# Dimensions and Space-filling curves

If a space-filling curve can continuously map a 1-dimensional interval onto a 2-dimensional region, then what actually makes the region 2-dimensional? Doesn't this mean that only 1-dimension is required to actually describe all the points?

Furthermore, couldn't a space-filling curve reduce a surface integral into a line integral?

I feel like I'm missing a core-concept here and I would like to know what it is.

• The core concept you're missing is "homeomorphism." Dimension is a homeomorphism invariant, and that's the point. The space-filling curve is not a homeomorphism. – Jesse Madnick Aug 19 '13 at 3:39
• en.wikipedia.org/wiki/Space-filling_curve : "A space-filling curve must be everywhere self-intersecting in the technical sense that the curve is not injective." In other words, the curve is not a bijection between the the interval and the region ; it's just a continuous surjection. (And its right-inverse is not even close to being continuous, it's actually quite chaotic.) Using it to do a integral would be not very wise. – Patrick Da Silva Aug 19 '13 at 3:42
• Is that because it isn't bijective? Also, what property of a space-filling curve prevents it from being so? – Grant Aug 19 '13 at 3:42
• Space-filling curves can be made to be Hölder continuous with exponent $1/n$, where $n$ is the dimension, but they are not 1-1. – robjohn Aug 19 '13 at 3:42
• @PatrickDaSilva: the part about the Hölder continuity? I actually worked out a function $\Phi_n:[0,1]\mapsto[0,1]^n$ that is Hölder-$1/n$. It is all in notes, but I have been meaning to write it up in a more presentable form. – robjohn Aug 19 '13 at 3:53

Each of these notions of dimension is motivated by some property, trying to formalize some aspect of the intuitive notion of dimension. It is then a theorem, for instance, that the covering dimension of $n$ dimensional Euclidean space is actually $n$. It is also the case that some of these notions of dimension (e.g., the Hausdorff dimension) can be fractional, and that it can be made to make sense.
For these notions of dimension, it is the case that the following is not a theorem: if $f:X\to Y$ is a continuous surjection, then $X$ and $Y$ have the same dimension. It simply is not true. Thus, the intuitive notion (at least intuitive to some) that the existence of such a function should entail the equality of dimensions does not agree with the formalisms above.
One can try to introduce a new notion of dimension, as follows. For spaces $X$ and $Y$ say that the dimensionality of $X$ is the same as the dimensionality of $Y$ if there exits a continuous surjection between the two spaces. This defines a relation on (ignoring set theoretic problems) the set of all spaces. The relation is clearly reflexive and symmetric, but not transitive. Its transitive closure is then an equivalence relations, and one may call the (again, ignoring set theoretic issues) equivalence classes 'dimensions'). Then the dimension of $X$ is simply $[X]$. It is then a trivial theorem that if $f:X\to Y$ is a continuous surjection, then $X$ and $Y$ have the same dimension. However, I doubt that this notion of dimension has any good properties.
• Yeah, I think that went you take the transitive closure you get too much stuff. For instance, $[0,1]$ and $\mathbb R^n$ would have the same dimension in this case because of space-filling curves. So it's not a very good notion. At least to me it seems. But it's good you worked it out, that was interesting to read :D +1 – Patrick Da Silva Aug 19 '13 at 14:45