📂Set Theory

Homotopy Type

Definition ¹

Let’s say a equivalence relation $R$ is defined on a set $X$. For $x \in X$, $x / R := \left\{ y \in X : y R x \right\}$ is called the equivalence class of $x$. The set of all equivalence classes given by $X$ is represented as $X / R := \left\{ x / R : x \in X \right\}$.

Explanation

Though the expression might look a bit messy, it’s not a difficult concept at all if you think of an example. On the set of natural numbers $\mathbb{N}$, let’s say that if the remainder when divided by $3$ is the same, they are equivalent, and if $x,y \in \mathbb{Z}$ are equivalent, let it be represented as $x \equiv y \pmod{3}$. Since $5,7$ have different remainders when divided by $3$, which are $2,1$ respectively, they are not equivalent but $11,17$, having the remainder $2$ when divided by $3$, can be written as $11 \equiv 17 \pmod{3}$. $$1 \equiv 4 \equiv 7 \equiv 10 \equiv \cdots \pmod{3} \\ 2 \equiv 5 \equiv 8 \equiv 11 \equiv \cdots \pmod{3} \\ 3 \equiv 6 \equiv 9 \equiv 12 \equiv \cdots \pmod{3}$$ As can be seen from the calculation above, the equivalence class of $1$ is $( 1 / \equiv) = \left\{ 1, 4, 7, 10, \cdots \right\}$, and the equivalence class of $2$ is $(2 / \equiv) = \left\{ 2, 5, 8, 11, \cdots \right\}$, and the equivalence class of $3$ is $(3 / \equiv) = \left\{ 3, 6, 9, 12, \cdots \right\}$. From numbers larger than $3$, these three equivalence classes repeat, and thus, $$(\mathbb{N} / \equiv) = \left\{ 1 / \equiv , 2 / \equiv , 3 / \equiv \right\}$$ As can be seen from the example above, equivalence classes have the following common-sense properties.

Basic Properties

• [1] $x / R \ne \emptyset$
• [2] $x / R \cap y / R \ne \emptyset \iff xRy$
• [3] $x/R = y/R \iff x R y$
• [4] $x / R \cap y / R \ne \emptyset \iff x/R = y/R$

Proof

[1]

Since $R$ is an equivalence relation on $X$, by reflexivity, for all $x \in X$, $x R x$, and $x \in x / R$ must hold.

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Strategy [2][3]: Take an arbitrary $z$, separate from $x,y$, and connect the equations using the symmetry and transitivity of the equivalence relation.

[2]

$x/R$ and $y/R$ are not empty sets and are equivalence relations on $X$, so $x/R \cap y/R \ne \emptyset$, which is equivalent to $$\begin{align*} & z \in x / R \land z \in y / R \\ \iff & z R x \land z R y \\ \iff & x R z \land z R y \\ \iff & x R y \end{align*}$$ for some $z$.

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[3]

$( \implies )$ If $x/R = y/R$, then $x/R \cap y/R \ne \emptyset$, so by [2], $x R y$

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$( \impliedby )$ If $x R y$, for all $z \in x / R$, $z R x$. Since $x R y$, by the transitivity of $R$, $z R y$ and $z \in y / R$. To summarize, $$z \in x / R \implies z \in y / R$$ If we change it to the form of a set inclusion relation, $$x / R \subset y / R$$ Similarly, we can obtain $y / R \subset x / R$, so $$x / R = y / R$$

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[4]

Following the principle of syllogism, from [2] and [3], $$x / R \cap y / R \ne \emptyset \iff xRy \iff x/R = y/R$$

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