1. ## Dimension of fractal

for a cantor middle-half set K, I think it is one-dimensional because it is on a real line. But the book say that it is 1/2-dimensional.
Consider KxK, I think it is 2-dimensional because it is a square(containing many small squares) on a plane. But it is a 1-dimensional set. Why?

2. Originally Posted by ldpsong
I think it is one-dimensional because it is on a real line.
As a matter of fact, the "dimension" is a precise (and delicate) notion that has a mathematical definition (in fact, several). Maybe you know what the dimension of a vector space is (which is an integer). For Cantor subsets, one obviously needs a different definition, that would account for their different self-similarity.

But the book say that it is 1/2-dimensional.
Maybe your book says how it defines dimension. Anyway, you can find lots of ressource on the internet: look for "fractal dimension". In the case of your subset, you can cut it in 2 parts that look like the initial subset up to a scale $\displaystyle \frac{1}{4}$. Similarly, if we cut a line segment (d=1) in 4 we get parts that look like the initial subset to a scale $\displaystyle \frac{1}{4}$, if we cut a square (d=2) in 16 we get smaller squares like the first one to scale 1/4, and if we cut the cube (d=3) in $\displaystyle 4^3$ we get smaller cubes like the first one to scale $\displaystyle \frac{1}{4}$, and in dimension $\displaystyle d$, we can cut a cube into $\displaystyle 4^{d}$ smaller cubes that look like half the initial cube up to a scale $\displaystyle \frac{1}{4}$. So for the middle-half Cantor subset, you would have $\displaystyle 2=4^d$, hence $\displaystyle d=1/2$. This is not a proof, since I gave no definition of dimension, but it gives a reason why it should be $\displaystyle 1/2$.

3. thank you very much