I'm looking at the <https://arxiv.org/abs/1904.09193> paper (version 2, from 2021) and think it has a few errors. I think I found three small places where the paper needs to be corrected (in the sense that the corrected version is valid - none of the key conclusions are undermined by this).


1. On page 5 it says "Lemma 3.5. Let Q ∈ $2^{N\_∞}$ be given. If Q(ω) = 1 and ∀n ∈ N. Q(n) = 1, then ∀p ∈ $2^{N\_∞}$ . Q(p) = 1." I believe the second $2^{N\_∞}$ should be $N\_∞$ (because Q ∈ $2^{N\_∞}$, you'd want to evaluate Q at an element of $N\_∞$)
2. A few lines down is the statement "Theorem 3.6 ([Esc13, Theorem 3.15]). There is a function ε : $2^{N\_∞}$ → $N\_∞$ such that for every
Q ∈ $2^{N\_∞}$ , if Q(ε(Q)) = 1, then ∀p ∈ $2^{N\_∞}$ . Q(p) = 1.". The last $2^{N\_∞}$ should be $N\_∞$ for the same reason.
3. At the end of the proof of that theorem, in "Hence
Q(p) = 1 for every p ∈ $2^N$", $2^N$ should be $N\_∞$ again for the same reason.


Perhaps this is a social question - how do I contact an author? - but I also wanted to post here to see whether what I am proposing is right and in case other people are hitting this when trying to read this paper.


P.S. I formalized these proofs at <https://us.metamath.org/ileuni/nninfall.html> and <https://us.metamath.org/ileuni/nninfsel.html> in case that helps.



Did I find a few (small) errors in the Pradic and Brown 2021 paper that Schroeder-Bernstein implies excluded middle?
It may interest you that Martín Escardó also formalized the proofs, see [`Cantor-Schröder-Bernstein-gives-EM`](https://www.cs.bham.ac.uk/%7Emhe/TypeTopology/CantorSchroederBernstein.CSB.html#6327) in his [TypeTopology](https://www.cs.bham.ac.uk/%7Emhe/TypeTopology/index.html) library.


Regarding contacting the authors, I am pretty sure this [Pierre Pradic](https://www.swansea.ac.uk/staff/p.r.a.pradic/) is the right one. Chad has always made it hard to be tracked down. If you search the [coq-club](https://sympa.inria.fr/sympa/info/coq-club) mailing list archives from 2011 or so, you'll find his email address.


