An odd proof of the uncountability of the reals So, in proving that the reals are uncountable you assume otherwise and attempt to make a list of them. Working in binary this list is of form $a_1,a_2,\ldots$ where $a_i = x_{i,1}x_{i,2},\ldots$ represents the binary expansion. Given any index $j$ some real number (actually, an infinite number of them) has its $j$th digit $1$. If $x_{j,j} \neq 1$ we can swap $a_j$ with $a_k$ such that $k > j$ and $x_{k,k} = 1$; by the remark above such an $a_k$ exists. Because $k > j$ we do not undo any previous work. Thus we can assume that $x_{j,j} = 1$ for all $j$. Then $0$ is not in our enumeration.
This is of course the diagonal argument but it is a little unsettling to me that specifying any decimal expansion in advance, we could have made the exact same argument to show that ${\it any}$ real number is not in our enumeration. Is there something wrong with this method or is uncountability just this strange?
 A: This doesn't prove that $\mathbb R$ is uncountable, to see why consider the following "proof" that $\mathbb N$ is uncountable.
Assume $\mathbb N$ is countable and we have a bijection $f\colon\mathbb N \to \mathbb N$ (for example, the identity!).  I want to show that there is a bijection that doesn't contain $0$, a contradiction.  Well, if $f(0) = 0$ then swap $f(0)$ and $f(1)$ so that we get a bijection $f\colon \mathbb{N \to N}$ such that $f(0) \neq 0$.
By induction if $f(x) \neq 0$ for all $x < n$ but $f(n) = 0$ then swap $f(n)$ and $f(n + 1)$ so that we can assume $f(n) \neq 0$.
When we make a swap we don't alter values below $n$, so "after" all these swaps we get a perfectly well defined function $f\colon\mathbb{N \to N}$ with the property that $0$ is not in the image.  Contradiction!
Except it's not a contradiction!  There's no reason why the resulting function $f$ should have been a bijection.  In fact the function you get from this procedure is simply $f(n) = n + 1$.  This is not a bijection, but that doesn't mean there doesn't exist some other bijection $\mathbb{N \to N}$, like the identity function that we started with!
A: So you are given a function $f:\ \Bbb N\to[0,1]$ which is hopefully surjective. By some involved, and by no means "finitary", process you then construct a new function $g:\ \Bbb N\to[0,1]$ that does not contain $0$ in its image. I don't see how this proves anything about the given $f$.
A: Although the argument is often presented as a proof by contradiction, it really isn’t one. Rather, it shows how, given any function $f:\Bbb N\to\Bbb R$, you can actually construct a specific real number $r_f$ that is not in the range of $f$. Thus, no function from $\Bbb N$ to $\Bbb R$ can be surjective, and by definition $\Bbb R$ must be uncountable. When you modify your original $f$ by swapping two values, you’re simply replacing $f$ by a different function $g:\Bbb N\to\Bbb R$, and the same argument shows that the range of $g$ is not all of $\Bbb R$. Since $g$ and $f$ are not the same function, it should not by surprising that $r_g\ne r_f$.
