Disjointness and Coverage of Vitali Translates

From Representatives to Translates

The selection of one representative from each rational-equivalence class gives a set with a useful geometric structure. Translating that set by carefully chosen rationals produces disjoint copies, while still reaching every point of the unit interval. The key is to keep track of two facts separately: representatives in the same class must be identical, and the shift needed to move between equivalent points is rational.

Let $x\sim y$ mean $x-y\in\mathbb{Q}$, for $x,y\in[0,1]$. This is an equivalence relation: $x-x=0\in\mathbb{Q}$; if $x-y\in\mathbb{Q}$, then $y-x=-(x-y)\in\mathbb{Q}$; and if $x-y,y-z\in\mathbb{Q}$, then $x-z=(x-y)+(y-z)\in\mathbb{Q}$. Let $V\subseteq[0,1]$ be a representative set: every equivalence class in $[0,1]$ contains exactly one element of $V$. In particular, $V$ contains one representative, not every element, from the class of each $x\in[0,1]$.

For $q\in\mathbb{Q}$, write $V+q=\{v+q:v\in V\}$. The shifts in this tutorial are indexed by $\mathbb{Q}\cap[-1,1]$. These translated sets need not be assumed measurable; the claims below concern membership, equality, and containment of sets.

Definition: Two sets in a family indexed by $q\in\mathbb{Q}\cap[-1,1]$ are pairwise disjoint if, for every two distinct indices $q_1$ and $q_2$, their intersection is empty. The union of the translates is $\displaystyle\bigcup_{q\in\mathbb{Q}\cap[-1,1]}(V+q)$.

The Representative Property Forces Uniqueness

The immediate consequence of the representative property is the step that will rule out overlap.

Lemma: If $v,w\in V$ and $v-w\in\mathbb{Q}$, then $v=w$.

Proof. Since $v-w\in\mathbb{Q}$, the points $v,w\in[0,1]$ belong to the same equivalence class under $\sim$. The set $V$ contains exactly one element of that class. Both $v$ and $w$ belong to $V$, so they must be that same element; hence $v=w$. $\square$

Notice the direction of this implication. A rational difference puts two points in the same class, and uniqueness in $V$ then identifies them. It does not say that arbitrary points of $[0,1]$ with rational difference are equal; that would be false. It applies specifically when both points are representatives.

Worked Example: Testing Two Specific Shifts for Overlap

Consider the indices $q_1=\tfrac13$ and $q_2=-\tfrac14$. Suppose, to test whether these translates can overlap, that a point belongs to both $V+\tfrac13$ and $V-\tfrac14$. Then there must be $v_1,v_2\in V$ such that $v_1+\tfrac13=v_2-\tfrac14$. Rearranging gives

v_1-v_2=-\frac14-\frac13=-\frac{3}{12}-\frac{4}{12}=-\frac{7}{12}.

The difference is rational, so the lemma gives $v_1=v_2$. Substituting that equality into the proposed overlap equation gives $\tfrac13=-\tfrac14$, which is false: in twelfths, $\tfrac13=\tfrac4{12}$ and $-\tfrac14=-\tfrac3{12}$. Therefore these two translates do not overlap. The calculation illustrates the general argument; it does not require knowing which elements belong to $V$.

Disjointness of Distinct Rational Translates

Theorem: If $q_1,q_2\in\mathbb{Q}\cap[-1,1]$ and $q_1\ne q_2$, then $(V+q_1)\cap(V+q_2)=\varnothing$.

Proof. Fix distinct $q_1,q_2\in\mathbb{Q}\cap[-1,1]$. Suppose, for contradiction, that $z\in(V+q_1)\cap(V+q_2)$. By the definition of a translate, there are $v_1,v_2\in V$ such that $z=v_1+q_1=v_2+q_2$. Subtracting the two expressions for $z$ yields

$$ v_1-v_2=q_2-q_1. $$

Because both shifts are rational, $q_2-q_1\in\mathbb{Q}$. Thus $v_1-v_2\in\mathbb{Q}$, and the lemma implies $v_1=v_2$. The equality $v_1+q_1=v_2+q_2$ now implies $q_1=q_2$, contrary to the assumption that the indices are distinct. No such $z$ exists, so the intersection is empty. This proves pairwise disjointness for every distinct pair of allowed shifts. $\square$

The restriction to distinct indices matters: a set is not disjoint from itself unless it is empty. The theorem asserts that different translates do not share points. It does not say that each translate is nonempty; if $V$ were empty, the conclusion about intersections would still hold, although the representative property ensures $V$ is not empty.

Coverage of the Unit Interval

To place a given $x\in[0,1]$ in one of the translates, use the representative of its equivalence class. If that representative is $v$, then $x-v$ is rational by the definition of the class. Since both $x$ and $v$ lie in the unit interval, their difference lies between $-1$ and $1$. Thus $x-v$ is one of the allowed shifts.

Theorem: For the representative set $V\subseteq[0,1]$, the rational translates indexed by $\mathbb{Q}\cap[-1,1]$ satisfy $$ [0,1]\subseteq\bigcup_{q\in\mathbb{Q}\cap[-1,1]}(V+q)\subseteq[-1,2]. $$

Proof. First let $x\in[0,1]$. There is exactly one $v\in V$ in the equivalence class of $x$. Hence $x-v\in\mathbb{Q}$. Since $0\le x\le1$ and $0\le v\le1$, subtracting gives $-1\le x-v\le1$. Set $q=x-v$; then $q\in\mathbb{Q}\cap[-1,1]$, and $x=v+q\in V+q$. This proves that every $x\in[0,1]$ belongs to the union.

For the reverse containment, take $z$ in the union. Then $z=v+q$ for some $v\in V$ and $q\in\mathbb{Q}\cap[-1,1]$. From $0\le v\le1$ and $-1\le q\le1$, addition gives $-1\le v+q\le2$. Therefore $z\in[-1,2]$. Both inclusions hold, as required. $\square$

Worked Example: Covering a Specific Point

Take $x=\tfrac25$. Let $r\in V$ be the representative of the equivalence class of $\tfrac25$. Because $r\sim\tfrac25$, the shift $q=\tfrac25-r$ is rational. Since $0\le r\le1$, subtracting $r$ from $\tfrac25$ gives

-\frac35=\frac25-1\le \frac25-r\le \frac25-0=\frac25.

Thus $-1\le q\le1$, so $q$ is an allowed index. Substituting its definition gives $r+q=r+(\tfrac25-r)=\tfrac25$, and hence $\tfrac25\in V+q$. As a check, the same identity rearranges to $q=\tfrac25-r$, exactly the shift used above. The argument determines the shift from the representative without needing to know its particular value.

Worked Example: Checking Coverage at an Endpoint

Take $x=1$, and let $s\in V$ be the representative of the class of $1$. Equivalence gives $1-s\in\mathbb{Q}$. Since $s\in[0,1]$,

0=1-1\le 1-s\le 1-0=1.

Therefore $q=1-s$ lies in $\mathbb{Q}\cap[-1,1]$. The translate with that index contains the endpoint, because $s+q=s+(1-s)=1$. Checking the other endpoint works in the same way: if $t\in V$ represents the class of $0$, then $q=-t\in\mathbb{Q}$, and $-1\le-t\le0$; consequently $t+q=t-t=0$. Both endpoints are covered, including the possibility that an allowed shift is an endpoint of $[-1,1]$.

Why the Outer Bounds Are Exactly the Useful Claim

Worked Example: Bounding Every Point of a Translate

Fix any $v\in V$ and any allowed shift $q\in\mathbb{Q}\cap[-1,1]$. The smallest possible sum permitted by these bounds is $0+(-1)=-1$; the largest is $1+1=2$. More explicitly,

-1=0+(-1)\le v+q\le1+1=2.

Thus every element $v+q$ of every allowed translate lies in $[-1,2]$. The endpoint checks are exact: the bounds on $v$ and $q$ alone allow the sums $-1$ and $2$. This calculation establishes containment in $[-1,2]$; it does not show that the union contains every point of that larger interval. The coverage argument established only that it contains $[0,1]$.

These two conclusions are designed to be used together: the translates are indexed by distinct rational shifts, do not overlap, and their union is trapped between a covered interval and a larger bounded interval. At this stage, the conclusions are purely set-theoretic. They do not yet assign a measure to $V$ or to its translates.

A Common Sign Error

To solve $x=v+q$ for the shift, the correct expression is $q=x-v$, not $q=v-x$. For a concrete arithmetic check, take the numbers $v=\tfrac15$ and $x=\tfrac25$. The incorrect choice $q=v-x=\tfrac15-\tfrac25=-\tfrac15$ gives $v+q=\tfrac15-\tfrac15=0$, not $\tfrac25$. The corrected choice is $q=x-v=\tfrac25-\tfrac15=\tfrac15$, and then $v+q=\tfrac15+\tfrac15=\tfrac25$. These numbers illustrate the sign in the algebra; they make no claim that $\tfrac15$ is the selected representative in $V$.

A separate overstatement is to replace the proved outer containment by equality with $[-1,2]$. The inequality $-1\le v+q\le2$ only bounds elements that occur. It does not show that every point between those endpoints occurs. The proof of coverage uses the representative property for $x\in[0,1]$, and no corresponding argument has been given for every $x\in[-1,2]$.

Check Your Understanding

For each question, keep the representative condition and the allowed range of shifts explicit.

If $v,w\in V$ and $v-w\in\mathbb{Q}$, which part of the representative property gives $v=w$?
Suppose a point belongs to both $V+q_1$ and $V+q_2$. Derive an expression for the difference of the two representatives, then explain why distinct indices are impossible.
For an arbitrary $x\in[0,1]$, define the shift using its representative $v$, and verify both that it is rational and that it lies in $[-1,1]$.
Why does $V\subseteq[0,1]$ together with $q\in[-1,1]$ imply $V+q\subseteq[-1,2]$?
Which part of the argument proves coverage of $[0,1]$, and why does it not prove equality of the union with $[-1,2]$?