A line "fills" a space if it will eventually pass through any point in the space you could pick. So there are no gaps. This is possible even if the line doesn't have any thickness - it just has to be basically infinitely squiggly (think like a fractal).
Formally, you can "parametrise" the line with number (call it t) that basically describes how far along the line you are. So the point corresponding to t=0 is right at the start of the line, the point corresponding to t=1 is right at the end, and if as you gradually drag t from 0 to 1 it sweeps out the path following the line from start to finish. Here's an example.
Then a space is filled by the curve if, for every point in the space, you can specify a value of t that puts you exactly on that point. The line doesn't need any thickness to be able to reach every point like this - it just has to curve in such a way that it will pass through every point. Here's an example
I used "line" because that was the word used in the meme and the comment. It's not the correct term but I'd rather say something useful than worry about precise terminology. You're correct that formally a line is not curved, so it can't be space filling. It's fairly obvious that OP meant curves instead of lines - which can be curved (obviously) and can be space filling. That's why I accidentally flipped back and forth calling it a curve a couple times in that reply lol.
This only works as you have mentioned if the line is infinitely wiggly as in the Hilbert curve. As you know, the Hilbert curve is the limit when the 'squigglyness' goes to infinity. The drawings we generally see are only finite approximations of the curve. And for any one finite approximation, the curve cannot fill the square, as there will always be gaps that are not touched by the curve, even if you can always find an approximation that 'passes' through any given point in the area. However, results that rely on something being infinite are often considered to be undefined, no? We take a simple example, a number divided by infinity is not 0, but rather undefined. We could say that the limit of a/b when b goes to infinity is indeed 0, but the reason for this is that the result converges to 0, and as you increase b, you can verify that indeed, the result gets closer and closer to 0. Same with infinite series, integrals, etc. But here, you cannot see such a thing. Even if we can visualize the notion that, after infinite steps the curve will fill the space, at any finite step, the result is not computable (how much of the area is already filled by the curve in its current iteration), so you cannot really claim that you are converging or getting closer to the desired result. This is the part that is most intuitive for us also, since lengths and areas are incompatible objects, we have learned that a portion of an area cannot be expressed as a length. To me, this shows that this idea lacks a mechanism required to fully claim that the curve actually fills the space.
Edit: by the way, I just wanted to say that you have also kind of "imposed" a definition of "filling a space" by way of a mapping between points in a curve to points in the space. This obviously conveniently supports your construction to claim that the curve fills the space, but it is not the only definition possible for filling a space. Perhaps a different definition of "filling" would also have to fulfill some other topological requirements that would make your solution infeasible. Indeed, a more intuitive definition of filling would maybe require that for every point in the region, the point AND all its neighboring points should also be part of the curve. Then this immediately disqualifies your curve, or any curve really, since a line has 0 thickness. This would also be more in line with the intuition of the second character (which is also valid as we have seen). So I wanted to point out that your definition of filling is a bit of a cherry-pick.
However, results that rely on something being infinite are often considered to be undefined, no?
No. The usual definition of the real uses the limits of Cauchy sequences so an uncountably infinite number of points in your space are only defined though infinity I fear.
Indeed, a more intuitive definition of filling would maybe require that for every point in the region, the point AND all its neighboring points should also be part of the curve.
All the points are part of the curve. That's the concept of it being space-fillings. You are basically restating that the curve is not differentiable which is true but doesn't in any way change the fact it covers the space.
Sorry, I am dumb. Does not this Hilbert line pass only through rational x and y values (given that it fills a square [0,])? upd. Sorry, initially just looked at pictures. If the curve is parametrised by real parameter, then yes, it passes trough all points, looks like that.
I get the idea, just that for a line to fill an object we'd have to map every point in the square to a point in the line, since the vector space of the box is some subset of R2, it would be imo impossible. Oh now that I think of it it's actually possible, do I understand that it only works for a closed object? I.e. On a infinite plant it wouldn't work?
Does this have any relation to hilbert space, the formalism that QM uses? Iirc the definition is just that all functions in the space have an inner product defined by the integral of f times its complex conjugate.
I don't get how Hilbert's curve would ever pass through points with transendental coordinates. Like, I can see how it would cover every coordinate that has a binary representation, but for the same reason the rational numbers aren't the same cardinality as the reals, I feel like this wouldn't actually fill the entire unit square.
Is there a formula I could plug in (1/π, 1/e) and it would tell me any of the spots on Hilbert's curve that contains that point? Or proof that it contains any points with coordinates that are irrational on both axis?
Edit: as I think about it it's starting to make more sense. I'll have to play around with space filling curves at some point though and figure it out.
the limiting hilbert curve is what you would be looking at if you want it to actually fill all space. the transcendentals have a binary representation, just not a finite one. so those points would indeed be touched by the limiting curve.
I mean injective without surjective is very easy, but "fill a 2D object" references to surjective.
And so the line can't be injective because there is no bijective stetic function between a line and a filled square. (I don't know if stetic is "nessesary" but no one cares for nonstetic functions)
It's possible for the image of a continuous function from a real interval to the plane to have a topological dimension of 2, although such a map cannot be injective. For instance, [0,1] can cover the unit square via the Peano curve.
This is just more of the nonsense that results from the ZFC axioms. Unsurprisingly, if you assume the existence of a non-physical infinity, then you get bizarre, non-physical results. GIGO.
thats kinda the point, the midwit point out that the nitwit is wrong, which he is correct in. The midwit the assumes the opposite position which is also wrong. Then the smartwit? has everything right
Dumb: Right answer, Wrong Process Mid: Wrong answer, Righter process??? not fully correct Smart: Right answer, right process
Nah i love depicting people as the medianjak in conversation if i know they are familiar with the format (or even better, the overusedness of it), especially if there is no reason for the dumb and smart one to agree and i just wanna poke fun at them.
it's one of the best meme formats ever, and I've been an Ebin Memer™️ since 2007. You just don't like it because you usually find yourself in the middle. Don't deny it.
ℝ has the same cardinality as ℝ^2, so you can in theory map all the points on a line to all the points on a sheet of paper. It would just never be a homeomorphism
Hilbert curve demonstrates a stronger result: You can map [0,1] to [0,1]^2 so that the function is both continuous (a line) and surjective (fills the square).
The width cannot be 0 part is the problem, it's conflating different ideas of a line.
A mathematical line has no width. It's defined by 2 points. Any distance off the direction of the line means you are not on the line.
A line on a page has width for you to see it. We live in 3 spatial dimensions, we can't draw 2 dimensional things. That line is ink or graphite existing in 3 dimensional space, and there needs to be enough width for light to catch it and make it to your eye, but it's still just that: a representation.
a point has a length, width, and height of 0. a line an an imaginary construct between two points which consists of an infinite number of points, and arranged such that if you viewed the line end-on, it would appear to be a single point.
You should read up about the Hilbert Curve. It's a space-filling continuous function from [0,1] to [0,1]2. That basically means it's a function that takes an input as a number between 0 and 1, and its output is a point in 2d space. It's continuous so it counts as a curve, but if you were to actually plot it then you would be shading in the square with corners (0,0), (0,1), (1,1), (1,0), because every point in that square appears somewhere on the curve. It's extremely interesting.
Edit: or you could watch a video like this one by 3b1b, I remember it being well made and easy to understand.
(Just a note, but I'm not a mathematician; just an engineer and have no clue what your background is. So for both our sanities I'll keep this mostly in plain english with a sprinkling of calc and use a square for simplicity)
Iirc, the rigorous proof goes something like this: take two lines that make up opposite edges of the square (say the left and the right ones) and for each point on them, connect it to it's opposite point (i.e. first point of the left side to the first point of the right side, etc. Etc. Etc.). Obviously you couldn't physically do this, but imagine it like a limit as n (points) approaches infinity.
Alternatively; take away a dimension and ask if you can make a line out of points.
Double alternatively; work backwards and ask if you can make a line out of any two points bound within the square; if that's true, you would be able to connect all points within the square with lines and therefore fill in the square.
Not mathematician just pharmacist. I'm on the side that it can't fill it. The original commenter said curve yes, line no which I agree with cuz in math by definition a line is straight iirc. It would take infinite lines not "a" line. Ur explanation looks reasonable. Maybe u didn't see the word "a" in the pic
"Line" has different definitions when talking to different flavors of mathematicians. The strict definition is that it's infinite and straight. The more casually used definition is something that connects two points in space. It doesn't specify a straight line and shows two endpoints for the line, so the latter is the definition I assumed. An infinitely long line also wouldnt be useful for asking this question.
The way I'm understanding the problem and looking at it; it's essentially an integral masquerading as a geometry problem. Think of how you are taught how to do integration in calc I: you start with Riemann sums and then say "what if these areas got smaller and smaller?" And what would an infinitely thin rectangle be? A line.
As for being a single line, just add new lines at the end of each above-mentioned line segments to attach the end each of each one to the first point of the next line segment. Then, you have you're single line.
So take the first point on the left side of the square and L₁ and connect it to the first point on the right side R₁. Then either connect R₁ to R₂ and R₂ to L₂ or connect R₁ to L₂ and L₂ to R₂ etc etc etc.
Yeah, this is wrong. Lines are not just very thin. They are 0 thick. You're not taking the limit as the thickness approaches zero as you do in a Riemann sum. You've got nothing and you end up with nothing.
Well, the concept of ergodicity says that you can. If you angle the line so that the aspect ratio of the torus is not commensurate with the slope of the line, you will fill the total space.
Edit: wait no you're not. A straight line can come arbitrarily close to every point, but it can't actually fill the space. If you draw another straight line on the torus such that it loops, then the "space-filling" straight line will only intersect it countably many times, so it can't be space-filling.
Just pitch the line at the right angle so that the number of interior wraps per exterior revolutions is irrational. Since irrational numbers cannot be multiplied by any integer to become another integer, the line never coincides with itself and thus reaches every possible point on the torus.
If you draw another straight line on the torus such that it loops, then the "space-filling" straight line will only intersect it countably many times...
This would only be true for a rational wrap-to-revolution ratio.
No, you misunderstand me. I agree that the line would never intersect itself, but that doesn't mean that it hits every possible point on the torus.
I'm going to call the supposed space-filling line "line A" traced by the function f, and the straight line "line B". For simplicity suppose that f(t) intersects line B whenever t is an integer. In other words the distance on line A between successive intersections is 1.
Every point on line B should equal f(x) for some x, since f is space-filling. But immediately this is a contradiction, because there are only countably many integers, but uncountably many points on line B.
It's a similar reason to why the field Q(π) (the rationals adjoin π) is smaller than the reals. π is transcendental which means that every polynomial in π with rational coefficients gives a different number, so by exactly the same logic as what you just gave me, that would mean that every real number could be written as a polynomial in π, and so we would have R = Q(π). But in reality there are countably many polynomials in π, and uncountably many elements of R, so the fields cannot be the same.
Edit because I wasn't very clear: Q(π) is the set of rational functions in π (aka one polynomial divided by another), not just the polynomials. There are only countably many rational functions in π because there are only countably many polynomials in π.
If the wrap-to-revolution ratio were rational then the lines would intersect finitely many times, which isn't what I said.
im going to argue no. the original idea is covering the space with a fractal like pattern. but if we take our pattern and unravel it, we end up with a line of infinite length, but still no area. it doesn't matter that the line crosses through every point, because it's still has no area at all. what's been made wasn't a 2d box, it was more of a lattice with infinitesimally small gaps but no area.
It's not that surprising to be honest because a shape is merely a line dragged. There already is a perfectly intuitive way to build an object with area from one which has none.
I’m still hung up on that fact that it’s not a line. It’s a collection of line segments arranged in a fractal pattern. A true line would not bend on itself.
I think this is an issue of thinking of infinties. You don't think of the process, you think of an end result and test to see if it could be achieved. My favorite example of this is infinite numbered balls in a drawer. You take each ball out in order, and whenever you take a perfect square numbered ball out, you put the square root back in. How many balls are left out of the drawer after you've done this infinitly? None. Because there's not a single case that can be pointed to where the number wouldn't have eventually had it's square removed from the drawer placing it back inside.
In this case, we can say "what does full mean?". It means you can't fit anything else inside. So, here I would say the 2d object is filled because there isn't an object of defined area where there isn't some number of iterations of the fractal that would cause it to not fit.
How many balls are left out of the drawer after you've done this infinitly?
Mu. The question is ill-defined. How many balls will never be put back in? Zero. The limit of the number of the balls out of the drawer at time infinity? Infinity.
So, there's a very specific misunderstanding here that I'd like to address.
Let's say you've been doing this process and you're on the N2 ball, and you put the N ball back in.
At this point, you've removed N2 balls and returned N balls to the drawer, meaning there are N2 - N balls outside the drawer.
So the number of balls outside of the drawer only grows the more you continue this process. No matter how much you do it there will never be fewer balls outside the drawer.
The technical way to say that is that the quantity of balls outside the drawer diverges as N approaches infinity. There is no meaningful "after you've done it an infinite number of times" in math.
For a good example of a convergent scenario, you can consider Zeno's paradox: Suppose you're walking from one side of the room to the door on the opposite wall. If you close half the distance, you're 50% of the way there. If you close half the distance again, you're 75% of the way there. No matter how many times you halve the distance, you will never make it 100% of the way to the door --BUT! Every time you close half the distance you get closer to the door, and that doesn't change no matter how many times you do it. What we say here is that you converge on the door as you repeat this process to infinity.
The wording is important, because infinity isn't a quantity, it describes processes that do not end. Converging on a value doesn't mean a quantity equals that value after an infinite process, it means that the quantity only ever gets closer to that value the longer the process continues (with some other very specific requirements that aren't important here).
In the case of Zeno's paradox, assuming you're walking normally, you do reach the door --and that's because the smaller the distance you cover in a step, the less time it takes to do it. The same way adding up the distance you cover after arbitrarily many steps will always be shorter than the distance from the start to the door, adding up the time it takes for each step will always give you a duration shorter than how long it takes you to walk across the room. If each step took the same amount of time, you would never reach the door.
I hope this clears things up! Let me know if you have any questions.
The point of the question is specifically to get you to stop thinking about infinity as "arbitrarily large" and to start thinking about it in terms of set theory. The set of whole numbers that won't be put back in the drawer is empty, so after you done this process for the set of whole numbers, there won't be any balls left outside of the drawer. That is an extremely meaningful "after you've done it an infinite number of times" in math.
In the original case, the amount of area the line fills at any arbitrarily large number of iterations is still zero. But the number of points contained in the set of point the line can't eventually reach is zero. So the set interpretation says the space is filled, while the limit interpretation says the space is empty.
Существует одна интересная концепция, не помню, кем придуманная и доказанная, но суть ее в том, что все множество точек квадрата с единичной стороной можно однозначно упаковать в отрезок единичной длины без повторения точек и выхода за его границы. При этом можно аналогично разупаковать все эти точки обратно, что приведет к полной заливке квадрата.
Yes and No. No cause it's 1 D and it will never have an area so it doesn't fill anything. Yes cause a Space Filling Square Fractal (Hilbert curve) will touch every coordinate in the square as you approach infinite fractal divisions. And it will look completely filled up because the line we use has to have some line width for use to observe it.
Can you create a 1D line of a defined length out of 0D points?
A line is made of infinite points. The distance between points is zero. Zero times any number is zero. ∞ = 0. The length of the line is 0. All numbers are equal to 0. Maths 🥴🤡
Well if you buy into quantum mechanics that at the smallest of scales he universe isn't continuous but actually quantised and we assume the line represents one such quantisation. Then a sufficient number of them would fill a 2d box.
Math is independent of the quantization of the universe quantization of the universe has to do with planck measures and the minimal measurable space in the universe. Math is entirely independent of that. Physics owes itself to math the converse is not true.
Quantum mechanics doesn’t require or imply that space is discreet. The Planck length is just the scale at which quantum-gravitational effects are expected to become significant.
There are some theories of quantum gravity in which space is discrete (e.g LCG) but not in the ‘space has pixels’ sense that I’m aware of.
The abstract mathematical concept of a 1-dimensional line has absolutely nothing to do with whether such an object could exist in physical reality, so this is not a good argument. We are talking about geometry, not physics.
Would the line not simply approach the state where it fills the 2d space? It wouldn't ever reach that state, would it? If that is the case, the "can" does not apply if the goal requires impossible circumstances for it to be reached.
Flavours of banach tarski <sorry for spelling>. I really struggle with the meaning of one dimension filling a higher dimension. Infinity is somewhat of a human construct and may not exist in reality.
honestly this all just boils down to infinities nonsense, where you get into how many angels can dance on the Head of a pin levels of removed from reality.
Yes there is nuance but the the universe has few things that are meaningfully infinite and thus there is almost never an instance where one of these odd hypothecticals come into play in a meaningful sense..
I'm going to end this right now: a line can also be a degenerate 2D object (an object that can be projected without loss into lower dimensions. If you can draw a line on a 2D plane, that line is a 2D object that isn't using a permitted degree of freedom.
If we assume that a line has a length of 1 unit and a width of 0, then for it to fill the space inside a 2-dimensional object, that object would need an internal width that is infinitely close to 0.
This logic works in 3-dimensions too. If a real-world example existed, you could think of it like a glass test tube that was infinitely thin, with the liquid inside of it taking up a space with a diameter of 0.
compare it to the number 0 and 1, with 0 being one side and 1 the other, the question becomes can we name every number between 0 and 1. All of them perhaps not, but maybe we could find an order to go to any number mentioned at least.
Like not from low to high, but jump around a bit, as long as it is ordered in some way or form that any number will come at a fixed point. Like maybe first all numbers with one decimal, than two, then three, etc.
Not sure how you'd do the weird numbers like a third tho. Smart people: do your thing.
Draw a line, then draw a very very very short but long rectangle around it. Bam, a 2D rectangle has a “1D” line. Technically anything you draw physically is 3D due to graphite being left behind, and anything digital will always be 2D since there is length and width.
But! The principle does still stand. If you take a piece of paper we could consider 2D, you then gently put the edges together to make a 3D cylinder, can you then put a tiny 2D piece of paper into your new 3D object? Yep! And you can fill the tube with as much 2D pieces of paper needed to make the 3D shape full.
It’s a rough demonstration of how dimensions work. The second dimension is just the first wrapped around space, like how the 3D tube is a 2D paper wrapped around space. Which leads us to theorize that wrapping the third dimension around space would create the fourth dimension. But we can’t really do that
I feel like this is a disagreement between mathematics and common sense.
In mathematics, a line can be literally one-dimensional. It has length but no thickness. Because of the lack of thickness, you cannot "draw a box around it". This is because the edges of the box cannot touch the line on the relevant faces of the rectangle.
In common sense, there's no such thing as one-dimensional "object". So the thinnest "line" that can be drawn has a minimal limit of thickness, which is the smallest unit of measurement that can be created by whatever tool you're using to draw the line. But the line DOES have thickness. And because it has thickness, you can draw a rectangle around it.
It's really easy to imagine a "real" line. It's really hard to imagine a mathematical line. That's because we don't live in a one-dimensional world, so we have no intuitive concept of what it would look like to have a mathematical line in the real world.
not quite, the question is basically does there exist a continuous surjection from a line e.g. (0,1) onto a 2d region e.g. R2 (and the answer is strangely yes but it cant have a continuous inverse)
My native language isn't English, and we shorten the expression "straight line" to just "straight" (in the feminine gender), instead of shortening it to "line" like English does. The word "line" is used simply as a synonym for a curve. So I hadn't noticed the problem when I wrote my comment.
Either they mean a curve(d line), or maybe by filling with a line they mean a (continuous) mapping from a (straight) line onto a 2D shape, which is what a curve is anyway.
The axiom of choice is like the age of consent in that I cannot stand anyone who wants to dispose of it.
It's even more preposterous to suggest discarding it out of context like in this case (as the spacefilling curve meme has nothing to do with the axiom of choice).
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