1
$\begingroup$

Gabriel's Horn has the interesting property that it is an infinite surface area bound within a finite volume.

I was wondering if there was an extension of this to 3D space in a higher dimensional space, specifically an infinite volume contained within a finite "hyper-volume" (is that a thing)?

More specifically, I'm looking for something that, in layman's terms, would be appropriate to describe as an infinite volume contained within a finite hyper volume, that like Gabriel's horn, could itself be a part of an infinite higher dimensional hyper volume.

Bonus points for showing a way to include an infinite 3D volume itself bounded "finitely" in another infinite 3D volume, even if you need to "fold" it somehow with an extra dimension, or showing that it can't be done.

By "finitely" bounded, I explain with the following example:

Assume I have an infinite 2D sheet, now I cut out some finite circle in that sheet and join it to the mouth of a Gabriel's horn (obviously this requires the addition of a 3rd dimension). Now I have an ant walk on that sheet. The ant can walk infinitely in any direction since the sheet is infinite, but it can also walk around the opening to the Gabriel's horn. If the ant walks inside the circle, it now has another "infinite" space to explore on the inside of the Horn, but the horn itself is in effect "contained" within the infinite sheet.

In this sense it has circumvented an "infinity" by simply walking around it, but you have the case of one infinity bound in a finite space within another infinity.

Essentially what I'm looking for is a 3D equivalent of this, where our ant is a human in 3D space who can infinitely explore one space, and then by passing through our "hole" explore another infinite space, but both spaces are joined in a finite manner, so the human could just as soon go "around" the other infinity.

(As an aside, yes I'm aware this starts looking a lot like black holes and event horizons, but no, I'm not asking the question for anything to do with black holes)

$\endgroup$
  • $\begingroup$ I would write it that the usual Gabriel's horn contains a finite volume in infinite surface area. The surface contains the volume. $\endgroup$ – Ross Millikan Aug 12 at 21:12
  • $\begingroup$ @RossMillikan That is a reasonable way of inverting things, but it doesn't really seem to provide an answer to the question in extending it to 3D. Are you saying I should think about the question more as an infinite volume containing a finite "hypervolume" inside it? I can see that approach, but it seems to violate the "spirit" of the question, as provided by my added ant example. $\endgroup$ – stix Aug 12 at 21:19
  • $\begingroup$ I think Ross's comment was meant primarily to address the exposition. We normally do think of a boundary as containing the region, not vice versa. I would ordinarily think I know what is meant by a higher-dimensional analogue of Gabriel's horn, but I can't tell if what you're describing is it. After all, I could take an infinitely long cylinder and attach it to the punctured plane too, and that would still be two infinite areas, one "contained" within a finite circle, but the cylinder doesn't have the defining property of Gabriel's horn. $\endgroup$ – Brian Tung Aug 12 at 21:40
  • $\begingroup$ At any rate, there is an article by Vincent Coll and Michael Harrison, in Mathematics Magazine, Vol. 87, No. 4, that might answer your question in any case, entitled "Gabriel's Horn: A Revolutionary Tale." (Yeah, I'm quite sure that was intentional.) $\endgroup$ – Brian Tung Aug 12 at 21:41
  • $\begingroup$ The problem with the infinite cylinder is that it doesn't have a finite volume in 3-space, so you're kind of cheating. But in thinking about this, another concern pops up: Gabriel's Horn has infinite surface area in a finite volume, but only because it sacrifices a dimension to do so. You can think of the horn as having infinite surface area, but only in one dimension, since it "rolls up" to allow it to fit in a finite volume. Our ant could travel infinitely in the horn, but only in one direction. If it travels orthogonal to that direction, it will return to its starting point. $\endgroup$ – stix Aug 12 at 21:45
2
$\begingroup$

The same approach as Gabriel's horn works, but we have two axes to revolve in, not just one. We imagine a function $f(x)$ with domain $[0,\infty)$ and revolve it around $x$. Each segment from $(x,0)$ to $(x,f(x))$ becomes a ball of radius $f(x)$. That ball has volume $\frac 43\pi(f(x))^3$ and surface $4\pi(f(x))^2$. The $4-$volume of the whole thing is $$\int_0^\infty\frac 43\pi(f(x))^3dx$$ and the $3-$volume of the boundary, which is like the surface area, is $$\int_0^\infty4\pi(f(x))^2dx$$ We want the first to be finite while the second is infinite. If we let $f(x)$ be $x^{-1/2}$ the second diverges logarithmically while the first is the integral of $x^{-3/2}$, which is finite. The contrast with the $3$ dimensionals case is the exponent in the volume integral is $3$ instead of $2$ and the exponent in the boundary integral is $2$ instead of $1$.

In $n$ dimensions the volume integral will have an exponent $n-1$ and the boundary integral $n-2$ If we make $f(x)=x^{\frac 1{n-2}}$ the boundary integral will diverge while the volume integral will converge. We can make the exponent on $f(x)$ anything up to but not including $\frac 1{n-1}$ and get the behavior we desire.

$\endgroup$
1
$\begingroup$

What you want is $\int f^{n-1}(x) dx \to \infty $ and $\int f^n(x) dx $ is bounded.

A solution to this is $f(x) = x^{-1/(n-1)} $.

For example, is 2-dimensional space, $f(x) = x^{-1} $, the traditional horn.

$\endgroup$
  • $\begingroup$ I think you want the integral of $f^{n-1}$ bounded and the integral of $f^{n-2} \to \infty$. In three dimensions it is that way. $\endgroup$ – Ross Millikan Aug 13 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.