I'm currently writing a short paper on Intuitionism for uni. The subject of this paper is the decay of the intermediate value theorem under intuitionism. I have found a proof for this but I have a question about a step the author makes in this proof. It is abbout the following:
The problem I have with using this in my paper is that in the introduction I proclaim that under intuitionism you abandon the law of the excluded middle and then proceed to give an implicit proof for this sequence being Cauchy where (at least in my head) you give the two possibilities where the Goldbach conjecture is true (the limit is 0) or false (the limit is 2^{-k0}) It could be that this is nothing to worry about, but the reasoning seems a little crooked to me.
If anyone could maybe clarify this to me it would be greatly appreciated!
There is no need to use the law of excluded middle here.
In general, consider some sequence $\{s_n \in \{0, 1\}\}_{n \in \mathbb{N}}$. We can define a corresponding Cauchy sequence
$$a_n = \begin{cases} 2^{-k_0} & k_0 \leq k \text{ is the smallest }k_0 \text{ such that } s_{k_0} = 1 \\ 2^{-k} & otherwise \end{cases}$$
We can prove by induction that for all $k \in \mathbb{N}$, there either is a least $k_0 \leq k$ such that $s_{k_0} = 1$, or there is no $k_0 \leq k$ such that $s_{k_0}$. Thus, $a$ is well-defined.
I claim that for all $i, j \geq k$, we have $|a_i - a_j| < 2^{-k}$. This means that $a$ is a Cauchy sequence.
To prove this, we suppose WLOG that $i \leq j$. We consider three cases.
Case 1: there is no $k_0 \leq j$ such that $s_{k_0} = 1$. If this is the case, then there is also no $k_0 \leq i$ such that $s_{k_0} \leq i$. So we have $|a_i - a_j| = 2^{-i} - 2^{-j} < 2^{-i} \leq 2^{-k}$.
Case 2: there is a least $k_0 \leq k$ such that $s_{k_0} = 1$. Then we have two sub cases:
Case 2a: $k_0 \leq i$. In this case, $a_i = a_j = 2^{-k_0}$, so $|a_i - a_j| = 0 < 2^{-k}$.
Case 2b: $i < k_0$. In this case, we have $|a_i - a_j| = 2^{-i} - 2^{-k_0} < 2^{-i} \leq 2^{-k}$.
This completes the proof that $a$ is Cauchy.
Your Goldbach situation is a special case of this general situation. We have
$$s_k = \begin{cases} 1 & k \text{ is a counterexample to the Goldbach conjecture}\\ 0 & otherwise \end{cases}$$
Finally, note that $\lim\limits_{n \to \infty} s_n > 0$ if and only if there exists some $k$ such that $s_k = 1$.