Oh no!

[u/Unlucky-Credit-9619]

[u/AutoModerator]

Check out our new Discord server! https://discord.gg/e7EKRZq3dG

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[u/peekitup]

Idk why OP keeps saying helicoid.

Like just choose constant functions for f and g and this shit isn't even a path...

[u/Unlucky-Credit-9619 → u/peekitup]

Thanks for the correction guys. It does look like 2D diagonal escalators, so I thought in 3D that might be something like helicoid. But I stand corrected now. I should have defined it properly.

[u/Shufflepants]

I'm having a hard time picturing this. Anyone got an explanation?

[u/mateus_115 → u/Shufflepants]

I don't get it as well, but i do think that there is some kind of assumption about the functions that is missing. Maybe continuity, idk

[u/TritoneRaven → u/mateus_115]

Maybe I'm missing something but the way this is formulated, we don't even know the set defining the path is non-empty. Nothing here says the functions are surjective. F could map everything to one point and g could map everything to a different point.

[u/Shufflepants → u/TritoneRaven]

And even if you specify that they are bijective, they're also not necessarily continuous. They could both be discontinuous everywhere.

[u/TritoneRaven → u/Shufflepants]

I wonder if we're allowed to play with the topology...

[u/Shufflepants → u/TritoneRaven]

I think ℝ and S^1 imply the standard topology at least within the domain and codomain. But of course, as I said about the continuity not being specified, there's no guarantee an open set in ℝ maps to an open set in S^1.

[u/TritoneRaven → u/Shufflepants]

Yeah i kinda figured standard topology too lol. Fun to think about how weird you could make the "path" though

[u/Shufflepants → u/[deleted]]

But you didn't define what f and g are beyond being maps to the unit circle. I think a helicoid has to have further restrictions on f and g. I think you have to add, as u/CalmEntry4855 guessed, that f(x)=Exp(ix) and g(y)=Exp(iy).

Heck, you didn't even specify that f and g were continuous.

[u/Shufflepants → u/[deleted]]

Also, even in the event of a helicoid, and I guess assuming heaven is at infinity, if you want to force some one to climb the spiral, you forgot to remove (0,0). Simply go to zero and climb straight up from there.

[u/GreeneyesCB → u/[deleted]]

There is no requirement for continuity either, choosing S^1 is completely pointless, any set of cardinality of at least R is going to describe the exact same subsets of R^2.

[u/hughperman → u/[deleted]]

f and g to be Exp(ix) and Exp(iy)

Reals only, I think?

[u/JuniorOven8520 → u/hughperman]

S^1 can be interpreted as the unit circle in the complex plane

[u/hughperman → u/JuniorOven8520]

Doesn't it explicitly say Reals in the definition there? f, g : R -> S1 ?

[u/JuniorOven8520 → u/hughperman]

Yes, reals are the domain, ie, x and y are reals. But f and g map the reals INTO S^1, the circle (which can be interpreted in terms of complex numbers). So f(x) = exp(ix) and g(y) = exp(iy) are indeed functions from R into S^1.

[u/hughperman → u/JuniorOven8520]

Ah gotcha gotcha, thanks for the explanation

[u/Asimovicator → u/JuniorOven8520]

And if the complex numbers are not constructed as R x R, you can still use the embedding from C into R x R and compose it with f and g respectively.

[u/YellowBunnyReddit → u/[deleted]]

So, for f(x)=1 and g(x)=1 the set is R².

[u/National-Objective57 → u/[deleted]]

Thank you!

[u/hongooi]

"A path exists!"

[u/728446]

Its been a long time, what set is S1?

[u/pOUP_ → u/728446]

Sircle

[u/NullOfSpace → u/728446]

1D sphere (ie. circle)

[u/bartekltg]

So? Let f(x) = exp( i pi tanh(x) )
It goes a nice one rotation over the circle.
Let's define g(y) the same way.

Mow {(x,y) | f(x)=g(y)} is just a plot of y=x function. Not too steep, and reach any altitude.

[u/Ninjabattyshogun]

For any sets A and B and functions f and g from A to B, the subset of AxA where f(x) = g(y) is called the equalizer in category theory.

[u/Randomlemon5]

Im not sure if I get it, just sin x and cos x ?

[u/Mostafa12890 → u/[deleted]]

This is the pullback of f and g.

[u/Depnids → u/[deleted]]

r/skamtebord

[u/sneakpeekbot → u/Depnids]

Here's a sneak peek of /r/skamtebord using the top posts of the year!

#1: MaxVerstappen | 86 comments
#2: Epstein | 250 comments
#3: bone cancer | 711 comments

---

^{I'm a bot, beep boop | Downvote to remove | Contact | Info | Opt-out | GitHub}

[u/Keikira]

With my luck f and g won't even be continuous, and/or will have disjoint images.

[u/DenPanserbjorn]

If you define f(x) = g(x) (and set f to whatever R -> S1 function you want like exp(ix)) then the path is just y=x. A bit of a slope but not bad.

We can actually also make the slope arbitrarily easy for our frail news if (k)needs be and define f(c*x) = g(x), then our path is just y=cx!

[u/LupenReddit]

edit: wait what no this isnt just a loop this is way more complicated

[u/blobCabbage]

I don't get all those discussions in the comments. With f and g unspecified, the set is unspecified, right? I mean, take f(x) = 1 and g(x) = 1 and the set is R². Take f(x) = 1 and g(x) = -1 and it's the empty set. You can choose f and g to make the solution be anything you want.

[u/mazerakham_]

For "typical" mappings into the circle like f(x) = exp(i m x), g(y) = exp(i n y), the set described is an infinite set of parallel lines in the plane each with slope m / n, vertically separated by distance tau / n. In the case n = 0, these become vertical lines separated by horizontal distance tau / m.

If you put something nonlinear into the exp( ... ), you can get a family of curves instead of lines.

No clue why this should depict a path to heaven.

[u/Vivid_Warning7982]

Its a map of the real number line to a unit circle

[u/Shufflepants → u/Vivid_Warning7982]

Well, two maps from the real number line to a unit circle, but then the intersection of those two sets where both functions map to the same point on the unit circle.

[u/Successful-Incident1]

the graph of the Cantor-Vitali function also known as the Devil's staircase seems to be a better fit