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When I first learned about limits (my senior year in high school), I visualized a "step" shape graphed (0, 0) to (0, 1) and then to (1, 1). Up one unit, right o

When I first learned about limits (my senior year in high school), I visualized a "step" shape graphed (0, 0) to (0, 1) and then to (1, 1). Up one unit, right one unit. (I know, not technically a function since it has a vertical component.) But now you subdivide both segments and put in an extra step. Up 0.5, right 0.5, up 0.5, right 0.5. Same starting point, same ending point. Even the same length of line segments as before — but the area bounded beneath is 3/4 the original. I reasoned that if you continued this subdivision, as you approach infinity the area under the "curve" approaches that of a half-unit right triangle (i.e. root-two). How is it, I asked my math teacher, that this shape can have the perimeter of a square but the area of a triangle? I don't recall the answer from the teacher. Maybe somewhat hand-wavy; ultimately unsatisfactory to me. I still don't really know the answer. I suppose the answer is that it is not in fact ever a triangle — just a thing with a fat edge.

aqme28 2y ago
Reminds me of Gabriel’s Horn, which has an infinite perimeter but finite area.
wat10000 2y ago
The answer is that the perimeter “at infinity” matches the triangle. There’s nothing that says the perimeter of the final “infinite” shape has to equal the limit of the perimeters as they go to infinity. In order for this to work, you have to prove that they’re equal. Typically you’d come up with two limits, one which you prove is always greater than the value in question, and one which is always less. Then you show that they’re equal in the limit. That squeezes the number you’re interested in, proving that it’s also equal to that number. For example, the typical proof of the circumference of a circle involves making two polygons, one inside the circle and one outside. The perimeter of the one inside is always less than the circumference, and the one outside is always greater. Then you take the limit of the polygon perimeters as the number of sides goes to infinity. They converge on the same number, 2 pi r.
- pif 2y ago
  Sorry, you are completely wrong. The perimeter is equal to the square one at every iteration, thus its limit towards infinity is still the square one.
  - wat10000 2y ago
    Huh? I never said otherwise. I said that the perimeter of the limit doesn’t have to equal the limit of the perimeter. In order for this to work, you have to bound the value from above and below and show that they meet at the limit. This only bounds it from above. Obviously you’ll never find a sequence that bounds the perimeter from below that has a limit equal to the square one, so you’ll never be able to prove this equality.
    - pif 2y ago
      > This only bounds it from above AND from below. At every iteration, the two perimeters are the same.
      - wat10000 2y ago
        What two perimeters? There’s only one. And it hasn’t been shown to bound the triangle’s perimeter from below. It is plainly greater than the triangle’s perimeter at every step, so it just establishes an upper bound. To derive the triangle’s perimeter, you also need another thing that’s always less than the triangle’s perimeter, then show that the lower and upper bounds converge on the same value.
- deleted 2y ago
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roenxi 2y ago
You may be referring to the https://en.wikipedia.org/wiki/Staircase_paradox https://en.wikipedia.org/wiki/Staircase_paradox. As for how a shape can have the perimeter of a square and area of a triangle, it just can. As you've identified, it is easy to construct. It might be more interesting to ask what combinations of perimeter and area aren't possible.
- JKCalhoun 2y ago
  Yes, this is what I "re-invented" (I didn't know until now that it had a name). I saw it later in a math book that I flipped through on a co-workers bookshelf (but they didn't name it there either). Also, very cool, followed your link and they mentioned a 3D analog of sorts: https://en.wikipedia.org/wiki/Schwarz_lantern https://en.wikipedia.org/wiki/Schwarz_lantern
- madcaptenor 2y ago
  You can always add more perimeter, so you want a lower bound on the perimeter given the area. The isoperimetric inequality (https://en.wikipedia.org/wiki/Isoperimetric_inequality https://en.wikipedia.org/wiki/Isoperimetric_inequality) says that the smallest perimeter for a given area is that of a circle.
lordnacho 2y ago
The thing is, each time you do a fold, you don't get any more of the triangle's perimeter covered by the staircase. Not only that, every point that ends up on the triangle's line segment is adjacent to a point that isn't: there's always some little epsilon that you can step along, and you are no longer on an overlapping point. You've made no difference to the perimeter. When it comes to the area, this is not true. Folding the triangular segments over the smooth line moves an area on which you can slightly move to one side or another, and you are still in an overlap. Gotta wonder whether the sawtooth is actually a triangle. Each step makes it... not any closer to a triangle. Why would it be a triangle when you keep going to infinity?
- JKCalhoun 2y ago
  > The thing is, each time you do a fold, you don't get any more of the triangle's perimeter covered by the staircase. Not disagreeing with that. > When it comes to the area, this is not true.… Not following your point. And this part is the stickler for me: I believe the area does in fact approach the area of the triangle as the steps approach infinity. Is this not the case as you see it? Perhaps my "stickler" is not really one. To your first point, it's not really a triangle in the end, even as you approach infinity (though I still believe the area becomes that of said triangle). And there is nothing in geometry that says you can't have a shape with the area of a triangle but with a different perimeter. It's just this "shape" looks an awful lot like a triangle but with a much larger perimeter.
  - lordnacho 2y ago
    > Is this not the case as you see it? Yeah I see it your way. Perimeter doesn't change, but the area does. But it's not a triangle. For instance, the internal angles don't add up to 180, and they never tend to either as you fold more and more.
  - sfink 2y ago
    Hm, it doesn't seem problematic for the area to be lower than expected for a given perimeter. You could do the same thing by having a triangle (an actual triangle), but then at some random point have the bounding line shoot off in some random outward direction, then come back (increasing the perimeter without enclosing any area at all). It's no more or less a triangle than your fractally jagged triangle, and it has the same perimeter and area. I think the lesson for me is that any part of any line, if you zoomed into it enough, could actually be found to be all wadded up and much longer than you would otherwise expect. A line from point A to point B must be at least as long as the straight line from A to B, but could be any length greater than that. The only thing that makes your jaggy line special is that the distance from any point along it to the nearest point on the straight (triangle's) line is zero.
praptak 2y ago
To me it's more interesting that every one of these curves has length 2 and yet the limit has length of sqrt(2). (many properties are not preserved under limit but it's still somewhat interesting)
constantcrying 2y ago
The perimeter (which is something very close to the derivative) is not continuous. If you think about it a bit, you can easily change a shape in a way that greatly increases its perimeter but leaves it's area (almost) unaffected. Which means that just because two shapes are extremely close in shape, they can be very different in perimeter. What is interesting is that it is semi continuous. The limit (this isn't quite correct) of the series is larger than the point. As the limit of the series is 2, but the actual distance is sqrt(2), which is smaller. So if you calculate the limit of the staircase perimeter you only know that it is larger or equal to the perimeter of the limit. >I suppose the answer is that it is not in fact ever a triangle — just a thing with a fat edge. It absolutely is a triangle. Because it converges in shape.
fiforpg 2y ago
Nice example. You can avoid the issue with these being non-functions (because of the vertical segments) by rotating the picture so that the (0,0)-(1,1) diagonal is horizontal. That way you are talking about a sequence of continuous functions F_n with slopes alternating between ±1 that approach the constant-zero function F_0=0. The reason your intuition was confused was that you felt that there was a way in which these functions with ±1 slope converge to zero function (it is called uniform convergence), but were unsure about the way in which their derivatives f_n=(F_n)' (which keep flipping between ±1) converge to zero. The resolution of the seeming difficulty is that there are different modes of convergence. The ±1 slope functions F_n converge uniformly to zero, but their derivatives f_n, which determine curve length, do not converge uniformly or even pointwise to any limit (uniform convergence is stronger — more restrictive — than pointwise). This is why it is reasonable that the F_n curve length stays 2 throughout the exercise. Note: if the derivatives f_n were converging to f_0=(F_0)'=0 pointwise while staying bounded, then the curve lengths of F_n would converge to sqrt(2), the curve length of F_0. This is called "dominated convergence theorem" for integrals — point being that the curve length of F_n is an integral of sqrt(1+f_n^2). But there is no pointwise convergence, and no such implication. Finally, you may be curious as to whether there is a sense in which f_n converge to f_0=0 — does all that flipping back and forth amount to some way of converging to zero? Turns out yes! It is called weak* (weak-star) convergence and it applies to f_n if you think of them as measures (or distributions) — that is, if instead of pointwise evaluation you characterize f_n by the way they act on test functions by integration: instead of computing f_n(x) you multiply f_n * phi and integrate. Phi has to be a continuous function. Under this notion of evaluation, f_n do converge to zero. Edit: typos.
whatshisface 2y ago
What you're describing has a simple explanation. It's possible to fit more edge length in less space by scrunching it up.
pavel_lishin 2y ago
3Blue1Brown actually made a video tangentially explaining this particular paradox: https://www.youtube.com/watch?v=VYQVlVoWoPY https://www.youtube.com/watch?v=VYQVlVoWoPY
deleted 2y ago
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aqme28 2y ago

Reminds me of Gabriel’s Horn, which has an infinite perimeter but finite area.

wat10000 2y ago

The answer is that the perimeter “at infinity” matches the triangle. There’s nothing that says the perimeter of the final “infinite” shape has to equal the limit of the perimeters as they go to infinity. In order for this to work, you have to prove that they’re equal. Typically you’d come up with two limits, one which you prove is always greater than the value in question, and one which is always less. Then you show that they’re equal in the limit. That squeezes the number you’re interested in, proving that it’s also equal to that number. For example, the typical proof of the circumference of a circle involves making two polygons, one inside the circle and one outside. The perimeter of the one inside is always less than the circumference, and the one outside is always greater. Then you take the limit of the polygon perimeters as the number of sides goes to infinity. They converge on the same number, 2 pi r.

pif 2y ago
Sorry, you are completely wrong. The perimeter is equal to the square one at every iteration, thus its limit towards infinity is still the square one.
- wat10000 2y ago
  Huh? I never said otherwise. I said that the perimeter of the limit doesn’t have to equal the limit of the perimeter. In order for this to work, you have to bound the value from above and below and show that they meet at the limit. This only bounds it from above. Obviously you’ll never find a sequence that bounds the perimeter from below that has a limit equal to the square one, so you’ll never be able to prove this equality.
  - pif 2y ago
    > This only bounds it from above AND from below. At every iteration, the two perimeters are the same.
    - wat10000 2y ago
      What two perimeters? There’s only one. And it hasn’t been shown to bound the triangle’s perimeter from below. It is plainly greater than the triangle’s perimeter at every step, so it just establishes an upper bound. To derive the triangle’s perimeter, you also need another thing that’s always less than the triangle’s perimeter, then show that the lower and upper bounds converge on the same value.
deleted 2y ago
[deleted]

roenxi 2y ago

You may be referring to the https://en.wikipedia.org/wiki/Staircase_paradox https://en.wikipedia.org/wiki/Staircase_paradox. As for how a shape can have the perimeter of a square and area of a triangle, it just can. As you've identified, it is easy to construct. It might be more interesting to ask what combinations of perimeter and area aren't possible.

JKCalhoun 2y ago
Yes, this is what I "re-invented" (I didn't know until now that it had a name). I saw it later in a math book that I flipped through on a co-workers bookshelf (but they didn't name it there either). Also, very cool, followed your link and they mentioned a 3D analog of sorts: https://en.wikipedia.org/wiki/Schwarz_lantern https://en.wikipedia.org/wiki/Schwarz_lantern
madcaptenor 2y ago
You can always add more perimeter, so you want a lower bound on the perimeter given the area. The isoperimetric inequality (https://en.wikipedia.org/wiki/Isoperimetric_inequality https://en.wikipedia.org/wiki/Isoperimetric_inequality) says that the smallest perimeter for a given area is that of a circle.

lordnacho 2y ago

The thing is, each time you do a fold, you don't get any more of the triangle's perimeter covered by the staircase. Not only that, every point that ends up on the triangle's line segment is adjacent to a point that isn't: there's always some little epsilon that you can step along, and you are no longer on an overlapping point. You've made no difference to the perimeter. When it comes to the area, this is not true. Folding the triangular segments over the smooth line moves an area on which you can slightly move to one side or another, and you are still in an overlap. Gotta wonder whether the sawtooth is actually a triangle. Each step makes it... not any closer to a triangle. Why would it be a triangle when you keep going to infinity?

JKCalhoun 2y ago
> The thing is, each time you do a fold, you don't get any more of the triangle's perimeter covered by the staircase. Not disagreeing with that. > When it comes to the area, this is not true.… Not following your point. And this part is the stickler for me: I believe the area does in fact approach the area of the triangle as the steps approach infinity. Is this not the case as you see it? Perhaps my "stickler" is not really one. To your first point, it's not really a triangle in the end, even as you approach infinity (though I still believe the area becomes that of said triangle). And there is nothing in geometry that says you can't have a shape with the area of a triangle but with a different perimeter. It's just this "shape" looks an awful lot like a triangle but with a much larger perimeter.
- lordnacho 2y ago
  > Is this not the case as you see it? Yeah I see it your way. Perimeter doesn't change, but the area does. But it's not a triangle. For instance, the internal angles don't add up to 180, and they never tend to either as you fold more and more.
- sfink 2y ago
  Hm, it doesn't seem problematic for the area to be lower than expected for a given perimeter. You could do the same thing by having a triangle (an actual triangle), but then at some random point have the bounding line shoot off in some random outward direction, then come back (increasing the perimeter without enclosing any area at all). It's no more or less a triangle than your fractally jagged triangle, and it has the same perimeter and area. I think the lesson for me is that any part of any line, if you zoomed into it enough, could actually be found to be all wadded up and much longer than you would otherwise expect. A line from point A to point B must be at least as long as the straight line from A to B, but could be any length greater than that. The only thing that makes your jaggy line special is that the distance from any point along it to the nearest point on the straight (triangle's) line is zero.

praptak 2y ago

To me it's more interesting that every one of these curves has length 2 and yet the limit has length of sqrt(2). (many properties are not preserved under limit but it's still somewhat interesting)

constantcrying 2y ago

The perimeter (which is something very close to the derivative) is not continuous. If you think about it a bit, you can easily change a shape in a way that greatly increases its perimeter but leaves it's area (almost) unaffected. Which means that just because two shapes are extremely close in shape, they can be very different in perimeter. What is interesting is that it is semi continuous. The limit (this isn't quite correct) of the series is larger than the point. As the limit of the series is 2, but the actual distance is sqrt(2), which is smaller. So if you calculate the limit of the staircase perimeter you only know that it is larger or equal to the perimeter of the limit. >I suppose the answer is that it is not in fact ever a triangle — just a thing with a fat edge. It absolutely is a triangle. Because it converges in shape.

fiforpg 2y ago

Nice example. You can avoid the issue with these being non-functions (because of the vertical segments) by rotating the picture so that the (0,0)-(1,1) diagonal is horizontal. That way you are talking about a sequence of continuous functions F_n with slopes alternating between ±1 that approach the constant-zero function F_0=0. The reason your intuition was confused was that you felt that there was a way in which these functions with ±1 slope converge to zero function (it is called uniform convergence), but were unsure about the way in which their derivatives f_n=(F_n)' (which keep flipping between ±1) converge to zero. The resolution of the seeming difficulty is that there are different modes of convergence. The ±1 slope functions F_n converge uniformly to zero, but their derivatives f_n, which determine curve length, do not converge uniformly or even pointwise to any limit (uniform convergence is stronger — more restrictive — than pointwise). This is why it is reasonable that the F_n curve length stays 2 throughout the exercise. Note: if the derivatives f_n were converging to f_0=(F_0)'=0 pointwise while staying bounded, then the curve lengths of F_n would converge to sqrt(2), the curve length of F_0. This is called "dominated convergence theorem" for integrals — point being that the curve length of F_n is an integral of sqrt(1+f_n^2). But there is no pointwise convergence, and no such implication. Finally, you may be curious as to whether there is a sense in which f_n converge to f_0=0 — does all that flipping back and forth amount to some way of converging to zero? Turns out yes! It is called weak* (weak-star) convergence and it applies to f_n if you think of them as measures (or distributions) — that is, if instead of pointwise evaluation you characterize f_n by the way they act on test functions by integration: instead of computing f_n(x) you multiply f_n * phi and integrate. Phi has to be a continuous function. Under this notion of evaluation, f_n do converge to zero. Edit: typos.

whatshisface 2y ago

What you're describing has a simple explanation. It's possible to fit more edge length in less space by scrunching it up.

pavel_lishin 2y ago

3Blue1Brown actually made a video tangentially explaining this particular paradox: https://www.youtube.com/watch?v=VYQVlVoWoPY https://www.youtube.com/watch?v=VYQVlVoWoPY

deleted 2y ago

[deleted]