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21:05
Take a Hamel basis of $\mathbb{R}$ and define a map $f\colon\mathbb{R}\rightarrow\mathbb{R}$ which sends each real number to the sum of the basis elements of which it has non-zero component. Is there a continuous $g\colon\mathbb{R}\rightarrow\mathbb{R}$ such that $f=g$ a.e.?
21:37
@Thorgott no. If $f=g$ a.e., then $f-g$ vanishes outside a null-set, so in particular $f-g$ is measurable, this implies that $f$ is measurable. But there's a theorem that under some assumptions (e.g. locally compact Polish group to Polish group) a measurable homomorphism is continuous. Evidently $f$ is discontinuous
I would appreciate a googlable name or a reference for that theorem if you know one @Lukas
Thanks
That paper looks both accessible and interesting, what a rarity
I've put it on my always lengthening list of things that I should read after I'm done with exams
$f-g$ is Lebesgue-measurable, but not necessarily Borel-measurable, no? I think the theorem need not apply in that case.
Can one have a vector field over the rationals?
21:51
Doesn't Lebesgue measurable imply Borel measurable? What am I missing
@Ultradark You can have vector spaces over any field
@AlessandroCodenotti okay I said vector field though?
I can't read so late into the evening
Lebesgue-measurable means the preimage of every Borel set is Lebesgue-measurable- Borel-measurable means the preimage of every Borel set is a Borel set, which is stronger.
Oh ok, you're changing the $\sigma$-algebra on the domain
Yeah, changing stuff in the codomain would ruin everything (the preimage of a Lebesgue-measurable set under a continuous map need not be Lebesgue-measurable).
21:55
@Thorgott the author explicitly writes before the theorem "Steinhaus eventually proved that any Lebesgue measurable solution is continuous and with the advent of abstract harmonic analysis in the 1930’s, this was extended by Weil to all locally compact Polish groups."
Oh, I was just looking at the theorem below
Thanks a bunch
22:40
The "state" of a Turing machine can be called its "intention"
if we feel like needlessly personifying it (for fun)
It "reacts to its environment" (reads from the strip) and "changes its intentions" (changes state)
Does turbulence happen outside of the navier stokes equations or is it just called chaotic behavior or something
Is $^{q}log(x)$ alternative notation for $log_{q}(x)$? Never seen that before
@Ultradark Don't know. @Micrified Never seen that before, looks weird
I've never seen it, either, @Micrified.
What's the context?
Who would be caring about $\log_q$?
It's the definition of code rate, for an alphabet of length q (GF(2)).
Hold on
22:54
Oh, $\log_q$ sounds plausible, then, but surely the author said he/she would be using that notation.
Yes I think given the context it must be log base.
Yeah, from the final sentence or so it's clear. But weird.
My second idea would've been some weird q-analog like $\dfrac{x^{1-q}-1}{1-q}$, something that becomes the usual log as $q\to1$, but that doesn't work in context
Perhaps the author was taught to write it in that form.
Just some obscure notation
22:57
I still say you should look earlier in the text.
${}^q_q\stackrel q\log^q|C|/n$
Gotcha
21
Q: Why did I learn to write the base of the logarithm differently from the rest of the world?

Carlo BeenakkerIt only occurred to me recently, in connection with this MO posting, that the way I write the base of the logarithm is not shared by the rest of the world. I am Dutch, and I learned at school to write the base as a superscript, $^{a}\!\log b$, rather than as a subscript, $\log_a b$. Searching the...

Well researched!
DogAteMy has too much time on his hands. Don't you have some good homework to do?
maybe can be todays fun fact.
We already had anakhro's fun puzzler (and mine) earlier. I still haven't solved his.
23:14
@anakhro I'm not sure I see why a circle it's the best plan
He's saying that going to the circle and then around is the canonical answer, but that it's not best.
I thought of that immediately and haven't yet figured out how to beat it.
*isn't
not it's
Oh.
You can try my pursuit question, too.
You go out in a spiral
There's an ever-expanding ring of "possible prisoner locations"
You mean for mine.
23:18
and your boat is riding around that ring
Yeah
You need to get just the right spiral, though.
The angular velocity will depend on the radius of the ring somehow
I'm mum.
Do you get something like $\|v\|=\sqrt{\omega^2+5^2}$?
Where $v$ is your velocity and $\omega$ is your angular velocity
I haven't worked it out in decades.
23:20
Which means $10=\sqrt{\omega^2+5^2}$
@tedshifrin look who got into the 100K club
Oh wait no whoops
So what do you mean by angular velocity for a spiral like $e^\theta$?
$\|v\|=\sqrt{r^2\omega^2+5^2}$ I mean
@TedShifrin Project onto unit circle
Hmmm.
Is there a new name somewhere there, skull?
23:22
And $r=5t$, right?
At the very bottom @TedShifrin
But you don't know when the prisoners left, DogAteMy.
OH
What??
OK hold on
Oh, cool, skull. I would have guessed Pedro'd gotten it ages ago.
Is it correct to say that a subspace of a vector space is always a subset, but not vice versa?
23:23
How's it doable then
Yes, @Micrified.
Wait so the set of points where the prisoners might be at any given time is an annulus?
Thank you!
How can we have a curve go through all of its points
Hmmm ... Maybe I'm misremembering the question. Maybe you know when they left, but it's before you can leave.
23:25
OK fine so $r=5(t+C)$ or something
Yeah, that must be right.
Right now I'm more intrigued by anakhro's.
$100=r^2\omega^2+25$ so $\omega^2=75/r^2$
and whatever you can continue that I guess
Hm
Inscribe the 1km-radius circle in a square
If you go in a line across the diagonal of that square, have you eliminated half the possible places the coast might be in?
I don't follow.
Hm never mind actually
I feel like the only possible solution would be to head towards the edge, go around the circle, and at some point move out tangentially from the edge
Does that work?
I think it does actually
Hold on so I end up with $\frac32\pi+2\approx 6.712$
as opposed to the naive $2\pi+1\approx7.283$
Oh, I thought about that earlier, but didn't wrap around the circle far enough.
I need to reread his question. Do I have to end up at the closest point on the shore?
23:31
Like what I'm doing is, starting at the center, going to the edge ($1$), going 3/4 around the circle ($\frac34(2\pi)=\frac32\pi$), and then going one unit away ($1$)
@TedShifrin I think you just need to hit the shore
Yeah, right.
5 hours ago, by anakhro
@Ultradark Suppose you get lost on your boat at sea. You have a device that indicates your distance from coast is 1km, but you have no idea the direction you should head to get to the shore. You know also that the coast is linear, being perpendicular to the line between your location and the coast itself.
Q: what is the minimal length of a path which guarantees you meet the coast?
Is it clear that no value between $\pi$ and $3\pi/2$ does better?
@anakhro $1+\frac34(2\pi)+1=\frac32\pi+2\approx 6.712$, final answer
Guess we need to compute.
23:43
@TedShifrin This is also true if the coast is $1+C$ away and you can see a distance of $C$ away
should we ever need to use this practically
So I did do the computation, but it should be clear somehow without it.
LOL @ use this practically.
I have had to drive in fog.
I guess that means $C\to 0$.
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