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There are many quite similar propositions. In some cases, we proved some of a collection of similar results, but not others, and so those would make good candidates. In others, we skipped over certain details, or gave a general result that you could work out in a more specific case. For example, you proved that: Proposition 1. Let C and D be subsets of a set A and $X_C$ and $X_D$ denote their characteristic functions. Then $X_{C \cup D}(a) = X_C(a) + X_D(a) - X_C(a)X_D(a)$ for all $a \in A$. I could give you a similar proposition involving intersection instead of union. We showed that: Proposition 2. Let $f: A \to B$ and $g: B \to C$ be surjective. Then $g \circ f: A \to C$ is surjective. But we did not work out the details of: Proposition 3. Let $f: A \to B$ and $g: B \to C$ be functions. If $g \circ f: A \to C$ is surjective, then g is surjective. Nor did we write out the details of: Proposition 4. For sets A and B, say $A \sim B$ iff there exists a bijection from A to B. Then $\sim$ is an equivalence relation on sets. Similarly, many of the homework exercises, or slight variants of them, would make good questions -- it might be worth reading them over.

          There are many quite similar propositions. In some cases, we proved some of a collection of similar results, but not others, and so those would make good candidates. In others, we skipped over certain details, or gave a general result that you could work out in a more specific case. For example, you proved that:
Proposition 1. Let C and D be subsets of a set A and $X_C$ and $X_D$ denote their characteristic functions. Then $X_{C \cup D}(a) = X_C(a) + X_D(a) - X_C(a)X_D(a)$ for all $a \in A$.
I could give you a similar proposition involving intersection instead of union.
We showed that:
Proposition 2. Let $f: A \to B$ and $g: B \to C$ be surjective. Then $g \circ f: A \to C$ is surjective.
But we did not work out the details of:
Proposition 3. Let $f: A \to B$ and $g: B \to C$ be functions. If $g \circ f: A \to C$ is surjective, then g is surjective.
Nor did we write out the details of:
Proposition 4. For sets A and B, say $A \sim B$ iff there exists a bijection from A to B. Then $\sim$ is an equivalence relation on sets.
Similarly, many of the homework exercises, or slight variants of them, would make good questions -- it might be worth reading them over.

There are many quite similar propositions. In some cases, we proved some of a collection of similar results, but not others, and so those would make good candidates. In others, we skipped over certain details, or gave a general result that you could work out in a more specific case. For example, you proved that:
Proposition 1. Let C and D be subsets of a set A and XC and XD denote their characteristic functions. Then XC ∪ D(a) = XC(a) + XD(a) - XC(a)XD(a) for all a ∈ A.
I could give you a similar proposition involving intersection instead of union.
We showed that:
Proposition 2. Let f: A → B and g: B → C be surjective. Then g ∘ f: A → C is surjective.
But we did not work out the details of:
Proposition 3. Let f: A → B and g: B → C be functions. If g ∘ f: A → C is surjective, then g is surjective.
Nor did we write out the details of:
Proposition 4. For sets A and B, say A ∼ B iff there exists a bijection from A to B. Then ∼ is an equivalence relation on sets.
Similarly, many of the homework exercises, or slight variants of them, would make good questions – it might be worth reading them over.

Added by Jeremy C.

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