If you want a single-sided surface (which in the jargon of mathematicians is called "non-orientable", because if you imagine livingin the surface rather thanon it, then you can't define clockwise and counterclockwise in any meaningful manner the way you could with a sphere), and you want it to be what you call "infinite" (which isn't really infinite per se, a sphere has very finite surface area, but it is without boundary), then you can't actually make one in our 3-dimensional space.

The closest you can get is probably aKlein bottle. It has only one side, and there is not supposed to be any edge. However, because we live in 3 dimensions rather than 4, the bottle has to "pass through" itself at some point.

Another possible candidate is the projective plane. The easiest way to (almost) make one is to take a Mobius strip, and roll it up into a tube. This is also impossible to actually complete in our 3 dimensional space, and it will have to pass through itself at some point. (It's a fun exercise to try to see where the tube-rolling goes wrong. It is intricately linked to the one-sidedness of it, and what happens when you go one "lap" around the loop as you do the rolling.)

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  $\begingroup$This is not an answer.$\endgroup$

  Moishe Kohan

  – Moishe Kohan

  2025-12-17 14:55:05 +00:00

  Commented13 hours ago
• 4
  $\begingroup$Actually it is a good answer that answers at the same level that the question was asked.$\endgroup$

  return true

  – return true

  2025-12-17 21:57:11 +00:00

  Commented6 hours ago
• 1
  $\begingroup$@MoisheKohan I do not think an actual proof of the non-embedability of non-orientable surfaces would be appreciated at this level. You are welcome to write your own and post it here if you think that's better suited than my answer.$\endgroup$

  Arthur

  – Arthur

  2025-12-17 22:01:06 +00:00

  Commented6 hours ago
• $\begingroup$An answer would at least point OP in the right direction (and indicate which tools are needed to resolve the issue, which sources OP should be reading to improve their understanding) and at least an attempt to clarify the concepts that OP is struggling with. Incidentally, there is a pretty intuitive proof in the case of smooth closed surfaces based on the intersection with rays argument (which can be made rigorous). You post does not attempt either one of these and amounts to handwaving. As for writing an answer: In my opinion, the current post does not meet the MSE standards.$\endgroup$

  Moishe Kohan

  – Moishe Kohan

  2025-12-17 22:09:22 +00:00

  Commented6 hours ago

What you are looking for, using standard mathematical parlance, seems to be anon-orientable closed surface without boundary.

• What you're meaning by "infinite" vs. "not infinite" is captured well by the idea of aclosed surface (without boundary) vs. aclosed surface with boundary, respectively.
  • Mathematically the distinction is as follows. Asurface (without boundary) is something that "looks like" a 2D plane if you zoom in; more specifically, each point on the surface has a neighborhood homeomorphic to an open subset of$\mathbb R^2$ (plus a few additional technicalities to avoid weird exceptions; seeWikipedia for more details). The sphere$S^2=\{x^2+y^2+z^2=1\}$ is an example of a surface without boundary; if you zoom in on any point, it looks like$\mathbb R^2$. When talking about surfaces (or more generally manifolds) this is usually the "default", as opposed to surfaces/manifolds with boundary.
  • Surfaces with boundary are similar, but the region around each point is also allowed to look like the half-plane$\{(x,y)\in\mathbb R^2:\, x\geq 0\}$.The Möbius strip is an example of this. Around points on the "edge" of the Möbius strip it looks like a half-plane, while around other points it simply looks like$\mathbb R^2$ (if you zoom in enough)
  • Aclosed surface is one that is compact as a topological space; loosely this means that it doesn't have any "open" edges, so all of the edges are closed ones. Both the sphere and Möbius strip are closed surfaces. Mostly we want to include this condition to exclude things like the open half-sphere$\{x^2+y^2+z^2=1:\,x>0\}$ (this is technically a manifold without boundary, but is not closed). This does have the unfortunate side-effect of excluding unbounded surfaces such as$\mathbb R^2$ however (not compact).
• The idea of "one-sided"/"two-sided" is usually thought of in terms oforientable/non-orientable surfaces.
  • A surface isorientable if it is possible to consistently define "counterclockwise" along the whole surface, which is referred to as anorientation of the surface.
  • Why does this correspond to one-sided-ness? If we have a notion of "counterclockwise", we can use this to consistently define an "outside" direction at each point of the manifold: choose a vector tangent to the point, rotate it 90 degrees counterclockwise, and then take the cross product of those two vectors. (Note: this does require assuming that we can fit the surface in$\mathbb R^3$; more on that in a minute.) If we take all these third vectors, this gives us a consistent outwards direction on the whole surface. Taking the negative gives us an "inside" direction.
  • To go in the other direction, if we can define an "outside" direction to the whole surface consistently, we can again use the cross product (again assuming our surface can be made to fit in$\mathbb R^3$): if we start with a "forwards" direction, the "outside" direction gives us an "upwards" direction, so we can use the cross product of these two vectors (plus some additional details) to get a "right" direction; then we can define "counterclockwise" using those.

So, what you are looking for are non-orientable closed surfaces (without boundary). The two classic examples of these are (as mentioned by Arthur) theKlein bottle and thereal projective plane. Both of these have the properties you want: they don't have any "edges" as they are closed without boundary, and they are one-sided as they are non-orientable. However, it is important to note that these cannot be embedded into$\mathbb R^3$, i.e. if you try to actually construct these in 3D space, the surface will self-intersect. However:

• These surfaces can beimmersed in$\mathbb R^3$, i.e. aside from the self-intersections you don't have to flatten/etc. them at all. In particular, the above-mentioned tricks with the cross product would work perfectly fine to convert between inside/outside and orientability.
• They can be embedded into$\mathbb R^4$ perfectly fine.
• It also turns out that a closed surface can be embedded into$\mathbb R^3$ if and only if it is orientable, and$\mathbb R^4$ suffices for any non-orientable surface. (The situation for general manifolds is much more complicated in general.)

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  $\begingroup$1. You are misusing the standard terminology: "A closed surface" usually means "compact without boundary." 2. You are not even attempting to prove that no closed nonorientable surface (in the usual sense) embeds in $R^3$: This is true but requires some algebraic topology or differential topology tools to prove. (One can find proofs in other MSE posts.)$\endgroup$

  Moishe Kohan

  – Moishe Kohan

  2025-12-17 22:00:18 +00:00

  Commented6 hours ago

• general-topology
• geometry
• 3d
• surfaces
• mobius-band