Some types of co-finite subsets of real numbers

Question

Is there any $A\subseteq \mathbb{R}$ such that $A-A$ is a non-trivial co-finite subset?

Note that $A-A=\{a_1-a_2: a_1,a_2\in A\}$, also see Some co-finite subsets of rational numbers ,
A type of integer numbers set.

Answer 1

Yes, such sets exist. But my example is not constructive, as it uses a well-ordering of the reals. We'll expose $A$, such that $A-A = \mathbb R \setminus \{\pm 1\}$.

Let $\{x_\alpha\}_{\alpha < \mathfrak c}$ be a enumeration of $\mathbb R \setminus \{\pm 1\}$ by the smallest ordinal $\mathfrak c$ of continuum cardinality. We'll construct by transfinite induction sets $\{A_\alpha\}_{\alpha \leqslant \mathfrak c}$, such that $$\{x_\beta\}_{\beta < \alpha} \subseteq A_\alpha - A_\alpha \subseteq \mathbb R \setminus \{\pm 1\}.$$ Then $A = A_{\mathfrak c}$ will work.

Base. $A_0 := \varnothing$.

Step. Let $0 < \alpha \leqslant \mathfrak c$. Suppose $\{A_\beta\}_{\beta < \alpha}$ is already constructed.

• If $\alpha$ is limit ordinal, simply define $A_\alpha := \cup_{\beta < \alpha} \, A_\beta$.
• If $\alpha = \gamma+1$, then pick $y \in \mathbb R \setminus (A_\gamma - \{\pm 1, \,x_\gamma\! \pm 1\})$, i.e. that $y - a \ne \pm 1, \,x_\gamma\! \pm 1$ for $a \in A_\gamma$. We can do that because $A_\gamma$ (and so $A_\gamma - \{\pm 1, \,x_\gamma\! \pm 1\}$) has cardinality strictly less than continuum (see below). Then define $A_\alpha := A_\gamma \cup \{y, y\,-x_\gamma\}$. By construction $A_\alpha - A_\alpha \subseteq \mathbb R \setminus \{\pm 1\}$.

On finite steps $\alpha$, $A_\alpha$ is clearly finite, so $A_\omega = \cup_{n=0}^\infty A_n$ is countably infinite. It's easy to see, hence (use induction again), that for every infinite ordinal $\alpha < \mathfrak c$ we have $\lvert A_\alpha \rvert = \lvert \alpha \rvert < \lvert \mathfrak c \rvert$.