There is a collaboration grant for a research group that I could apply for the master. But I think I will have enough with a suddenly first exposure to proof-based math courses, and I should study as much as possible to get good grades. It will be too much time to spend on both things. Maybe the next year I can think about it again
Specifically, today I've learned the equivalence between monomorphisms and injections
Okay, it is an errata, this edition isn't the newest tho
Well I did it but it was a very simple typechecker for expressions. I gave up when I tried to implement definitions
announcement (boosts appreciated)
We're delighted to announce the HoTTEST Summer School, which will take place online everywhere in the world during the months of July and August 2022.
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Prove that if $\sim$ is a relation on a set $S$, then the corresponding family $\mathscr P_\sim$ defined in 1.5 is indeed a partition of $S$: that is, its elements are nonempty, disjoint, and their union is $S$.
We will previously prove a lemma to use in the disjoint proof.
**Lemma 1**. $\forall a,b \in S (a \in [b]_\sim \implies [a]_\sim = [b]_\sim)$.
Proof: Suppose $a,b\in S$ and $a \in [b]_\sim$. To prove $[a]_\sim \subseteq [b]_\sim$., suppose an arbitrary $a' \in [a]_\sim$. We have $a' \sim a$ and $a \sim b$. Since $\sim$ is transitive, we have $a' \sim b$, therefore $a' \in [b]_\sim$. To prove $[b]_\sim \subseteq [a]_\sim$, suppose an arbitrary $b' \in [b]_\sim$. Then we have $b' \sim b$ and $a \sim b$. By symmetry, $b \sim a$, and by transitivity $b' \sim a$, therefore $b' \in [a]_\sim$. Therefore $[a]_\sim = [b]_\sim$, and by our assumptions the lemma is true. $\square$
**Proof of exercise 1.2**.
To prove that $\mathscr P_\sim$ is a partition of $S$, we need to prove that its elements are nonempty, disjoint, and their union is $S$.
(Non-empty) Suppose an equivalence class $A \in \mathscr P_\sim$. So there must be an $a \in S$ such that $A = [a]_\sim$. It can't be nonempty, since $\sim$ is reflexive so $a \in [a]_\sim$. Since $a$ was arbitrary, then no equivalence class is empty, hence the elements of $\mathscr P_\sim$ are nonempty.
(Disjoint) Suppose two equivalence classes $A, B \in \mathscr P_\sim$ such that $A \neq B$. Then we have $a, b \in S$ such that $A = [a]_\sim$ and $B= [b]_\sim$. Now suppose that it exists an element $x \in [a]_\sim \cap [b]_\sim$, therefore $x \in [a]_\sim$ and $x \in [b]_\sim$. By Lemma 1, we have $[a]_\sim = [x]_\sim = [b]_\sim$, but we assumed that $A \neq B$, so we get a contradiction. Therefore it doesnt exists any element in $[a]_\sim \cap [b]_\sim$.
(Union) Finally we will prove $\bigcup \mathscr P_\sim = S$.
($\subseteq$) Suppose an arbitrary element $x \in \bigcup \mathscr P_\sim$. Then it must exist a set $X \in \mathscr P_\sim$ such that $x \in X$. This must be its equivalence class $[x]_\sim \in \mathscr P_\sim$, which is indeed a subset of $S$.
($\supseteq$) Suppose an $x \in S$. We can construct the equivalence relation $[x]_\sim$, and it will be in $\mathscr P_\sim$. Since $x \in [x]_\sim$ and $[x]_\sim \in \mathscr P_\sim$ we have that $x \in \bigcup \mathscr P _\sim$.
We can finally conclude that $\mathscr P_\sim$ is a partition of $S$. $\square$
Interested in logic, type theory, proof assistants and formal methods
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