1.2. $\triangleright$ Prove that if $\sim$ is an equivalence relation on a set $S$, then the corresponding family $\mathscr{P}_\sim$ defined in $\S1.5$ is indeed a partition of $S$: that is, its elements are nonempty, disjoint, and their union is $S$. $[\S1.5]$
Solution
First recall that $\mathscr{P}_\sim$ is defined as the set of all equivalence classes, where for each $a \in S$ we have the equivalence class
$$[a]_\sim \coloneqq \{b \in S \mid b \sim a\} .$$
Each $[a]_\sim$ is nonempty as $a \in [a]_\sim$. For some $a, b \in S$ if $x \in [a]_\sim \cap [b]_\sim$ then we have both $x \sim a$ and $x \sim b$ so by symmetry and transitivity $a \sim b$, so our equivalence classes are disjoint. Finally, since we have an equivalence class for each $a \in S$ of course the union of all the elements is $S$. $\square$

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1.3. $\triangleright$ Given a partition $\mathscr{P}$ on a set $S$, show how to define a relation $\sim$ on $S$ such that $\mathscr{P}$ is the corresponding partition. $[\S1.5]$
Solution
Define $\sim$ such that for $a, b \in S$ we have $a \sim b$ if and only if for some $A \in \mathscr{P}$ we have $a, b \in A$. We need to show that $\mathscr{P}_\sim = \mathscr{P}$. Let $[a]_\sim \in \mathscr{P}_\sim$, so for some $A \in \mathscr{P}$ we have $a \in A$, and our construction of $\sim$ makes it clear that $[a]_\sim = A$, so $[a]_\sim \in \mathscr{P}$. Conclude that $\mathscr{P}_\sim \subseteq \mathscr{P}$. Now let $A \in \mathscr{P}$. Take any $a \in A$ and observe that $A = [a]_\sim$. Conclude that $\mathscr{P} \subseteq \mathscr{P}_\sim$, which combined with $\mathscr{P}_\sim \subseteq \mathscr{P}$ gives the desired $\mathscr{P}_\sim = \mathscr{P}$. $\square$

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1.6. $\triangleright$ Define a relation $\sim$ on the set $\R$ of real numbers by setting $a \sim b \iff b - a \in \Z$. Prove that this is an equivalence relation, and find a 'compelling' description for $\R / \sim$. Do the same for the relation $\approx$ on the plane $\R \times \R$ defined by declaring $(a_1, a_2) \approx (b_1, b_2) \iff b_1 - a_1 \in \Z$ and $b_2 - a_2 \in \Z$.
Solution
Let's start by proving that $\sim$ is an equivalence relation:

• reflexivity: $\forall a \in \R$ we have $a - a = 0 \in \Z$ so $a \sim a$;
• symmetry $(\forall a \in \R)$ $(\forall b \in \R)$ $a \sim b \implies b - a \in \Z \implies a - b \in \Z \implies b \sim a$;
• transitivity $(\forall a \in \R)$ $(\forall b \in \R)$ $(\forall c \in \R)$, $(a \sim b \text{ and } b \sim c) \implies \left(b - a \in \Z \text{ and } c - b \in \Z \right) \implies \left(c - a \in \Z\right) \implies a \sim c$.
  The argument for $\approx$ is similar (should just amount to repeating the above twice). I am lazy and also I don't know what 'compelling' description Mr. Aluffi is looking for. $\square$

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