Is $[0;1]$ equinumerous to $\mathbb{R}^{\omega}$? This article on Wikipedia states the following:

Cardinal arithmetic can be used to show not only that the number of points in a real number line is equal to the number of points in any segment of that line, but that this is equal to the number of points on a plane and, indeed, in any finite-dimensional space.

I was wondering why is it the case for finite-dimensional spaces, but not for $\mathbb{R}^{\omega}$. Can't a PMI proof of a bijection between $[0;1]$ and 
$\mathbb{R}^{\omega}$ be established, or would that proof be showing that one can establish a bijection between $[0;1]$ and $\mathbb{R}^n$ for any given $n\in \mathbb{R}/\mathbb{N}$ but not $\mathbb{R}^{\omega}$? Is there something special about infinitely-dimensional $R$ that makes it have more points than $\mathbb{R}$ or any segment on $\mathbb{R}$? Thanks a lot.
 A: The cleanest way of seeing the bijection from $\mathbb{R}^{\omega}$ to $\mathbb{R}$ is probably by taking advantage of your favorite bijection from $\mathbb{R}$ to $\left[0,1\right]$, then treating the numbers in that range as a sequence of decimals, 'stacking' them to get a $\omega\times\omega$ grid of digits, and using a diagonal sweep to compose them into a single decimal; so far instance, if the first three elements of your sequence of reals were .314159, .271828, and .161803, then take the first digit of the first, then the second digit of the first and the first digit of the second, then the third digit of the first, the second digit of the second, and the first digit of the first, continuing to interleave this way: .312471...  It's easy to convince yourself that this is a one-to-one bijection; you can go the other way by writing the digits of your single number one by one diagonally into a grid, and then taking the result as a series of reals.
A: Whoever told you that $[0, 1]$ is smaller than $\mathbb{R}^{\omega}$ is wrong; they both have the same cardinality as $\{ 0, 1 \}^{\omega} = \{ 0, 1 \}^{\omega \cdot \omega}$.  By the way, something about your post suggests to me that you are not convinced that this set has a rigorous definition, and it does: $\mathbb{R}^{\omega}$ is precisely the set of functions from $\omega$ to $\mathbb{R}$.  
There are several other ways one might imagine putting infinitely many copies of $\mathbb{R}$ together.  The above is the infinite direct product; there is also the infinite direct sum $\bigoplus_{n=1}^{\infty} \mathbb{R}$, which consists of all sequences of real numbers which are eventually zero; unlike the infinite direct product, the infinite direct sum is countable-dimensional.  And also one can consider the sequences $c_0$ of real numbers which converge to zero; this is not countable-dimensional but has nice topological properties.
