Is the class $\mathcal{S}$(difference set containing no intervals) a $\sigma$-ideal?

Question

For any $A\subset \mathbb{R}$ , define $D(A) =\{x-y:x,y\in A\}$.

Let, $\mathcal{S}=\{A\subset \mathbb{R}: I\subset D(A) \implies I \text{ is not an interval}\}$

Q. Is the class $\mathcal{S}$ a $\sigma$-ideal ?

1. $\emptyset\in \mathcal{S} $

2. $B\subset A $ and $A\in \mathcal{S}$ implies $D(B) \subset D(A) $ and $B\in \mathcal{S}$

Now I have to show $A=\bigcup_{n\in\mathbb{N}}A_n\in\mathcal{S}$ where $A_n\in \mathcal{S}$ for all $n\in\mathbb{N}$.

I have no clue .How to proceed ?

Answer 1

$\mathcal{S}$ is not a $\sigma$-ideal. I'll give an example where $A_1,A_2 \in \mathcal{S}$, but $A_1\cup A_2 \notin \mathcal{S}$.

Each real number has a decimal representation. For rationals $\neq 0$ with a finite representation there are actually two infinite ones, one with a period of $0$, the other with a period of $9$, e.g.

$$\frac12 = 0.5 = 0.5\bar{0} = 0.4\bar{9}.$$

In the following we will allow both such representations.

So for a real number $d$ in the interval $[0,1]$, let $d=0.d_1d_2d_3\ldots$ be one of the potentially two infinite decimal representations.

We define

$$A_1=\{d \in [0,1]: d \text{ has a decimal representation with } 0=d_1=d_3=d_5=\ldots\},$$

$$A_2=\{d \in [0,1]: d \text{ has a decimal representation with } 9=d_2=d_4=d_6=\ldots\}.$$

$A_1$ is in $\mathcal{S}$.

Proof: The positve difference of any two elements of $A_1$ will, at each odd position after the decimal point, have either a $9$ or a $0$, depending on if there is a "borrow" coming from the next even position, or not. Taking into account the non-uniqueness of decimal representations, that still means the difference cannot be a number that has a representation with a $5$ in odd positions.

But since any interval, however small, needs to include one full interval of the sort $[\frac{n}{10^k}, \frac{n+1}{10^k}]$ for some (possibly large) $k$, $D(A_1)$ does not contain any interval! $\blacksquare$

The proof that $A_2 \in \mathcal{S}$ is totally analogous.

Now we prove that $[0,1] \subseteq D(A_1 \cup A_2)$, showing $A_1 \cup A_2 \notin \mathcal{S}:$

Let $x=0.x_1x_2x_3x_4\ldots$ be an number in $[0,1]$. Then we have, defining for each decimal digit $d$ the complement $\dot{d} = 9-d$ (which is again a decimal digit):

$$x = 0.x_19x_39\ldots - 0.0\dot{x_2}0\dot{x_4}\ldots$$

The difference never uses a borrow, and we have $0.x_19x_39\ldots \in A_2$ and $0.0\dot{x_2}0\dot{x_4}\ldots \in A_1$.

This concludes the proof that $[0,1] \in D(A_1 \cup A_2)$, and so we know that $\mathcal{S}$ isn't a $\sigma$-ideal.