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Cantor's proof of the uncountability of $\mathbb{R}$ seems well-established. I am trying to figure out why I am wrong here. If I have a function $B_n$:

$$ B_n = \left\{ \begin{array}{ll} \pi & \mbox{if } n = 0\\ e & \mbox{if } n = 1\\ 194/3 & \mbox{if } n = 2\\ -37 \pi & \mbox{if } n = 3\\ \vdots \end{array} \right. $$

Why can't this be a valid bijection between the naturals and the reals? If $B_n$ is constructed such that the value at $n$ is not a duplicate of any other values for preceding $n$s.

In other words, why can't it be:

$$ B_n = \left\{ \begin{array}{ll} \pi & \text{if } n = 0\\ \text{Pick any element in } \mathbb{R} \setminus \{ B_{m} | m \in \mathbb{N} \land m < n \} & \text{else} \end{array} \right. $$

1 Answers1

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Let's see it as a conversation. You are trying to convince me that there exists a list of all real numbers, and I'm skeptical of your proof, but I'm willing to give you a single chance to show me that it is correct.

Note that if your proof is indeed correct, then you will only need one chance: either you show me the complete list of real numbers, or if your proof doesn't work your list is not complete.

Me: "I give you the challenge to give me a list containing all the real numbers. You only get one chance to give me the list, so you better make sure that it's complete!"
You: "Ok, here's my list. Its elements are called $B_n$ and it contains all the real numbers, because I picked all of them one by one."
Me: "So, just to check, you claim all the real numbers are present on this list? For every real number $R$ there is some $n\in\mathbb N$ such that $R=B_n$ ?"
You: "Yes."
Me: "So, I cannot find a real number that is absent from your list?"
You: "No, you can't, because my list contains all the real numbers."

Me: "Ok, let me try. What about the number $R=0.c_0c_1c_2c_3c_4\cdots$? Here's how I defined it:

Let's write the decimals of $B_n$ as $d_0^nd_1^nd_2^nd_3^nd_4^nd_5^n\cdots$.

So for example:
$B_0=\pi$ has decimals $14159265\cdots$, so we have $d_0^0=1$, $d_1^0=4$, $d_2^0=1$, $d_3^0=5$, $d_4^0=9$ and so on,
$B_1=e$ has decimals $71828182\cdots$, so we have $d_0^1=7$, $d_1^1=1$, $d_2^1=8$, $d_3^1=2$, $d_4^1=8$ and so on,
$B_2=194/3$ has decimals $66666666\cdots$, so we have $d_0^2=6$, $d_1^2=6$, $d_2^2=6$, $d_3^2=6$, $d_4^2=6$ and so on,
$B_3=-37\pi$ has decimals $23892818\cdots$, so we have $d_0^3=2$, $d_1^3=3$, $d_2^3=8$, $d_3^3=9$, $d_4^3=2$ and so on...

Now let $c_n=d_n^n+1$ if $d_n^n<5$ and $c_n=d_n^n-1$ if $d^n_n\geq5$. For example, since $d_0^0=1$ we have that $c_0=2$, and since $d^1_1=1$ we have $c_1=2$, and since $d^2_2=6$ we have $c_2=5$, and since $d^3_3=9$ we have $c_3=8$, and so on.

Now, my real number is $R=0.2258\cdots=0.c_0c_1c_2c_3c_4\cdots$. Is it on your list? If so, $R=B_n$ for some $n\in\Bbb N$, so which $n$ is it?"

You: "Let's see, it cannot be $n=0$, since the first decimal of $B_0=\pi$ is $d_0^0=1$, and the first decimal of $R$ is $c_0=2$, and they are different. It can also not be $n=1$, since the second decimal of $B_1=e$ is $d_1^1=1$ and the second decimal of $R$ is $c_1=2$ and they're also different."

You keep going, but you soon find out that $R$ is different from every number in your list: the $n$-th decimal of $B_n$ is always different from the $n$-th decimal of $R$!

Me: "See, $R$ is not on your list! So your list was not complete."
You: "But let me add $R$ to my list, and then try again!"
Me: "That's not how it works: you told me your list is complete, and I told you you only get a single chance. Your list wasn't complete, and your single chance has failed."

Vsotvep
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