Does it relate to the uncountability of irrationals? Recently I am reading Measure, Integration & Real Analysis by Sheldon Axler, and the author claimed that the outer measure is non-additive by constructing a set $\tilde{a}$ generated by $a \in [-1,1]$, and $\tilde{a} = \left\{c\in [-1,1]: a - c \in \mathbb{Q}\right\}$. I found some properties of $\tilde{a}$ quite interesting.

*

*For any two set $\tilde{a}, \tilde{b}$, if $\tilde{a} \cap \tilde{b} \neq \emptyset$, then $\tilde{a} = \tilde{b}$.

This means that for any two such sets, they would have either every element in common or no element in common.


*For $a \in [-1,1]$, for $\forall t \in \mathbb{Q}, a+t \in [-1,1]$, we have $\tilde{a} = \widetilde{a+t}$.


*If $a \in [-1,1]\cap\mathbb{Q}$, and hence $\tilde{a} = \left\{c\in[-1,1]: a-c\in\mathbb{Q}\right\}$, then $\forall c_{1}, c_{2} \in \tilde{a}, c_{1} \neq c_{2}: c_{1}-c_{2} \in \mathbb{Q}$.
As far as I am concerned, I think 2 and 3 imply that, sets generated by a group of numbers which have rational differences between any two of them are the same. Hence imply that, for any $a \in [-1,1]\cap\mathbb{Q}$, the generated set $\tilde{a}$ are the same, which are $[-1,1]\cap\mathbb{Q}$. However, for any irrational numbers $b_{1}, b_{2} \in [-1,1]-\mathbb{Q}$, $\tilde{b_{1}} = \tilde{b_{2}}$ only if $b_{1}-b_{2}\in\mathbb{Q}$.
I think the first-looked reason of this difference is, a rational number plus an irrational number would always outcome an irrational number, while adding up two irrational numbers may result in a rational number. My thought ended up here, intuition tells me that this is also relevant to the uncountability of $[-1,1]-\mathbb{Q}$, would anyone please reject me or give some hints?
Moreover, $\forall a\in [-1,1]$, there will be a corresponding $\tilde{a}$. As stated above, when $a\in\mathbb{Q}$, there will only be one kind of $\tilde{a}$ which is $[-1,1]\cap\mathbb{Q}$. Results would be different if $a\in[-1,1]-\mathbb{Q}$, although there would also be some identical sets since they are generated by irrational numbers which any two of them are having rational difference. Actually this is where my intuition came from, Is the number of distinct sets generated by $a\in[-1,1]-Q$ countable or uncountable?
 A: You are right to observe that, if $a$ and $b$ are reals such that $a - b \in \mathbb Q$, then $\tilde a = \tilde b$. The reason it works is not because $[-1,1]\setminus \mathbb Q$ is uncountable, but because $\mathbb Q$ has the property [if $p$ and $q$ are any rationals, then $p - q$ is rational]. We say that $\mathbb Q$ is closed under the $-$ operation, because the result of the $-$ operation can never "leave" $\mathbb Q$.
Here is a similar example: suppose we start with $\mathbb Z$ instead of $\mathbb Q$. That is, for any real $a$, let $[a] = \{c \in \mathbb R : a - c \in \mathbb Z\}$. Then this also has the property [if $a$ and $b$ are reals such that $a - b \in \mathbb Z$, then $[a] = [b]$.] And moreover, if $a$ and $b$ are integers, then $[a] = [b] = \mathbb Z$. It is also true for the same reason: $\mathbb Z$ is closed under the $-$ operation.
Here is a situation where it does not work: consider starting with $\mathbb N = \{0,1,2,\ldots\}$ and, for each real $a$, let $\mathring{a} = \{c\in\mathbb R : a - c \in \mathbb N\}$. Then by inspection, $\mathring{a} = \{a, a-1, a-2, a-3, \ldots\}$.
So consider $a = 6$ and $b = 5$. Then $a - b = 1 \in \mathbb N$, but $\mathring a = \{6, 5, 4, 3, 2, 1, 0, -1, -2, \ldots\}$ while $\mathring b = \{5, 4, 3, \ldots\}$ and they are not the same. It fails because $\mathbb N$ is not closed under the $-$ operation: $5$ and $6$ both belong to $\mathbb N$, but $5 - 6 = -1$ does not.
These ideas are codified and generalized by the notion(s) of a ring, subrings, and quotient rings, which are studied in algebra. $\mathbb Q$ and $\mathbb Z$ are subrings of $\mathbb R$, but $\mathbb N$ is not.

Your second question, on the other hand, does involve the fact that $[-1,1]\setminus \mathbb Q$ is uncountable.

Is the number of distinct sets generated by $a \in [-1,1]\setminus \mathbb Q$ countable or uncountable?

It is uncountable. The proof comes from these properties:

*

*Every real number belongs to one of the sets in $\{\tilde a : a\in [-1,1]\}$.


*For each real $a$, the set $\tilde a$ is countable (to see this, observe $\tilde a = (a + \mathbb Q) \cap [-1,1]$).
Short proof: you cannot express $[-1,1]$ as the countable union of countable sets, so there must be uncountably many sets.
Long proof: suppose to the contrary, there were only countably many sets. Let $\tilde a_1, \tilde a_2, \ldots$ be an enumeration of all of them. Each real $b$ belongs to some class $\tilde a_i$, where $i = i(b)$ depends on $b$. This demonstrates that
$$[-1,1] = \bigcup_{n=1}^\infty \tilde a_n$$
and because $\tilde a_n$ is countable for every $n$, the set $[-1,1]$ must be countable as well. But this is a contradiction, because $[-1,1]$ is uncountable. Therefore, there must be uncountably many distinct sets in $\{\tilde a : a \in [-1,1]\}$.
