Ten boxes are given with $a_1,a_2,a_3,a_4,a_5 ......a_{10}$ number of balls in them respectively .These boxes are randomly ordered but $a_1,a_2 .....a_{10}$ is told.We can arbitrary select a box and guess number of balls in it.If our guess is greater or equal to number of balls in it then we win ...
@MikeMiller I asked the question on mathoverflow and seem to have gotten two conflicting answers: https://mathoverflow.net/questions/336073/possible-isometry-groups-of-open-manifolds
although it appears the one providing a counter example is correct
Problem: If $g$ is strictly monotone on $\Bbb{R}$, and $f$ is a measurable function on the measurable set $E \subseteq \Bbb{R}$, then $g \circ f$ is measurable.
Proof: Since $g$ is strictly monotone, there is a set $D \subseteq \Bbb{R}$ set of measure $0$ such that $g$ is continuous on $\Bbb{R} \setminus D$ (WLOG, assume that $D \subseteq E$). Since $m(D)=0$, $g \circ f$ is measurable on $D$. Now restrict the functions to $E \setminus D$. Let $U \subseteq \Bbb{R}$ be open. Then $g^{-1}(U)$ is open in $E \setminus D$, since $g$ is continuous on $E \setminus D$.
Hence, $(g \circ f)^{-1}(U) = f^{-1}(g^{-1}(U)$ is measurable, because $f$ is measurable by assumption and $g^{-1}(U)$ is an open set (in $E \setminus D$). Hence, $g \circ f$ is measurable on $E \setminus D$, so it is measurable on $(E \setminus D) \cup D = E$.
@user193319 Actually you don't need to worry about the discontinuities. Recall that for measurability it suffices to check if $f^{-1}((a,\infty])$ is measurable for all $a\ge -\infty$. But for a monotone (increasing) function, $f^{-1}((a,\infty])$ is always an interval.
So the preimage by $g$ gives an interval, then use the measurability of $f$...
I don't even think if my security clearance required me to answer questions about criminal record. But probably there was something in it about allowing them to check that themselves.
A security clearance is a status granted to individuals allowing them access to classified information (state or organizational secrets) or to restricted areas, after completion of a thorough background check. The term "security clearance" is also sometimes used in private organizations that have a formal process to vet employees for access to sensitive information. A clearance by itself is normally not sufficient to gain access; the organization must also determine that the cleared individual needs to know specific information. No one is supposed to be granted automatic access to classifi...
TS is above "Secret"
the next level is CTS, compartmentalized top secret
that's technically the highest level but it's stratified from within (indicated by compartmentalized)
@RyanUnger Probably. The level below it is called NATO secret (I think I have that clearance. At least I applied for it, and they have now given me access to the systems I needed it for, so presumably the clearance went through)
for NATO secret it was not that bad. But it did require a full list of all current debt, all stays abroad the past 5 years (not counting short vacations) and past 10 years of employment history, including all periods without work
@RyanUnger Because lie detectors have been scientifically proven to not work properly and be beatable
@s.harp Yes, Misha's answer is correct. Peter Michor was trying the approach I suggested yesterday, and it doesn't work for reasons I explained then about needing closed orbits.
@RyanUnger they do in fact use them sometimes, they legally can’t force you to undergo it, but police do things to subtly violate people’s rights during investigations literally constantly
for the open house they gave us a final list of admits w our chosen fields for the ppl who’d gotten in at that point and only me and 1 other person had GA as their field so there aren’t many of our kind
literally everyone i met at all my open houses was a number theorist
@ÉricoMeloSilva So look at any smooth path $\gamma : I \to \Bbb R^2$ which is short, in the sense $\|\gamma'\| < 1$. I imagine this as a non-vertical path in $\Bbb R^3$ with it's tangent vector lying inside a fixed cone field along the path
You can also say it's the trajectory of a particle through spacetime idk
Roughly speaking this happens because $\|\gamma'\| < 1$ is a convex differential relation (cones are convex). Gromov proved every ample differential relation satisfies the h-principle; every solution which lies in the differential relation can be approximated by solutions which are tangential to the differential relation (goes along the boundary, of which it's the convex hull of)