From 610d27d2fb4a0d1d872e2f0517eb9ae71b4cf54d Mon Sep 17 00:00:00 2001 From: andyeisenberg Date: Sat, 19 Sep 2026 12:44:09 +0000 Subject: [PATCH] Set Theory solutions --- source/sec-Set-Theory.ptx | 348 +++++++++++++++++++++++++------------- 1 file changed, 232 insertions(+), 116 deletions(-) diff --git a/source/sec-Set-Theory.ptx b/source/sec-Set-Theory.ptx index f0e3be5..c36556c 100644 --- a/source/sec-Set-Theory.ptx +++ b/source/sec-Set-Theory.ptx @@ -1,73 +1,84 @@
- Set Theory + Set Theory -

- When we perform an experiment, there are many results that we might see. - We want to be able to quantify the likelihood of seeing certain results. - For this, we need to develop some mathematical terminology. -

+

+ When we perform an experiment, there are many results that we might see. + We want to be able to quantify the likelihood of seeing certain results. + For this, we need to develop some mathematical terminology. +

- - -

- The sample space, often denoted \Omega, is the set of all possible results of an experiment. - A single result is called an outcome, while a collection of results is called an event. -

- - + + +

+ The sample space, often denoted \Omega, is the set + of all possible results of an experiment. + A single result is called an outcome, while a collection of + results is called an event. +

+ + - - -

- An experiment consists of rolling a standard 6-sided die. - The sample space is \Omega = \{1, 2, 3, 4, 5, 6\}. - One possible event is A = \{2, 4, 6\}, i.e., the event that the result of the roll is even. -

+ + +

+ An experiment consists of rolling a standard 6-sided die. + The sample space is \Omega = \{1, 2, 3, 4, 5, 6\}. + One possible event is A = \{2, 4, 6\}, i.e., the event that the + result of the roll is even. +

-

- What would the sample space look like if we roll the die two times and recorded the results? -

- +

+ What would the sample space look like if we roll the die two times and + recorded the results? +

+ - -

- - \Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), - \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), - \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), - \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), - \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), - \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} - - Note that, for example, (1, 2) is a different outcome from (2, 1). -

- - + +

+ + \Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), + \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), + \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), + \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), + \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), + \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} + + Note that, for example, (1, 2) is a different outcome from + (2, 1). +

+ + - - -

- Let A be a set. - The symbol \in means "is an element of", as in a \in A. - Given another set B, we say A is a subset of B, written A\subset B, to mean that every element of the set A is also an element of the set B. -

- - + + +

+ Let A be a set. + The symbol \in means "is an element of", as in a \in A. + Given another set B, we say A is a subset of + B, written A\subset B, to mean that every element of the + set A is also an element of the set B. +

+ + -

- The subset symbol includes the possibility that A and B are equal sets, i.e., that they contain precisely the same elements. -

+

+ The subset symbol includes the possibility that A and B are + equal sets, i.e., that they contain precisely the same elements. +

- - -

- Consider sets A and B, each contained inside \Omega. - We can combine sets in a variety of ways:

+ + +

+ Consider sets A and B, each contained inside + \Omega. + We can combine sets in a variety of ways: +

  • Union

    - The union of A and B is the set A \cup B = \{x \mid x \in A \text{ or } x \in B\}. + The union of A and B is the set + A \cup B = \{x \mid x \in A \text{ or } x \in B\}.

    • @@ -75,7 +86,8 @@ Intersection

    - The intersection of A and B is the set A \cap B = \{x \mid x \in A \text{ and } x \in B\}. + The intersection of A and B is the set + A \cap B = \{x \mid x \in A \text{ and } x \in B\}.

    @@ -83,7 +95,8 @@ Difference

    - The set difference A-B is the set A - B = \{x \mid x \in A \text{ and } x \notin B\}. + The set difference A-B is the set + A - B = \{x \mid x \in A \text{ and } x \notin B\}.

    @@ -91,7 +104,8 @@ Complement

    - The complement of A is the set A^c = \{x \in \Omega \mid x \notin A\}. + The complement of A is the set + A^c = \{x \in \Omega \mid x \notin A\}.

    @@ -99,27 +113,30 @@ Empty Set

    - The empty set, usually written \emptyset or \{\}, is the set which contains no elements. + The empty set, usually written \emptyset or + \{\}, is the set which contains no elements.

    -
    +
    +

    +     +     + +

    + It's useful sometimes to draw pictures called Venn diagrams + representing sets: +

    + +
    + Example Venn Diagram + + +

    + Venn diagram showing sets A, B, C with the region representing + (A\cup B\cup C) - (A \cap C) shaded.

    - - - -

    - It's useful sometimes to draw pictures called Venn diagrams representing sets: -

    - -
    - Example Venn Diagram - - -

    - Venn diagram showing sets A, B, C with the region representing (A\cup B\cup C) - (A \cap C) shaded. -

    -     - + + \begin{tikzpicture} \def\firstcircle{(90:1.75cm) circle (2.5cm)} \def\secondcircle{(210:1.75cm) circle (2.5cm)} @@ -137,35 +154,38 @@ \draw \thirdcircle node [text=black,below right] {$C$}; \node at (0, -4.5) {$(A\cup B\cup C) - (A \cap C)$}; \end{tikzpicture} - - -
    + + +
    - - -

    - Two sets A and B are disjoint if A \cap B = \emptyset. -

    -     -     + + +

    + Two sets A and B are disjoint if + A \cap B = \emptyset. +

    +     +     - - -

    - Given a finite set A, the cardinality of A, written |A|, is the number of elements in A. -

    -     -     - - + + +

    + Given a finite set A, the cardinality of A, + written |A|, is the number of elements in A. +

    +     +     +

    - Consider the sets A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, B = \{2, 4, 9, 10, 12, 14, 19\}, and C = \{9, 10, 11, 14, 16, 17, 20\}, which are all subsets of \Omega = \{1, 2, 3, \dotsc, 20\}. + Consider the sets A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, + B = \{2, 4, 9, 10, 12, 14, 19\}, and + C = \{9, 10, 11, 14, 16, 17, 20\}, which are all subsets of + \Omega = \{1, 2, 3, \dotsc, 20\}.

    -

    @@ -178,24 +198,78 @@ \{1, 2, 3, 4, 5, 6, 7, 8\}

    -     + +

    + The set A - (B \cap C) consists of elements in the set + A which are not in the overlap of B and C. + The overlap is B \cap C = \{9, 10, 14\}, and of these + elements, 9 and 10 are in A. + So A - (B \cap C) = \{1, 2, 3, 4, 5, 6, 7, 8\}. +

    +     +

    - Find |A|, |B|, |C|, |A\cup B|, |A \cap B|, |B\cap C|, |A\cap C|, and |A\cup B\cup C|. - Is it true that the size of the union of sets is equal to the sum of the sizes of the individual sets? + Find |A|, |B|, |C|, |A\cup B|, + |A \cap B|, |B\cap C|, |A\cap C|, and + |A\cup B\cup C|. + Is it true that the size of the union of sets is equal to the sum of + the sizes of the individual sets?

    - |A| = 10, |B| = 7, |C| = 7, |A \cup B| = 13, |A \cap B| = 4, |B\cap C| = 3, |A\cap C| = 2, |A\cup B\cup C| = 17. In particular, note that |A\cup B| = 13 \neq 10 + 7 = |A| + |B|, so it is not true in general that the size of the union of sets is the sum of the sizes of the individual sets. + |A| = 10, |B| = 7, |C| = 7, + |A \cup B| = 13, |A \cap B| = 4, |B\cap C| = 3, + |A\cap C| = 2, |A\cup B\cup C| = 17. + In particular, note that + |A\cup B| = 13 \neq 10 + 7 = |A| + |B|, so it is not true in + general that the size of the union of sets is the sum of the sizes + of the individual sets.

    -     + +

    + |X| counts the number of elements in a finite set X. + We can quickly count the elements in sets A, B, C to see that + |A| = 10, |B| = 7, and |C| = 7. +

    + +

    + The union of two sets includes all elements from either set, so: + + A \cup B \amp = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 19\} + A \cup B \cup C \amp = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 17, 19, 20\} + + Counting the elements, we see |A \cup B| = 13 and + |A \cup B \cup C| = 17. +

    + +

    + The intersection of two sets is the overlap, consisting of elements + which show up in both sets simultaneously, so: + + A \cap B \amp = \{2, 4, 9, 10\} + B \cap C \amp = \{9, 10, 14\} + A \cap C \amp = \{9, 10\} + + Counting the elements, we see |A\cap B| = 4, + |B \cap C| = 3, and |A\cap C| = 2. +

    + +

    + In particular, note that + |A\cup B| = 13 \neq 10 + 7 = |A| + |B|, so it is not true in + general that the size of the union of sets is the sum of the sizes + of the individual sets. +

    +     + @@ -206,19 +280,33 @@

    - A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}, (A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}. + A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}, + (A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}.

    + + +

    + A^c consists of elements of \Omega which are not in + A, thus + A^c = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}. + Similarly, (A\cup B)^c consists of elements that are not in + A nor in B. + Therefore, (A\cup B)^c = \{11, 13, 15, 16, 17, 18, 20\}. +

    +

    - Suppose we have a 6-sided die that's weighted to roll a 6 half of the time. + Suppose we have a 6-sided die that's weighted to roll a 6 half of the + time. We roll the die two times. List the set of all possible results. - [Note: the result (2, 4)---rolling a 2 and then a 4---is different from the result (4, 2)---rolling a 4 and then a 2.] + [Note: the result (2, 4)---rolling a 2 and then a 4---is different + from the result (4, 2)---rolling a 4 and then a 2.]

    @@ -232,10 +320,25 @@ \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} - Note that \Omega simply lists outcomes with no reference to the probabilities. - So the answer here is the same as in .

    + + +

    + + \Omega = \{\amp (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), + \amp (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), + \amp (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), + \amp (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), + \amp (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), + \amp (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)\} + + Note that \Omega simply lists outcomes with no reference to the + probabilities. + So the answer here is the same as in + . +

    +     @@ -243,7 +346,9 @@

    Suppose we flip a coin two times. List the set of all possible results. - What about flipping three times? Four times? If we flip the coin 10 times, how many possible results will there be? + What about flipping three times? + Four times? + If we flip the coin 10 times, how many possible results will there be?

    @@ -253,7 +358,11 @@

    - For three flips: \Omega = \{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \}. + For three flips: + + \Omega = \{ \amp HHH, HHT, HTH, THH, + \amp HTT, THT, TTH, TTT \} +

    @@ -276,16 +385,23 @@

    - If we roll a 6-sided die ten times, how many possible results will there be? + If we roll a 6-sided die ten times, how many possible results will + there be?

    - Each additional roll will multiply the number of outcomes by 6. - So, with 10 rolls, we'll have |\Omega| = 6^{10}. + |\Omega| = 6^{10}.

    + + +

    + Each additional roll will multiply the number of outcomes by 6. + So, with 10 rolls, we'll have |\Omega| = 6^{10}. +

    +     -     -
    \ No newline at end of file + + \ No newline at end of file