Motivation. In computer science, addition of integers $a+b$ can be approximated by a very fast operation: $(a,b)\mapsto (a\oplus b) \oplus ((a\land b) \ll 1)$, where $\oplus$ denotes bitwise XOR, $\land$ is bitwise AND, and $\ll 1$ is the left shift by one position. The purpose of $\ldots \oplus ((a\land b) \ll 1)$ is to simulate carry propagation. This mechanism can be generalized to an operation $2^\omega \times 2^\omega \to 2^\omega$, as described below.
Formal setting. Let $2 = \{0,1\}$ and let $\omega$ denote the first infinite ordinal. If $A, B$ are sets, by $B^A$ we denote the set of functions $f:A\to B$.
We define the following operations on $2^\omega$:
- $\ll 1\; :\; 2^\omega \to 2^\omega$: $f \mapsto s(f)$ where $s(f)(0) = 0$ and $s(f)(n+1) = f(n)$ for all $n\in\omega$.
- $\oplus : 2^\omega \times 2^\omega \to 2^\omega$: $(f,g) \mapsto (f\oplus g)$ where $(f\oplus g)(n) = f(n) + g(n)$ for all $n\in \omega$, and $+$ denotes addition modulo $2$.
- $\land : 2^\omega \times 2^\omega \to 2^\omega$: $(f,g) \mapsto (f\land g)$ where $(f\land g)(n) = \inf\{f(n), g(n)\}$ for all $n\in \omega$.
Finally we define $+'\;:\; 2^\omega \times 2^\omega \to 2^\omega$ by $(a,b)\mapsto (a\oplus b) \oplus ((a\land b) \ll 1)$.
Unfortunately, $+'$ is not an associative binary operation on $2^\omega$. (The second comment in the comment section below contains an example showing non-associativity.) It has a neutral element, though (the constant $0$-function), as well as inverses.
Question. Is there an uncountable subset $S\subseteq 2^\omega$ such that $S$ is closed under $+'$, and $+'$ restricted to $S$ is associative?
