What is wrong with this infinity proof?

Question

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?

Answer 1

Your mistake is that a vector is "a full string". But in fact a linear combination is by definition finite. So if $x$ is an element of an $\aleph_0$-dimensional space, then $x$ has only finitely many non-zero coordinates.

This means that the cube of an $\aleph_0$-dimensional space has $\aleph_0$ vertices.

On the other hand, you are thinking about the space of infinite strings, whose dimension is indeed $2^{\aleph_0}$.

(Also, $\aleph_1$ is the first uncountable cardinal, which may or may not be equal to $2^{\aleph_0}$, depending on your choice of set theory.)

Answer 2

"This integer can have an infinite number of digits."

No it can't.

There's no such thing as an integer with an infinite number of (non-zero) integers.

Such a construct is not an integer and has no finite value.

More thouroughly, this construct will be an infinite sum (all non-negative summands) with an infinite number of terms greater or equal to $1$ (or, actually greater than or equal to any positive power of 2). Such a sum is divergent, infinite, and certainly not an integer.

Answer 3

As mentioned elsewhere, integers don't have an infinite number of digits.

Something that should be mentioned as well, though:

At the end of your proof, you also say that $\aleph_0 = 2^{\aleph_0} = \aleph_1$ (where by $\aleph_1$ you mean the cardinality of $[0, 1)$) is "obviously false". Er, no it's not... it's what you're trying to disprove! This is like trying to prove $\sqrt 2$ is irrational by saying "Suppose $\sqrt 2=\frac a b$. But this would mean $\sqrt 2$ is rational, which is obviously false. Therefore $\sqrt 2$ is irrational."