From 6e4612430b51c92f16c02dafc02b79098b30da5b Mon Sep 17 00:00:00 2001 From: tomit4 Date: Wed, 20 May 2026 22:05:24 -0700 Subject: [PATCH] :construction: Almost done 1 --- chapter_1/1_5/additional_exercises.md | 283 +++++++++++++++++++++++++- chapter_1/1_5/practice_problems.md | 49 +++++ chapter_1/1_5/reading_questions.md | 4 + leftoff.txt | 2 +- 4 files changed, 333 insertions(+), 5 deletions(-) diff --git a/chapter_1/1_5/additional_exercises.md b/chapter_1/1_5/additional_exercises.md index fda45b8..2d8f788 100644 --- a/chapter_1/1_5/additional_exercises.md +++ b/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 diff --git a/chapter_1/1_5/practice_problems.md b/chapter_1/1_5/practice_problems.md index e5668d9..d3e6185 100644 --- a/chapter_1/1_5/practice_problems.md +++ b/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 $f^{-1}(B_1) \subseteq f^{-1}(B_2)$. diff --git a/chapter_1/1_5/reading_questions.md b/chapter_1/1_5/reading_questions.md index 77b0130..d55c5ef 100644 --- a/chapter_1/1_5/reading_questions.md +++ b/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. diff --git a/leftoff.txt b/leftoff.txt index b16e5f7..ee977b5 100644 --- a/leftoff.txt +++ b/leftoff.txt @@ -1 +1 @@ -104 +115