LLPO implies $x\leq0$ or $x\geq0$ in $\mathbb R$

Question

LLPO:If $a^{n}$ is a binary sequence containing at most one $1$, then either $a^{2n}=0$ for each $n$, or else $a^{2n+1}=0$ for each $n$.

Prove LLPO implies $x\le 0$ or $x\ge 0$ in $\mathbb R$.

Answer 1

Let x be some real number.

Assume that x > 0. assume that x is greater than 0. This is the first step in proving that x is either less than or equal to 0 or greater than or equal to 0.

Let a = x^2. We let a be equal to x^2. This is to ensure that a is a binary sequence containing at most one 1.

We know that a^2n+1=0 since x > 0. Since x is greater than 0, a^2n+1 must equal 0. This is because x^2 is a binary sequence containing at most one 1.

By LLPO, a^2n=0 for each n. By the Law of Large Numbers Principle, since a^2n+1=0, a^2n must equal 0 for each n.

Therefore, x^4=0 for each n. Since a^2n=0 for each n, x^4 must equal 0 for each n.

Therefore, x=0. Since x^4=0 for each n, x must equal 0.

Therefore, x≤0. Since x=0, it follows that x must be less than or equal to 0.

Therefore, x≤0 or x≥0 in R. Since we have proven that x is either less than or equal to 0 or greater than or equal to 0, this implies that x≤0 or x≥0 in R.

Answer 2

Suppose $r$ is a real number, and $f : \mathbb N → \mathbb Q$ a means of approximating it (or choose comparable data if you prefer some other definition of the reals). Construct binary sequences as follows:

• $α(n)$ is $1$ for the least $n$ such that $f(n) > 2^{-n}$, and 0 otherwise. I.E. $α(n) = 1$ if and only if $n$ is the coarsest approximation required to conclude that $r > 0$.
• $β(n)$ is the same, except for $r < 0$
• $γ$ is the interleaving of $α$ and $β$, so $γ(2n) = α(n)$ and $γ(2n+1) = β(n)$

$γ$ has at most one $1$, because:

• if both $α$ and $β$ contained a $1$, we would have both $r > 0$ and $r < 0$
• $α$ and $β$ cannot each contain multiple 1s, because one of them would not be at the minimal index
• if $γ$ contained multiple 1s, then one of the above cases would hold, based on the indices

So, by the LLPO for $γ$, either $∀ n. α(n) = 0$ or $∀ n. β(n) = 0$.

• if $∀ n. α(n) = 0$, then $r \leq 0$, because $f(n) \leq 2^{-n}$ for all $n$
• if $∀ n. β(n) = 0$, then $r \geq 0$, because $f(n) \geq 2^{-n}$ for all $n$

Ergo, we have $∀ r. r \leq 0 ∨ r \geq 0$