# What is the logical contradiction to producing a finitely thick Koch snowflake?

In another question a user asked about how to come to grasp with the idea of fractals bounded by infinite area containing only finite space. I'm familiar with this property but I am befuddled by one of the comments given to the post. The specific comment is:

"There's another fundamental problem here: if you give the Koch snowflake finite thickness, then it cannot exist. Each arm of the snowflake has infinitely many sub-arms, ad infimum. This implies that if you have a non-zero thickness $T$, then each arm has a minimal width at it's base of 2$T$ which would imply the snowflake has finitely many arms"

This does not makes sense to me, what is to stop each of the Koch's curves infinite arms from having finite thickness. How does one describe the surface obtained by continuously translating the koch's curve $1$ unit through the $z$-axis? or the surface described by the equation of a Koch's curve with free but bounded $z$.

• – user535339 Mar 8 '18 at 0:46
• The comment you are referring to suffers from the same problem as the linked question: Lack of a precise definition what a "thickened Koch snowflake" could mean. Once you make a precise definition (and there are several!), the problem is likely to go away very quickly. For instance, with one definition that I know such an object exists and with another definition, it does not. – Moishe Kohan Mar 8 '18 at 1:23
• This is about what I expected, I noticed I was forced to make the abstraction of "the 2d formula for a Koch's" curve" and assumed it was in this ignorance the true problem lie, I took that comment at face value and assumed there was some absolute truth to it. Thanks! – Brian Mar 8 '18 at 4:12