What are some of the consequences of Cantor's famous "I see it, but don't believe it" result showing a bijective mapping from $[0,1]$ to $[0,1]^n$? Cantor wrote in a letter to Dedekind "I see it, but don't believe it!" While discussing his discovery of a bijection between the interval $I=[0,1]$ and $I^n$.  While this is certainly a neat result, I can't think of where this becomes a wholly useful fact.  In what mathematical context has Cantor's result proven useful?  Are there are any notable cases where Cantor's discovery was utilized in a proof?  Are there any particular math problems which require knowing $|I|=|I^n|$ to solve?
Sorry if this is question is vague.  I recently read an article on Cantor's letter to Dedekind but cannot find any specific examples of his discovery being cited.
 A: It's not a useful fact. 
It is an important fact; it shows that we can't distinguish $[0, 1]$ from $[0, 1]^n$ just by their cardinality, so if we want to make precise the intuition that they are really different (e.g. because the former has dimension $1$ and the latter has dimension $n$) we need to do something else (e.g. invent topological dimension, which involves inventing topology). 
A: Cantor's formula for the bijection is itself important, because it is used to prove an important theorem in probability theory: 

Theorem: $\mathbb R$ with its Borel sigma algebra is isomorphic to $\mathbb R^n$ with its Borel sigma algebra. 

Cantor's function defines such an isomorphism. 
Take a look at the early passages of the Wikipedia page on standard probability spaces, in particular:

Rokhlin showed that the unit interval endowed with the Lebesgue measure has important advantages over general probability spaces, yet can be effectively substituted for many of these in probability theory. The dimension of the unit interval is not an obstacle, as was clear already to Norbert Wiener. 

Implicit in that last quote is that $\mathbb R^n$ with its Borel sigma algebra is also a standard probability space; presumably Wiener was aware of how to apply Cantor's formula to prove this fact.
A: I'm not fully certain that you'll count this as an answer, but it's worth point out that this fact is really a specific instance of a more general one: 

Any infinite set is in bijection with any of its finite powers. 

It's impossible to overstate the importance of this more general fact in set-theoretic combinatorics. Practically any result in the area invokes it somewhere. 
Now this is an application within logic and of a more general principle, so I don't know if it will be fully satisfying; but I think the takeaway is that it can be viewed as a tiny piece of a foundational fact in a whole avenue of study, which is pretty cool.

OK fine, technically the fact above requires the axiom of choice; indeed, Zermelo showed that it's equivalent (over ZF) to the axiom of choice. But we actually don't need choice for the point to stand, and this is worth explaining:
Even without choice, we can prove that any infinite well-orderable set is in bijection with each of its finite powers. Moreover, we can also prove (again without choice) that if an infinite set $X$ is in bijection with each of its finite powers then its powerset $\mathcal{P}(X)$ is in bijection with each of its (= the powerset's) finite powers. Since $I$ is (up to cardinality) $\mathcal{P}(\omega)$, Cantor's observation is indeed a special case of this one, even without choice.
(The first part falls out of the choice-y proof: just look at how we invoke choice. To prove the second part, given such an $X$ note that the map $$\mathcal{P}(X)^2\rightarrow\mathcal{P}(X): (U,V)\mapsto\{\langle a,b\rangle: a\in U, b\in V\},$$ where $\langle\cdot,\cdot\rangle$ is a bijection from $X^2$ to $X$, is clearly injective - now apply Cantor-Bernstein, which does not require choice.) 
