# equivalence relation partitioning for the set $[0,1]$

I have a question regarding equivalence partitioning in the interval of $[0,1]$.

Define an equivalence relation on $[0,1]$ by: $x \sim{y}$ if and only if the difference $y-x$ is rational. This relation partitions the interval $[0,1]$ into a disjoint union of equivalence classes. Let H be a subset of $[0,1]$ consisting of precisely one element from each equivalence class. Now since $H$ contains an element of each equivalence class, we see that each point in $[0,1]$ is contained in the union $\bigcup_{r\in [0,1] r rational} (H \bigoplus r)$ of shifts of $H$.

where $H \bigoplus r$ is defined as $H \bigoplus r = \{ a+r; a\in{H}, a+r \leq 1\} \cup \{a+r-1; a\in H, a+r>1\}$

My question is specifically:

how each point in $[0,1]$ is contained in the union of $H \bigoplus r$ and also how the equivalence relation $x \sim y$ if and only if the difference $y-x$ is rational" would partition the interval $[0,1]$.

Thank you

There is a flaw in the Q. We cannot have $0=a+r$ with $a+r\leq 1$ because $a\geq 0$ and $r\in [0,1]$ \ $\mathbb Q \implies r>0.$ We cannot have $0=a+r-1$ with $a+r>1$ because $0=a+r-1\implies a+r=1.$ I will amend this by replacing "$\{a+r-1:a\in H \land a+r>1\}$" with "$\{a+r-1:a\in H\land a+r\geq 1\}$" .

Generalities about equivalence relations and partitions:

(1). For any equivalence relation $\sim$ on a set $S,$ the set $\{[x]_{\sim}:x\in S\}$ of equivalence classes (...where $[x]_{\sim}=\{y:y\sim x\}$...) is a partition of $S$:

(i). Each $x\in S$ belongs to its equivalence class $[x]_{\sim}$. So $\cup S_{/\sim}=S.$

(ii). Distinct members of $S_{/\sim}$ are disjoint, for if $z\in [x]_{\sim}\cap [y]_{\sim}$ then $x\sim z \land z\sim y$, implying $x\sim y$. So for any $w$ we have $w\sim x\iff w\sim y.$ Hence $$[x]_{\sim}\cap [y]_{\sim}\ne \emptyset\implies [x]_{\sim}=[y]_{\sim}.$$

(2). Let $H\subset S$ where $H$ has exactly one member in each $[x]_{\sim}.$ For each $x\in S$ there is exactly one $h\in H$ such that $x\sim h.$ Because

(i'). There exists exactly one $h(x)\in [x]_{\sim}\cap H$ , and $x\sim h(x)$.

(ii'). If $h_1\sim x \sim h_2$ for $h_1,h_2 \in H$ then $$h_1\sim x \sim h_2\implies \{h_1, h_2\}\subset H\cap [x]_{\sim} =\{h(x)\}\implies h_1=h_2=h(x).$$

IN YOUR Q: For brevity let $J=[0,1]$ \ $\mathbb Q.$ $$\text {Let }\; K_1=\{a+r:a\in H\land r\in J\land a+r\leq 1\}.$$ $$\text { Let }\; K_2=\{a+r-1:a\in H\land r\in J \land a+r\geq 1\}.$$ For $x\in [0,1]$ take $h(x)\in H\cap [x]_{\sim}$ . Take any $y\in [0,1]$ such that $x-y \not \in \mathbb Q.$ Then $x-h(y)\not \in Q$ because $$x-h(y)\in Q\implies x\sim h(y) \sim y \not \sim x\implies x \not \sim x.$$ (Or,if you prefer, $x-h(y)=(x-y)+(y-h(y))$ is the sum of the irrational $x-y$ and the rational $y-h(y)$.)

If $x-h(y)>0$ let $r=x-h(y).$ Then $r\in J$ and $x= h(y)+r\in K_1.$

If $x-h(y)<0$ let $r=1+x-h(y).$ Then $r\in J$ and $x+1=h(y)+r\geq 1$ so $x=h_y+r-1\in K_2.$

Footnote: A partition $P$ of a set $S$ can define an equivalence relation on $S$: For $x,y\in S$ let $x\sim y$ iff $x,y$ belong to the same member of $P.$ Along with (1) above, we see that partitions of a set correspond to equivalence relations on the set.

• Hi user254665, thanks for your explanations here. But I have not able to fully understand everything you said here yet since my math level requires more time to digest all the notation and explanations here. I will thoroughly go through every single sentence and post questions here within couple days based on your comments. Thanks a lot for your guidance here. – john_w Jan 9 '17 at 16:46

First, see if your understanding or agreement breaks down on any one of the following statements:

• Some numbers in the interval $[0,1]$ are rational.
• Some numbers in the same interval are not rational.
• Some irrational numbers in the interval $[0,1]$ are a rational distance apart on a number line.
• Example: $\pi/4$, and $\pi/4-0.5$. These numbers are both irrational, but the difference between them is obviously rational. (And a calculator should convince you that both are in the interval $[0,1]$.)
• Some irrational numbers in this interval are an irrational distance apart.
• Example: $\sqrt2/2$ and $\pi/4$.

So as to how the equivalence relation "$x \sim y$ if and only if the difference $y-x$ is rational" would partition the interval $[0,1]$, it's very simple:

All rational numbers in this interval would be in a single equivalence class (since the difference between any two rational numbers is also rational), and the irrational numbers in the interval would be in uncountably infinite other equivalence classes, each with countably infinite members.

If you're not familiar with equivalence classes, but you are comfortable reading the following symbols:

$\forall, \exists, \in, \neg$

Read page 11 of this PDF which defines reflexivity, irreflexivity, symmetry, antisymmetry, asymmetry, and transitivity.

Then (in the same PDF) read section 7.4 "Equivalence Relations" (starting on page 10), including section 7.4.1 "Partitions."

If you run into any trouble go earlier. If you need to go even earlier than the beginning of that particular PDF, all the readings for that course are available online as well.

• Hi when you say all "rational number would be in a single equivalence class", is it also true that some irrational number would fall into that class also? because say if i have a irrational number denoted as $y_1$ and another number say k (rational or irrational), and $y_1$ - k = rational number. Then does it mean $y_1$ also falls in the equivalence class as all rational numbers? Also could you could explain little bit more about the "other equivalence classes" in your very last sentence of "the irrational numbers in the interval would be in uncountably infinite other equivalence classes"? – john_w Dec 29 '16 at 8:43
• Hi, does it mean some irrational number would fall into the same equivalence class as the equivalence class that consists of rational numbers? because say I have an irrational number denoted as $y_1$ and $y_1 - k$ = $r_3$ which is rational. Is it correct? – john_w Dec 29 '16 at 8:45
• @john_w, it is impossible to add two rational numbers and get an irrational number. So no, there are no irrational numbers in the equivalence class which includes all the rational numbers. – Wildcard Dec 29 '16 at 23:24
• Hi, but if I have two irrational numbers say $r_1$ and $r_2$, and $r_1$-$r_2$ = a rational number, then those two irrational number ($r_1$, $r_2$) would fall in the equivalence class that contain the rational number, is it correct? because I was reading that the difference between two irrational numbers can also be a rational number. – john_w Dec 30 '16 at 2:42
• @john_w, $r_1$ and $r_2$ would fall into the same equivalence class as each other, but that equivalence class would not contain any rational numbers at all. Do you have a clear grasp on the term "equivalence class"? – Wildcard Dec 30 '16 at 7:12