2
$\begingroup$

As a metric space, the figure eight is a subset of the plane. I think that means it's 2D. This person I'm arguing with says it is 1D. I think that must be wrong because a disc containing the intersection point is not an interval which is a requirement for a 1D (connected) metric space, as far as I know. The counter argument is that the figure eight can be parameterized in $t$ and the interval $t\in[0,1]$ is 1D so the countable union of its diffeomorphisms (the figure eight) must also be 1D. Who is right? Why?

$\endgroup$
3
  • 1
    $\begingroup$ I am puzzled by this. You give an argument that the dimension is one, then give a "counter argument" that then concludes that the dimension is one! So how is that a counter argument? The dimension is one! $\endgroup$
    – user247327
    Commented Apr 16, 2021 at 14:13
  • 2
    $\begingroup$ There are a number of definitions of "dimension" in topology, even one which is called Hausdorff dimension and allows for fractional values. So the right answer may depend on whose definition is used. $\endgroup$
    – hardmath
    Commented Apr 16, 2021 at 14:13
  • $\begingroup$ @user247327 He means it's an argument countering his argument that the dimension is not $1$. $\endgroup$
    – saulspatz
    Commented Apr 16, 2021 at 14:35

1 Answer 1

2
$\begingroup$

Judging by what you wrote, you are trying to apply the definition of the dimension of a manifold, while your friend is talking about Hausdorff dimension.

  • Definition of a 1-dimensional manifold: Every point has a neighborhood homeomorphic to an (open) interval.

  • Definition of Hausdorff dimension: Hausdorff dimension of a metric space $X$ is $\le \alpha$ if an only if for each $\beta>\alpha$, the $\beta$-dimensional Hausdorff measure of $X$ is zero.

The trouble with your attempt on computing the dimension of the figure 8 is that the definition of dimension used for manifolds simply does not work when you try to apply it to spaces which are not manifolds, and figure 8 is not a manifold. In contrast, what your friend explained is that the standard drawing of figure 8 in the plane results in a space which is a countable (actually, finite) union of subsets each of which is diffeomorphic to an interval. From this, with a bit of work, it follows that Hausdorff dimension is exactly 1, when one uses the restriction of the Euclidean metric from the plane to the figure 8.

There are many other definitions of dimension in geometry and topology, so, before trying to argue about dimensions one should settle on the precise definition. The most commonly used definition of dimension for topological spaces is the Lebesgue covering dimension. My suggestion is to read the definition and then verify that figure 8 is indeed 1-dimensional in this sense.

Edit. One more thing: The terminology "Hausdorff space" used in the title of your question normally means "Hausdorff topological space." In the context of the figure 8 embedded in the plane, Hausdorfness is automatic. What you should specify instead is if you consider a topological space or a metric space. This (while insufficient) would indicate different notions of dimension. In the case of the usual drawing of the figure 8, however, both notions give you the same answer.

$\endgroup$
1
  • $\begingroup$ Thanks for taking the time to write up such a nice exposition. I suspect the OP has a genuine interest in the conflicting notions of dimension, but informed more by the limited intuition they can bring to bear, and so giving them more than they know how to ask for is a good approach. $\endgroup$
    – hardmath
    Commented Apr 16, 2021 at 16:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .