Fractal dimension of a spring whose wire is actually a spring, etc. Inspired by this video: https://www.youtube.com/watch?v=gB9n2gHsHN4
At a high level, he talks about how shapes can have different dimensionality depending on the scale they're observed from, an example being a spring. Far away, it looks like a line. Get closer, and it looks like a tube. Even closer, it looks like a (twisted) line again. So the guy talks about the limit of the dimension as the length scale goes to 0. But what if the twisted line itself was a spring, so that the dimension isn't convergent? What types of "measure" would we use that case? 
 A: That is a very nice video - thanks for sharing! 
A key stretch of the video to address your question is from about 17 minutes in until about the 18 1/2 minute mark. At the beginning of that stretch, he lists four specific definitions of fractal dimension, including:


*

*Box-counting dimension

*Hausdorff dimension

*Packing dimension


The box-counting dimension is the simplest of these and is defined by
$$\text{dim}(E) = \lim_{\varepsilon\to0^{+}}\frac{\log(N_{\varepsilon}(E))}{\log(1/\varepsilon)},$$
where $N_{\varepsilon}(E)$ is the number of $\varepsilon$ mesh squares that intersect the set $E$. 
Around the middle of that time stretch the video makes a compelling case that it doesn't really make sense to take the limit as $\varepsilon\to0$ but, in the words of the video, we should "look at a sufficiently wide range of scales, from very zoomed out up to very zoomed in, and compute the dimension at each one. In this applied perspective, a shape is considered to be fractal only when the measured dimension stays approximately constant across multiple scales." What this indicates is that fractals should be sets that display some degree of regularity and that, from that perspective, your set that oscillates back and forth from one dimensional to two dimensional should maybe not be considered as a fractal at all.
Having said that, I think it's pretty clear that the narrator's perspective is that of a programmer interested in applied mathematics, as opposed to that of a pure mathematician. To be clear, I think that is a valid perspective and that his narrative is spot on - from that perspective. As a pure mathematician, though, it is certainly feasible to compute a specific fractal dimension of a set like yours. They might not all agree, though. In the case of box-counting dimension, the limit might not exist, but the $\limsup$ and $\liminf$ certainly will. Thus, you can obtain an upper and a lower box-counting dimension. 
For concreteness sake, let's say that a set is regular if its upper box-counting dimension is equal to its lower box-counting dimension. From my perspective (that of a pure mathematician) there's nothing wrong with an "irregular" set; it just might not enjoy some of the properties that regular sets do. For example, the dimension of the Cartesian produce of regular sets is the sum of their dimensions, ie:
$$\dim(A\times B) = \dim(A)+\dim(B).$$
This is not necessarily true for irregular sets.
But "irregular" sets still live squarely in this domain of analysis. An example of a set that is irregular in this sense is explored in the question Minkowski Dimension of Special Cantor Set. In that question, there is a set $C'$ defined by
$$
C' = \left\{\sum_{i=1}^{\infty} a_i4^{-i} : a_i \in \{0,3\} \, \, \text{if} \, \, (2k)! \leq i \leq (2k+1)! \,\, \text{and arbitrary otherwise} \right\}.
$$
As it turns out, the upper box-counting dimension of $C'$ is $1$ while the lower box-counting dimension is $1/2$. The reason is that the set is specifically constructed so over scales in the interval
$$\left[\frac{1}{4^{(2k)!}}, \frac{1}{4^{(2(k-1))!}}\right]$$
it looks to have one dimension but over scales in the interval
$$\left[\frac{1}{4^{(2k+1)!}}, \frac{1}{4^{(2k)!}}\right]$$
it looks to have a different dimension.
