Whilst learning about infinities, I attempted to construct a proof by contradiction that the continuum of real numbers ($\aleph_1$) could not be represented by the set of positive integers ($\aleph_0$). It is as follows (simplified):
Let $x$ be a real number where $0 \leq x < 1$. Interpret $x$ as a base-2 string of the form $0.d_1d_2d_3d_4\dots$. Let there be an $\aleph_0$-dimensional cube. Each value of $x$ can be represented as the coordinate $(d_1, d_2, d_3, d_4 \dots)$ which is a vertex of the cube, therefore there exists a mapping of each real number onto an $\aleph_0$-dimensional cube. An $n$-dimensional cube has $2^n$ vertices, therefore an $\aleph_0$-dimensional cube has $2^{\aleph_0}$ vertices, therefore $\aleph_1 = 2^{\aleph_0}$.
If the coordinates of point $x$ are treated as a big-endian bit-string representing a number (i.e. $d_1$ is the $2^0$ digit, $d_2$ is the $2^1$ digit etc.) then $x$ can be mapped to an integer. This integer can have an infinite number of digits. There are $\aleph_0$ integers.
If there is a mapping between the reals and the integers then $\aleph_0 = 2^{\aleph_0} = \aleph_1$. This is obviously false, and is what I was trying to disprove. What part(s) of my proof is / are wrong?
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