Math 360, Fall 2021, Assignment 1

By one of those caprices of the mind, which we are perhaps most subject to in early youth, I at once gave up my former occupations; set down natural history and all its progeny as a deformed and abortive creation; and entertained the greatest disdain for a would-be science, which could never even step within the threshold of real knowledge. In this mood of mind I betook myself to the mathematics, and the branches of study appertaining to that science, as being built upon secure foundations, and so, worthy of my consideration.

- Mary Shelley, Frankenstein

Read:

1. Section 0.

Carefully define the following terms, then give one example and one non-example of each:

1. Containment (of sets).
2. Equality (of sets).
3. Property.
4. Ordered pair.
5. Cartesian product (of two sets).

Carefully state the following theorems (you do not need to prove them):

1. Russell's paradox (this is not really a theorem, but it is an important fact).
2. Basic counting principle (relating the size of $A\times B$ to the sizes of $A$ and $B$).

Solve the following problems:

1. Section 0, problems 1, 2, 3, 4, 5, 6, 7, 8, and 11.
2. Prove that for any set $S$, it must be the case that $\emptyset\subseteq S$. (Hint: begin with "Suppose not. Then by the definition of set containment, there must be some member of $\emptyset$ which is not a member of $S$. But...".)
3. Now suppose $S$ is any set with $S\subseteq\emptyset$. Prove that $S=\emptyset$. (Hint: use the previous result together with the definition of set equality.)
4. The previous two exercises show that $\emptyset$ is the "smallest of all sets." Is there a "largest of all sets?" (For more information on this question as well as on Russell's Paradox and its resolutions, see Wikipedia: Universal Set.)

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Solutions:

1. Containment: Set A is contained in set B if each element in A is also an element in B. This is also the definition of a subset. For example: $\{1,2\} \subseteq \{1,2,5\}$, $\{1,2,3\} \not\subseteq \{1,2,5\}$.
2. Equality: two sets are equal if they are mutually subsets of each other. If there are two sets $A \textrm{ and } B$. $A \subseteq B \wedge B \subseteq A \Leftrightarrow A = B$
3. Property: a statement that must be able to be evaluated as true or false about an object. For example $\{ x| x \textrm{ is even } \wedge x \leqslant 8\}$. However, "x is almost an integer" is not a property, as it is not well-defined.
4. Ordered Pair: two associated objects, the order of which is significant. In an ordered pair $(a,b), a \textrm{ is the first element and } b \textrm{ is the second}. (a,b) \neq (b,a) \textrm{ if } a \neq b.$ For example, $(1,2) = (1,2) \textrm{ because } 1 = 1, 2=2$ and $(1,2) \neq (2,1) \textrm{ because } 1 \neq 2, 2 \neq 1.$
5. Cartesian Product of Two Sets: $\textrm{if } A, B \textrm{ are two sets:}$ $A \times B = \{(a,b) | a \in A \wedge b \in B\}$. For example, $\{1,2\} \times \{3,4\} = \{(1,3),(1,4),(2,3),(2,4)\}$. As a non-example, $\{1,2\} \times \{3,4\} \neq \{(1,2),(3,4)\}$
6. Russell's paradox: if $R \textrm{ is a set and } R = \{x|x \neq x \}$, can one say $R \in R$ or not? Assume that $R \in R, \textrm{ then } R \notin R \textrm{ according to set R's property. However, if } R \notin R, \textrm{ according to R's property, } R \in R$ which creates paradox.
7. Basic Counting Principle: if there are a elements in set A and b elements in set B, there should be $a \times b$ pairs (or elements) in the set which represents Cartesian Product of set A and B. $|A \times B| = |A| \times |B|$, $(|A|$ is the cardinality of set A).

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Section 0

1. $\{x \in \mathbb R | x^{2} = 3\}$ $x \in R \rightarrow x \textrm{ are elements contained in the real number set}$, $x^{2} = 3 \rightarrow \textrm{ elements in the real number set which have their square } = 3$. $x^{2} = 3$, $x = \pm \sqrt{3}$, therefore, this set is $\{\sqrt{3} , -\sqrt{3} \}$
2. $\emptyset$
3. $\{1,2,3,4,5,6,10,12,15,20,30,60\}$
4. $\{m \in \mathbb Z | m^{2} - m < 115\}$, m are all integers. $m^{2} - m < 115 \rightarrow m^{2} - m - 115 < 0$, $m = -10.24 \textrm{ and } m = 11.24 \textrm{ when } m^{2} - m - 115 < 0$, $(m^{2} - m - 115)' = 2m - 1 = 0, m = \frac{1}{2}, m^{2} - m - 115 < 0 \textrm{ when } m = \frac{1}{2}$. Therefore, $-10.24 < m < 11.24$, the set is $\{-10, -9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$
5. Not well-defined, because "large number" is not unique value.
6. $\emptyset$
7. $\emptyset$
8. Not well-defined, because "almost an integer" is not unique value.

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Solve Problems

1. Suppose that $\emptyset \nsubseteq S$, $\exists a \in \emptyset \wedge a \notin S$. However, $\forall a, a \notin \emptyset$. Therefore, $\forall a \notin S, a \notin \emptyset$. Proved by contradiction that $\forall S, \emptyset \subseteq S$
2. Suppose that $S \subseteq \emptyset \wedge S \neq \emptyset$, which means that $\emptyset \nsubseteq S$. From the previous proof, we know that $\forall S, \emptyset \subseteq S$. $\emptyset \subseteq S \wedge S \subseteq \emptyset$ Proved by contradiction that $\forall S \subseteq \emptyset, S = \emptyset$.
3. Suppose S is the largest set in the whole mathematical world. S should have a power set, the power set $P(S) \textrm{of} S$. Power sets $P(S)$ always have higher cardinality than the set $S, \forall S$. $P(S) > S$, proved by contradiction that S is not the largest set. There is no largest set.

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Section 0

1, {-√3,√3}

2, ∅

3, {-60,-30,-20,-15,-12,-10,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6,10,12,15,20,30,60}

4, {m ϵ Z | -10.2 < m < 11.2}

5, Not well defined

6, ∅

7, ∅

8, Not well defined

11, {(a,1),(a,2),(a,c),(b,1),(b,2),(b,c),(c,1),(c,2),(c,c)}

Problem 2:

Containment def: We say A ⊆ B (read A is contained in B) whenever each element of A is also an element of B.

Solution 1: So it's also true that not single element of A is not an element of B. ∅ don't have any elements, so it's true that ∅ is contained in S.

Solution 2: Suppose that ∅ ⊈ S, then there is an element of ∅ that is also an element of S. But the ∅ don't have any elements, so it's false that there is an element of ∅ that is also an element of S. Thus, it's true that ∅ ⊆ S.

Problem 3:

Containment def: We say A ⊆ B (read A is contained in B) whenever each element of A is also an element of B.

Solution 1: So it's also true that not single element of A is not an element of B. ∅ don't have any elements, so if S ⊆ ∅ then not single element of S is not an element of ∅, which means S don't have any elements. Thus, S = ∅.

Solution 2: Suppose that S ≠ ∅, then there is at least one element in S. But then S ⊈ ∅, so it’s false that there is at least one element in S. Thus, S = ∅.

Problems 4:

Set A is "the largest set of all sets". Set S is any set of all sets. Then A ⊈ S. Suppose not A ⊈ S, then A ⊆ S. If A ⊆ S, then |S| ≥ |A|, which means A is not "the largest set of all sets". Thus, there is not set A that is "the largest set of all sets".

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