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Definition (Limit Point Of A Set Of Real Numbers)

Let $S$ be a set of real numbers. We say that a point $p$ is a limit point of $S$ if every open interval $I=(a,b)$ that contains $p$ also contains a point $x$ in $S$ with $x\neq p$.

Problem 7: (A Set With One Limit Point)

Let $\ds S=\left\{\frac{1}{n}\mid n\in \mathbb{N}\right\}$, the collection of fractions of the form $\frac{1}{n}$ where $n$ is a natural number. Prove that $p=0$ is a limit point of $S$.

Definition (Set, Subset, Equality Of Sets)

A set $S$ is a collection of elements that have been grouped together.

• We use brackets $\{$ and $\}$ to enclose elements of sets.
• We'll write $x\in S$ to say that $x$ is an element of $S$ or $x$ is in $S$. Similarly, we'll write $x\notin S$ to say that $x$ is not in $S$.
• We say that a set $B$ is a subset of the set $S$, and we write $B\subseteq S$, if every element in $B$ is also an element of $S$. We also read $A\subseteq B$ as "$A$ is contained in $B$." We'll often write $B\supseteq A$ instead of $A\subseteq B$, and read $B\supseteq A$ as either "$B$ is a super set of $A$" or "$B$ contains $A$."
• We say that $B$ is a proper subset of $S$ if $B\subseteq S$ but there is an element of $S$ that is not in $B$.
• We say that two sets $A$ and $B$ are equal if $A\subseteq B$ and $B\subseteq A$.

Problem 9: (Second Proof That Two Sets Are Equal)

Let $I=(5,9] $. Consider the sets $T=\{x\mid x \text{ is a lower bound of } I\}$ and $B=\{x\mid x\leq 5\} $. Prove that $T=B$.

An important set that will show up often throughout the semester is the set with nothing in it, which we call the empty set.

Definition (Empty Set)

The empty set is the set $\emptyset = \{\}$ that contains no elements. If we think of a set as a box with elements in it, then the empty set is a box with nothing in it.

Here is another axiom that you have probably used many times in your life without ever realizing it.

Axiom (Well Ordering Principle)

Every nonempty subset $S$ of the natural numbers has a least element. By least element, we mean that there is a natural number $m$ which is an element of $S$ such that $m\leq x$ for every $x$ in $S$.

Problem 10: (Which Dominoes Remain Standing)

Suppose that Jon has set up an infinite number of dominoes, with the dominoes numbered $1,2,3, \ldots$. The dominoes are set up so that if the $k$th domino falls, then the $(k+1)$st domino will also fall. So if the 7th domino falls, then the 8th must fall as well. Jon knocks down the first domino, which starts causing other dominos to fall. Which dominos fall? Which dominoes remain standing? Make sure you prove your result. The well ordering principle will come in handy.

Suggestion: Use set builder notation to help you, so let $F=\{n\in \mathbb{N}\mid \text{domino $n$ fell}\}$ and $S=\{n\in \mathbb{N}\mid \text{domino $n$ remains standing}\}$. Then make some claims and prove they are correct.

Definition (Implication)

If $P$ and $Q$ are statements or open sentences, then an implication, written symbolically as $P\implies Q$, is the sentence "If $P$, then $Q$" or equivalently "$P$ implies $Q$". There are several equivalent ways to express this sentence such as "$Q$ if $P$" or "$P$ only if $Q$." The implication $P\implies Q$ is true unless $P$ is true and $Q$ is false.

Definition (Converse, Inverse, And Contrapositive)

Consider the implication $P\implies Q$. From this implication we can define 3 other implications.

• The converse of $P\implies Q$ is the implication $Q\implies P$.
• The inverse of $P\implies Q$ is the implication $(\sim P)\implies (\sim Q)$.
• The contrapositive of $P\implies Q$ is the implication $(\sim Q)\implies (\sim P)$.

Problem 15: (Practice With Converse Inverse And Contrapositive)

Let $A=[3,7)$. Consider the implication "If $x\geq 8$ then $x$ is an upper bound of $A$."

1. Is this implication true or false?
2. Write the converse of this implication and determine the truth value of the converse.
3. Write the inverse of this implication and determine the truth value of the inverse.
4. Write the contrapositive of this implication and determine the truth value of the contrapositive.

Remember to always justify any claims you make.

Problem 16: (What Is Logically Equivalent To An Implication)

Consider the implication $P\implies Q$.

1. Construct a truth table that contains the possible values for this implication, the converse, the inverse, and the contrapositive. Feel free to use the table at the end of this problem to complete your work.
2. Which of these sentences are logically equivalent?

$$ \begin{array}{c|c|c|c|c|c|c|c} P&Q&P\implies Q&Q\implies P &\sim P&\sim Q &(\sim P)\implies (\sim Q) & (\sim Q)\implies (\sim P)\\\hline T&T&&&&&&\\ T&F&&&&&&\\ F&T&&&&&&\\ F&F&&&&&& \end{array} $$

Problem 17: (Creating Examples Of Implications)

Give an example of each of the following, or explain why it cannot be done. Make sure you justify your claims (as always).

1. An implication that is true, and the converse is true.
2. An implication that is false, but the converse is true.
3. An implication that is true, but the contrapositive is false.

Problem 18: (The Negation Of An Implication Is A Conjunction)

In this problem we want to find the negation of $P\implies Q$.

1. In a truth table for an implication $P\implies Q$, how many of the 4 rows contain the truth value $T$?
2. In a truth table for a conjuction $P\wedge Q$, how many of the 4 rows contain the truth value $T$?
3. In a truth table for a disjunction $P\vee Q$, how many of the 4 rows contain the truth value $T$?
4. In a truth table for the negation of the implication $P\implies Q$, how many of the 4 rows contain the truth value $T$? Based off this answer alone, explain why you expect the negation of an implication to be a conjunction.
5. Complete the truth table below, and use your answer to determine which statement is logically equivalent to $\sim(P\implies Q)$.

$$ \begin{array}{c|c|c|c|c|c|c|c} P&Q&P\implies Q&\sim(P\implies Q) &P \wedge Q &P \wedge (\sim Q) & (\sim P) \wedge Q& (\sim P) \wedge (\sim Q) \\\hline T&T&&&&&&\\ T&F&&&&&&\\ F&T&&&&&&\\ F&F&&&&&& \end{array} $$

For more problems, see AllProblems