3D - 1D = 2D? Doing arithmetic with dimension 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:


*

*Do any such spaces immediately come to mind?

*Are there suggestions for the types of spaces (or perhaps "removal" processes) other than $\mathbb{R}^n$ that might be more accomodating?

*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.
 A: 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.)
