seems that $M(x)=f$, but is clearly not continuous from the left at 1. Perhaps you have flipped the directions?