Confusion with Brouwer's Counterexample for Points on a Continuum Being Ordered I'm reading about intuitionism, and I've come across an argument put forth by L. E. J. Brouwer that claims to be a counterexample to the statement that
$\mbox{The points of the continuum form an ordered point species.}$
Independent of intuitionism's truth or usefulness, I just don't understand how what he proposes functions as a counterexample. He proposes the following:
Let $d_v$ be the $v^{\text{th}}$ digit of $\pi = 3.1415 \ldots$ to the right of the decimal point.
Let $m = k_n$ if, starting at $d_m$, the sequence $d_m, d_{m+1}, \cdots, d_{m + 9}$ is the sequence $0123456789$.
Let $c_v = (\frac{-1}{2})^{k_1}$ if $v \ge k_1$ and $c_v = (\frac{-1}{2})^{v}$ otherwise.
Then, the infinite sequence $c_1, c_2, \cdots$ defines a real number $r$ for which none of the conditions $r >0, r <0,$ or $r = 0$ is true.
My understanding of this is that $r$ is the concatenation of the points $c_1, c_2, \cdots$, where you only include the decimal point from the first one (so as to avoid having more than one decimal in your $r$). I'm not sure if something similar is supposed to occur with respect to the negatives in some of the $c_i$'s.
Can anyone explain how this is supposed to be a counterexample? Is Brouwer saying that $r$ will meet none of these conditions because it will have more than $1$ decimal point inside it, which isn't allowed? Or is there something more nuanced happening?
 A: In intuitionistic logic saying that $r > 0$ or $r < 0$ or $r = 0$ means that you have to give me a proof for one of them and you have to tell me which one. By how we constructed $r$ this would give us information about the digits of $\pi$. As mentioned in the comments, this particular information about $\pi$ is not completely unthinkable now that we computers to calculate $\pi$ to absurd precision. However, it is the idea of the construction that counts. I will try to give an explanation, show how the idea can be applied in general and fix a few imprecisions in your version.
Recall that in this setting a real number is a sequence of rational numbers that converges (i.e. a Cauchy sequence). Let for example $P_{99}(n)$ be the statement that the $n$th decimal of $\pi$ is the first place where there is a string of $99$ times the same digit. Even though for big $n$ this takes a long time to check, we could (theoretically) do it. So we can define the sequence
$$
c_n = \begin{cases}
\left( \frac{-1}{2} \right)^n & \text{if not } P_{99}(n) \\
\left( \frac{-1}{2} \right)^m & \text{if } P_{99}(m) \text{ for some } m \leq n
\end{cases}
$$
Regardless of $P_{99}$, this defines a Cauchy sequence. The sequence converges to $0$ if there is no string of 99 times the same digit in $\pi$. If there is such a string, say starting at place $m$, then the sequence will take the constant value $\left( \frac{-1}{2} \right)^m$, where it depends on the parity of $m$ whether or not this is $< 0$ or $> 0$. So if we call $r$ the real number defined by the sequence $(c_n)_{n \in \mathbb{N}}$, then knowing about $r < 0$ or $r > 0$ or $r = 0$ means that we know about $P_{99}$. In particular:

*

*$r < 0$ implies that a string of $99$ times the same digit exists and happens at an odd index,

*$r = 0$ implies that no string of $99$ times the same digit occurs in $\pi$,

*$r > 0$ implies that a string of $99$ times the same digit exists and happens at an even index.

We do not know which one of these is true, so we do not know $r < 0$ or $r = 0$ or $r > 0$.
Of course, it could be that at some point we find an $n$ such that $P_{99}(n)$ is true, and then for this particular real number we have solved what it exactly is. The point is that there is nothing special about $P_{99}$ here. We could replace it by any statement. For example, we could let $H(n)$ mean "$n$ encodes a proof or disproof of the Riemann Hypothesis". Now base a real number $t$ on $H$ in the same way as before, and "$t < 0$ or $t = 0$ or $t > 0$" solves the Riemann hypothesis!
