How can $3^{-n}= 0$ be true for any $n$? There is an answer that has 7 up votes (so it is probably right) that says, 

There are no segments of positive length left (this may be seen by,
  for instance, looking at the lengths of the segments at each
  iteration; after the $n$th iteration, the segments have length
  $3^{-n}$).

My understanding is that  for every $n \in \mathbb N : 3^{-n} \not= 0$
There may be some rule that allows us to say that $ 3^{-n} = 0$ under some condition, but it is not at all obvious to someone (like me) who has not studied math. After all topology is a very strange branch of mathematics.
Can someone explain the thinking behind this answer?
 A: The Cantor set is the intersection of all its iterations.  What the author is trying to say is: the Cantor set contains no interval.  Why not?  Let $I$ be an interval of length $L>0$.  Then there is some $n$ such that $3^{-n}<L$.  Therefore, $I$ cannot be contained in the $n$-th iteration towards the Cantor set, and therefore cannot be contained in the Cantor set itself.  
You are correct that $3^{-n}\neq 0$ for all $n$.  
A: The point is that when constructing the Cantor set, at each step the segments have size $3^{-n}$, so at each step there are intervals of positive length. But the construction goes through all the natural numbers, so after we exhausted all the natural numbers, we end up with a set that contains no (nontrivial) intervals.
A: It's a limit process.  If you aren't familiar with the concept of limits, the Cantor set is not a good introduction to it, in my opinion.
It's true that $3^{-n} \ne 0$ for any $n \in \Bbb N$ (and really for any $n \in \Bbb R$).
However, the limit as $n \to +\infty$ of $3^{-n}$ is zero.  We denote this by writing $\displaystyle \lim_{n \to +\infty} 3^{-n} = 0$.
In general, $\displaystyle \lim_{n \to c} f(n)$ describes what $f(n)$ looks like or how $f(n)$ "behaves" as $n$ gets extremely close to $c$.  Or, in the case that $c$ is $+\infty$, it describes what $f(n)$ looks like or how $f(n)$ "behaves" as $n$ gets extremely huge, i.e., as we send $n$ off in the direction of $+\infty$ and never let it stop.
And in this case, we can see that as $n$ gets larger and larger, the value of $3^{-n}$ always remains positive but continues to decrease, getting smaller and smaller.  As $n$ gets infinitely larger, $3^{-n}$ gets infinitely closer to zero.  So its limit is zero, but $3^{-n}$ never takes on the value zero for any $n$.  But limits don't care about actual function values, only what the function looks like it's going to do.
