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chapter_1/1_5/additional_exercises.md

@@ -7,6 +7,40 @@ $A \cup B = B$.
 
 A:
 
+First, let's prove that for any two sets $A$ and $B$, $A \subseteq B$ if
+$A \cup B = B$.
+
+$$ A \cup B = B \to A \subseteq B $$
+
+- Suppose that $A \cup B = B$.
+
+- Let $x$ be in $A$, $x \in A$.
+
+- Then $x \in A \cup B$.
+
+- Since $A \cup B = B$, then $x \in B$.
+
+- Therefore $A \subseteq B$
+
+Second, let's prove that for any two sets $A$ and $B$, $A \cup B = B$, if
+$A \subseteq B$.
+
+$$ A \subseteq B \to A \cup B = B $$
+
+- Suppose $A \subseteq B$
+
+- Let $x \in A \cup B$
+
+- Then $x \in A$ or $x \in B$ by definition of a union.
+
+- If $x \in B$, done.
+
+- If $x \in A$, then since $A \subseteq B$, we get $x \in B$.
+
+- So, in all cases $x \in B$.
+
+- Therefore $A \cup B \subseteq B$.
+
 2.
 
 Q: The **intersection** of sets $A$ and $B$, denoted $A \cap B$, is the set of
@@ -17,6 +51,34 @@ $A \cap B = A$.
 
 A:
 
+First, let's prove:
+
+$$ A \subseteq B \to A \cap B = A $$
+
+- Suppose that $A \subseteq B$
+
+- Let $x$ be some element such that $x \in A \cap B$
+
+- Since $A \subseteq B$, and $x \in A \cap B$, then $x \in A$ by definition of
+  an intersection
+
+- Therefore $A \cap B \subseteq A$.
+
+Secondly, let's prove:
+
+$$ A \cap B = A \to A \subseteq B $$
+
+- Suppose that $A \cap B = A$.
+
+- Let $x$ be some element such that $x \in A$.
+
+- Since $A \cap B = A$ and $x \in A$, then $x \in A \cap B$.
+
+- $x \in A \cap B$ means $x$ \in A$ and $x \in B$ by definition of an
+  intersection
+
+- Since every element of $A$ is an element of $B$, therefore $A \subseteq B$
+
 3.
 
 Q: Prove that for any sets $A$, $B$, and $C$, if $A \cup B \subseteq C$, then
@@ -24,6 +86,34 @@ $A \subseteq C$ and $B \subseteq C$.
 
 A:
 
+First, let us prove:
+
+$$ A \cup B \subseteq C \to A \subseteq C $$
+
+- Suppose $A \cup B \subseteq C$
+
+- Let $x$ be an element such that $x \in A$
+
+- Since $x \in A$, we know that $x \in A \cup B$, by definition of a union.
+
+- Since we know that $A \cup B \subseteq C$, it follows that $x \in C$.
+
+- Therefore $A \subseteq C$
+
+Second, let's prove that $B \subseteq C$:
+
+$$ A \cup B \subseteq C \to B \subseteq C $$
+
+- Suppose that $A \cup B \subseteq C$
+
+- Let $x$ be an element such that $x \in B$
+
+- Since $x \in B$, it follows that $x \in A \cup B$, by definition of a union
+
+- Since we know that $A \cup B \subseteq C$, it follows that $x \in C$.
+
+- Therefore $B \subseteq C$
+
 4.
 
 Q: Prove that for any sets $A$, $B$, and $C$, if $A \subseteq C$ and
@@ -31,23 +121,100 @@ $B \subseteq C$, then $A \cup B \subseteq C$.
 
 A:
 
+Let's prove:
+
+$$ (A \subseteq C) \wedge (B \subseteq C) \to A \cup B \subseteq C $$
+
+- Suppose that $A \subseteq C$
+
+- Suppose that $B \subseteq C$
+
+- Let $x$ be an element such that $x \in A \cup B$
+
+- $x \in A \cup B$ means that $x \in A$ or $x \in B$ by definition of a union
+
+- If $x \in A$, and $A \subseteq C$, then $x \in C$
+
+- If $x \in B$, and $B \subseteq C$, then $x \in C$
+
+- In both cases $x \in C$
+
+- Therefore $A \cup B \subseteq C$
+
 5.
 
-Q: The **difference** of sets $A$ and $B$, written $A$ \ $B$, is the set of all
-elements that are in $A$ but not in $B$.
+Q: The **difference** of sets $A$ and $B$, written $A \setminus B$, is the set
+of all elements that are in $A$ but not in $B$.
 
 The **empty set**, written $\emptyset$, is the set that contains no elements.
 
-Prove that if $A$ \ $B = A$ then $A \cap B = \emptyset$.
+Prove that if $A \setminus B = A$ then $A \cap B = \emptyset$.
 
 A:
 
+Let's prove:
+
+$$ A \setminus B = A \to A \cap B = \emptyset $$
+
+- Suppose $A \setminus B = A$
+
+- Let $x$ be an element such that $x \in A \cap B$
+
+- Since $x \in A \cap B$, this means that $x \in A$ and $x \in B$ by definition
+  of an intersection
+
+- Since $A \setminus B = A$ and $x \in A$, it follows that $x \in A \setminus B$
+
+- But if $x \in A \setminus B$, then it follows that $x \notin B$
+
+- This is a contradiction since we established earlier that $x \in B$ by
+  definition of an intersection, but now we also established that $x \notin B$.
+
+- Therefore $A \cap B = \emptyset$
+
+A small note is worthwhile here. This is not a proof by contradiction in the
+traditional sense. The contradiction does not start at the premise/argument
+level, it only occurs at a single step of the proof.
+
 6.
 
-Q: Prove that if $A$ \ $B = B$ \ $A$ then $A = B$.
+Q: Prove that if $A \setminus B = B \setminus A$ then $A = B$.
 
 A:
 
+Let's prove:
+
+$$ (A \setminus B = B \setminus A) \to A = B $$
+
+- Suppose that $A \setminus B = B \setminus A$
+
+- Let $x$ be an element such that $x \in A \setminus B$
+
+- Since $x \in A \setminus B$, this means that $x \in A$ and $x \notin B$ by
+  definition of a difference
+
+- Since $x \in A \setminus B$ and $A \setminus B = B \setminus A$, it follows
+  that $x \in B \setminus A$
+
+- Since $x \in B \setminus A$, this means that $x \in B$ and $x \notin A$ by
+  definition of a difference
+
+- This is a contradiction since we've established that $x \in A$, $x \notin A$
+  (and also $x \in B$ and $x \notin B$).
+
+- Therefore since $x$ cannot exist, we can conclude that
+  $A \setminus B = \emptyset$.
+
+- And since $A \setminus B = \emptyset$ and $A \setminus B = B \setminus A$, we
+  can further conclude that $B \setminus A = \emptyset$.
+
+- Since $A \setminus B = \emptyset$, we can conclude that $A \subseteq B$
+
+- And also since $B \setminus A = \emptyset$, we can further conclude that
+  $B \subseteq A$
+
+- Therefore $A = B$
+
 7.
 
 Q: Let $f:X \to Y$ be a function, and let $A$ and $B$ be subsets of $X$.
@@ -59,6 +226,69 @@ $f(A \cap B) \neq f(A) \cap f(B)$.
 
 A:
 
+(a) Prove that $f(A \cap B) \subseteq f(A) \cap f(B)$.
+
+- Let there be some output $y$ such that $y \in f(A \cap B)$.
+
+- Since $y \in f(A \cap B)$, there exists $x \in A \cap B$ such that $f(x) = y$
+  by the definition of an image.
+
+- Since $x \in A \cap B$, this means that $x \in A$ and $x \in B$ by definition
+  of an intersection.
+
+- It then follows that $f(x) \in f(A)$ and $f(x) \in f(B)$.
+
+- Since $f(x) = y$, it further follows that $y \in f(A)$ and $y \in f(B)$, which
+  is expressed as $y \in f(A) \cap f(B)$.
+
+- Therefore $f(A \cap B) \subseteq f(A) \cap f(B)$
+
+(b) Find an example of a function and two sets $A$ and $B$ such that
+$f(A \cap B) \neq f(A) \cap f(B)$.
+
+A: This can happen depending on exactly what the function $f$ is in this case
+and what $A$ and $B$ are defined as.
+
+Let's say that $f$ in this case is:
+
+$$ f(x) = 0, x \in \mathbb{R} $$
+
+Then let's say our sets are:
+
+$$ A = \{0\} $$
+
+$$ B = \{1\} $$
+
+Then for $f(A \cap B)$, we'll first need $A \cap B$:
+
+$$ A \cap B = \emptyset $$
+
+So:
+
+$$ f(A \cap B) = f(\emptyset) = \emptyset $$
+
+A small note here: when applying a function to a set via image notation, if the
+set is empty, then there are no elements to map under the function, so the image
+is the empty set.
+
+Then consider $f(A) \cap f(B)$:
+
+$$ f(\{0\}) = \{0\} $$
+
+$$ f(\{1\}) = \{0\} $$
+
+So:
+
+$$ f(A) \cap f(B) = \{0\} $$
+
+And then when we make our comparison:
+
+$$ f(A \cap B) = \emptyset \neq \{0\} = f(A) \cap f(B) $$
+
+So this is an example where:
+
+$$ f(A \cap B) \neq f(A) \cap f(B) $$
+
 8.
 
 Q: Let $f: X \to Y$ be a function, and let $A$ and $B$ be subsets of $X$.
@@ -71,6 +301,51 @@ Q: Let $f: X \to Y$ be a function, and let $A$ and $B$ be subsets of $X$.
 
 A:
 
+(a) Prove that $f(A \cup B) \subseteq f(A) \cup f(B)$.
+
+- Let there be some output $y$ such that $y \in f(A \cup B)$.
+
+- Since $y \in f(A \cup B)$, there exists $x \in A \cup B$ such that $f(x) = y$
+  by the definition of an image.
+
+- Since $x \in A \cup B$, this means that $x \in A$ or $x \in B$ by definition
+  of an union.
+
+- It then follows that $f(x) \in f(A)$ or $f(x) \in f(B)$.
+
+- Since $f(x) = y$, it further follows that $y \in f(A)$ or $y \in f(B)$, which
+  is expressed as $y \in f(A) \cup f(B)$.
+
+- Therefore $f(A \cup B) \subseteq f(A) \cup f(B)$
+
+(b) Prove that $f(A) \cup (B) \subseteq f(A \cup B)$
+
+- Let there be some output $y$ such that $y \in f(A) \cup f(B)$.
+
+- Since $y \in f(A) \cup f(B)$, this means that $y \in f(A)$ or $y \in f(B)$ by
+  definition of a union.
+
+- Since $y \in f(A) \cup f(B)$, there exists $x \in A$ or there exists $x \in B$
+  such that $f(x) = y$ by the definition of an image.
+
+- If $x \in A$, $x \in A \cup B$.
+
+- If $x \in B$, $x \in A \cup B$.
+
+- In both cases $x \in A \cup B$, so $y \in f(A \cup B)$
+
+- Since $f(x) = y$, it further follows that $y \in f(A \cup B)$.
+
+- Therefore $f(A) \cup f(B) \subseteq f(A \cup B)$
+
+\(c\) What can you conclude from the two proofs above?
+
+Since in part (a), we found that $f(A \cup B) \subseteq f(A) \cup f(B)$ and in
+part (b) we found that $f(A) \cup f(B) \subseteq f(A \cup B)$, we can conclude
+that:
+
+$$ f(A \cup B) = f(A) \cup f(B) $$
+
 9.
 
 Q: Given a function $f: X \to Y$ and a set $B \subseteq Y$, we define the

chapter_1/1_5/practice_problems.md

@@ -19,6 +19,8 @@ D. Suppose there is an element $b$ in $B$ that is not in $A$.
 
 A:
 
+B would be the best way to start.
+
 2.
 
 Q: Suppose you wanted to prove that for all sets $A$ and $B$ that
@@ -35,6 +37,8 @@ D. Let $a$ be an element of $A \cap B$.
 
 A:
 
+B would be the best way to start.
+
 3.
 
 Q: Arrange some of the statements below to form a correct proof of the following
@@ -62,6 +66,15 @@ $B \subseteq A$".
 
 A:
 
+- Suppose $B \subseteq A \cap B$, then let $b$ be an element of $B$.
+
+- Then $b$ is an element of $B$ since $B \subseteq A \cap B$.
+
+- Since $A \cap B$ contains all the elements that are in both $A$ and $B$, $b$
+  is an element of $A$.
+
+- Therefore $B \subseteq A$.
+
 4.
 
 Q: Prove that for any sets $A$ and $B$, $(A \cap B) \cup A$ = A$. Arrange the
@@ -91,6 +104,28 @@ statements below to form a correct proof.
 
 A:
 
+- First we will prove that $(A \cap B) \cup A \subseteq A$.
+
+- Let $x$ be an element of $(A \cap B) \cup A$.
+
+- Then $x$ is an element of $A \cap B$, or $x$ is an element of $A$.
+
+- So in particular, $x$ is an element of $A$.
+
+- Therefore $(A \cap B) \cup A \subseteq A$.
+
+- Second, we will prove that $A \subseteq (A \cap B) \cup A$.
+
+- Let $x$ be an element of $A$.
+
+- Then $x$ is an element of $(A \cap B) \cup A$, since $x$ is in $A$ or in the
+  other set.
+
+- Therefore $A \subseteq (A \cap B) \cup A$.
+
+- Since $(A \cap B) \cup A \subseteq A$ and $A \subseteq (A \cap B) \cup A$, we
+  have $(A \cap B) \cup A = A$.
+
 5.
 
 Q: Let $f: X \to Y$ be a function and let $B \subseteq Y$ be a subset of the
@@ -121,3 +156,17 @@ form a correct proof.
 - Let $a$ be an element of $f^{-1}(B_1)$.
 
 A:
+
+---
+
+- Suppose $B_1 \subseteq B_2$.
+
+- Let $a$ be an element of $f^{-1}(B_1)$.
+
+- This means that $f(a)$ is an element of $B_1$.
+
+- Since $B_1 \subseteq B_2$, $f(a)$ is an element of $B_2$.
+
+- This then means that $a$ is an element of $f^{-1}(B_2)$.
+
+- Therefore $f^{-1}(B_1) \subseteq f^{-1}(B_2)$.

chapter_1/1_5/reading_questions.md

@@ -15,6 +15,8 @@ D. The codomain $B$ is no smaller than the domain $A$.
 
 A:
 
+Answer A is the definition of a function $f: A \to B$ being injective.
+
 2.
 
 Q: When would you most likely use element chasing as part of a proof?
@@ -28,3 +30,5 @@ C. When proving that a relation is transitive.
 D. When proving that a graph has an odd number of edges.
 
 A:
+
+A is the correct answer.

leftoff.txt

@@ -1 +1 @@
-104
+115