I read that if a fractal is a union of N copies of itself, each copy shrunk by a factor of $1/z$; this yields a Hausdorff dimension of $D=ln(N)/ln(z)$
If a finite segment of (straight) line $[a,b]$ is scaled down by $n$, you need $n^1$ copies of the scaled version to match the original line segment, so the Hausdorff dimension of the line is $D=log(n^1)/log(n)=1$. The dimension is $D=1$
But if instead of a finite segment, it is a semi infinite line $[a,\infty]$, then scaling it down by any positive real number produces again a semi infinite line, then you only need 1 copy of the scaled version to fill the original. $D=log(1)/log(n)=0$
Does that means that the Hausdorff dimension of the semi infinite line is D=0?
If not, then what is the right answer, and why?