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Consider, for example the following sets:

  1. ${x,y,z} \in \mathbb{R}^3$ for which $x^2+y^2<1$
  2. ${x,y,z} \in \mathbb{R}^3$ for which $x^2<1$
  3. $\mathbb{R}^3$

Of course, the cardinality is $2^{\aleph_0}$ of all. Also their volume is infinite.

But, I intuitive feel, somehow there should some... measurement, some terminology exist, which would somehow differentiate some like a "1-dimensional infinity" of (1), the "2-dimensional infinity" of (2) and the "3-dimensional infinity" of 3.

I suspect, it may have more to do with analysis as with set theory.

Does any similar, well-defined terminology exist to make a distinction between (1), (2) and (3)?

Edit: the "dimensionality" of the points closer as 1 to the $x^2=y$ parabola should be also 1.

peterh
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  • The cardinality is $2^{\aleph_0}$ - that's not necessarily the same thing as $\aleph_1$. The statement that $2^{\aleph_0}=\aleph_1$ is the continuum hypothesis, and is independent - it can neither be proved or disproved from the usual axioms of set theory. – Noah Schweber Mar 12 '17 at 23:31
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    you can do intersections with affine subspaces and check the measure. if there is a subspace of a certain dimension so that the measure of the intersection is finite then you have some thing to categorize. – Nathanael Skrepek Mar 12 '17 at 23:31
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    Convex sets have a "recession cone" consisting of all directions that you can follow "to infinity" without leaving the convex set. Maybe you could look at the dimension of the affine hull of the recession cone or something. – littleO Mar 12 '17 at 23:38
  • @NathanaelSkrepek Oh, thanks. I've seen it always from a practical, engineer viewpoint. :-) I have made already a question about a similar topic (can we define a set with $\aleph_1 < 2^{\aleph_0}$ cardinality if we have ZFC + not-CH). I think we can assume CH from a practicality viewpoint in the sense of this question. What is the case for the set of the points closer as 1 distance from the $x=y^2$ parabola? The subspace-based definition probably won't work, I think it should be "1-dimensional" like the set (1). – peterh Mar 12 '17 at 23:41
  • This doesn't really have anything to do with set theory, as you suspected, so I've removed the set-theory tag. – Eric Wofsey Mar 12 '17 at 23:50
  • @EricWofsey Ok, thanks. – peterh Mar 12 '17 at 23:51
  • @peterh: Why does your intuition tell you that $(1)$ is one-dimensional? This is an infinite cylinder without it's boundary and to describe a point in this cylinder, you need to give me three parameters - the height ($z$), the distance $\rho$ from the $z$-axis and the angle which the point makes with respect to the positive $x$-axis. From this point of view ("freedom degrees count") it should be a three-dimensional object and indeed it is a three-dimensional manifold (being an open subset of $\mathbb{R}^3$). – levap Mar 12 '17 at 23:51

1 Answers1

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My first question would be: what is the purpose of distinguishing these three sets? From the point of topology, the three sets in (1), (2) and (3) are homeomorphic and from the point of differential calculus they are also diffeomoprhic.

Distinguishing them via their boundary is also difficult. The boundary of (1) is a cylinder, which has dimension 2 and hence codimension 1. This would correspond to the "1-dimensional infinity". However, the boundary of (2) are two planes, which have also codimension 1. On the other hand $\mathbb R^3$ has no boundary, which would be 0-dimensional and hence have codimension 3.

Most invariants in mathematics have been created to distinguish objects with some application in mind. Topological dimension, for example, answers the question which open subsets of euclidean space can be locally homeomorphic. It is difficult to find notions for distinguishing objects, without a use in mind.

Martins Bruveris
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  • It relates to my this question. I think, some "color number" could be defined also for 3 dimensions if we would reduce the allowed regions to the sets like (2). – peterh Mar 13 '17 at 00:06
  • Via the map $(x,y,z) \mapsto (\tan x, y, z)$, the set $x^2 < \pi/2$ is diffeomorphic to $\mathbb R^3$. And I believe this map would preserve convex sets, although I have not checked the details. Would this not present a problem? – Martins Bruveris Mar 13 '17 at 00:14