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If you consider $\mathbb{R}^3$ and a one-dimensional space curve, by "removing" the curve from $\mathbb{R}^3$ you are left with a space that is still three dimensional, for an appropriate definition of dimension (perhaps the one from linear algebra?). Thus the naive equation 3D - 1D = 2D does not hold.

I am interested in spaces or sets where such an equation would hold. That is, by removing a set of a given dimension, the total space must be arbitrarily reduced by that dimension.

My question has three parts:

  1. Do any such spaces immediately come to mind?
  2. Are there suggestions for the types of spaces (or perhaps "removal" processes) other than $\mathbb{R}^n$ that might be more accomodating?
  3. Am I using too naive a definition of dimension? I am most familiar with dimension as the cardinality of the set of basis vectors in a vector space, but am aware of a Hausdorff dimension for topological spaces, if that's where I should turn...

My apologies if this question makes no sense, or is too poorly thought out to receive adequate response.

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Maybe you should not remove spaces, but quotient them? – Thomas Rot Jul 8 '11 at 17:23
Yes, I had thought to quotient instead (and probably should have mentioned it above). I guess that removes it from the purview of "arithmetic"? Qiaochu touches on this in his great answer below... – Greg L Jul 8 '11 at 17:31
up vote 7 down vote accepted

This should never be true for a reasonable definition of dimension (for example the dimension of a manifold). A lower-dimensional thing should have measure zero in a higher-dimensional thing, so removing it shouldn't change the dimension of the higher-dimensional thing.

The correct version of the "naive equation" is that the Cartesian product of an $m$-dimensional thing and an $n$-dimensional thing should be an $m+n$-dimensional thing. Putting two things together (the opposite of removing a thing from another thing) doesn't add their dimensions; instead, the disjoint union of an $m$-dimensional thing and an $n$-dimensional thing should be considered to have dimension $\text{max}(m, n)$.

This algebraic structure (with addition and max instead of multiplication and addition) happens to have a name: it's called the max-plus semiring.

(If you really want to think about subtraction of dimensions instead of addition, then I guess the appropriate thing is to take the quotient by the free action of a Lie group. In nice cases, it should be true that the quotient of an $m$-dimensional manifold by an $n$-dimensional Lie group acting freely has dimension $m-n$. The easiest case of this occurs when we take the quotient of a vector space by a subspace.)

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Thanks, this is a great answer with lots of information to follow up on! Is it fair to say that the quotient procedure acts as an inverse to the Cartesian product, in analogy to the +,- operations? Perhaps only in certain situations... – Greg L Jul 8 '11 at 17:34
@Greg: I suppose so, in the sense that the projection map $M \times N \to M$ should be a quotient map. – Qiaochu Yuan Jul 8 '11 at 17:45
Also, I've only head the term "faithful" in regards to representations, so maybe I'm just misinterpreting. – Jason DeVito Jul 8 '11 at 17:52
@Jason: ah, you're right, I want a free action. – Qiaochu Yuan Jul 8 '11 at 17:56

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