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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 $T=B$.
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).
- An implication that is true, and the converse is true.
- An implication that is false, but the converse is true.
- 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 $P\implies Q$.
- In a truth table for an implication $P\implies Q$, how many of the 4 rows contain the truth value $T$?
- In a truth table for a conjuction $P\wedge Q$, how many of the 4 rows contain the truth value $T$?
- In a truth table for a disjunction $P\vee Q$, how many of the 4 rows contain the truth value $T$?
- 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.
- Complete the truth table below, and use your answer to determine which statement is logically equivalent to $\sim(P\implies Q)$.
$$ \begin{array}{c|c|c|c|c|c|c|c} P&Q&P\implies Q&\sim(P\implies Q) &P \wedge Q &P \wedge (\sim Q) & (\sim P) \wedge Q& (\sim P) \wedge (\sim Q) \\\hline T&T&&&&&&\\ T&F&&&&&&\\ F&T&&&&&&\\ F&F&&&&&& \end{array} $$
Problem 19 (Proof By Contrapositive Versus Proof By Contradiction)
Let $a$ and $b$ be real numbers with $a<b$. Consider the set $S=(a,b)=\{x\in \mathbb{R}\mid a<x<b\}$. We know $b$ is an upper bound for this set as if $x\in S$, then by definition of $S$ we have $x<b$, which clearly implies $x\leq b$. To show that $b$ is the supremum of $S$, we must show "If $m$ is an upper bound of $S$, then $b\leq m$." This is an implication of the form "If $P$, then $Q$."
- State the contrapositive of this implication.
- State the negation of this implication.
- Prove that the implication is true by proving that the contrapositive is true.
- Now, instead, prove that the implication is true by proving that the negation is false.
As a corollary to this problem, similar reasoning shows that if $S$ is an interval (of the form $(a,b)$, $ [a,b)$, $(a,b] $, or $ [a,b] $), then we have $\sup S=b$ and $\inf S=a$. You may use those facts now without proof.
Definition (Union And Intersection Of Sets)
Let $A$ and $B$ be sets.
- The intersection of $A$ and $B$, written $A\cap B$, is a new set whose elements are those that are in $A$ and in $B$. Using set builder notation, we can write the intersection as $$A\cap B = \{x\mid x\in A \text{ and } x\in B\}.$$
- The union of $A$ and $B$, written $A\cup B$, is a new set whose elements are those that are in $A$ or in $B$ (or both). Using set builder notation, we can write the union as $$A\cup B = \{x\mid x\in A \text{ or } x\in B\}.$$
Problem 20: (Intersection Of Two Intervals)
Suppose that $a,b,c,d\in \mathbb{R}$ and that $a<b<c<d$. Consider the intervals $ A=(a,c) $ and $ B=[b,d] $. Prove that $A\cap B = [b,c) $. (Remember this means you need to prove that $A\cap B \subseteq [b,c) $ and $ [b,c)\subseteq A\cap B$.
Problem 21: (Union Of Two Intervals)
Suppose that $a,b,c,d\in \mathbb{R}$ and that $a<b<c<d$. Consider the intervals $ A=(a,c) $ and $ B=[b,d]$. Prove that $ A\cup B = (a,d] $.
Problem 22: (The Empty Set Is A Subset Of Every Set)
Prove or disprove. The empty set is a subset of every set. In other words, If $S$ is a set, then we have $\emptyset\subseteq S$.
Here are three more limit point problems to work on, one with a solution provided.
Exercise (Points In An Intervals Are Limit Points)
Let $a,b\in \mathbb{R}$, with $a<b$. Let $M=[a,b]$. Prove that if $p\in M$, then $p$ is a limit point of $M$.
Click to see a solution.
I've given two solutions below. The first solution uses the supremum and infimum. The second solution doesn't use these words at all, but rather uses the same ideas needed to prove facts about the infimum and supremum of a set. Please read both proofs. Come with questions if you have any.
Solution using infimum and supremum of $(a,b)$
Pick $p\in M = [a,b] $. There are three cases to consider, namely $p=a$, $p=b$, and $p\in (a,b)$. We first suppose $p=a$. Let $I=(c,d)$ be an open interval that contains $p=a$. Since $a$ is the infimum of $(a,b)$, we know that any number larger than $a$ cannot be a lower bound of $(a,b)$. This means that $d$ is not a lower bound for $(a,b)$, so we can pick a number $x$ between $a$ and $d$ such that $x\in (a,b)$. Since $a<x<d$, we know $x\in I$ and $x\neq a$. Since $x\in (a,b)\subseteq [a,b] $, we know that $x\in M$. This completes that proof that $p=a$ is a limit point of $M$.
To prove that $p=b$ is a limit point of $M$, we use similar reasoning as above. Given an interval $I=(c,d)$ that contains $b$, we use the fact that $b$ is the supremum of $(a,b)$ to obtain a number $x\in (a,b)$ that lies between $c$ and $b$ (possible since $c$ is not an upper bound of $(a,b)$. We know $x\in M$ since $x\in (a,b)$. We also know $c<x<b$ which means $x\in I$ and $x\neq b$. This proves $p=b$ is a limit point of $M$.
To finish the proof, we now assume that $p\in (a,b)$ and must prove that $p$ is a limit point of $M$. Let $I=(c,d)$ be an open interval that contains $p$. We must produce a number $x$ such that $x\in I$, $x\in M$, and $x\neq p$. There are lots of ways to proceed, so what follows is not the only option. Let's look to the right of $p$. We know that both $b$ and $d$ are greater than $p$. All we need to do is pick a value for $x$ that is larger than $p$ but less than both $b$ and $d$. How do we do this? We use the fact that between any two real numbers, there is another real number. If $b<d$, then we pick $x\in (p,b)$. Otherwise, we know $d\leq p$ and we pick $x\in (p,d)$. So basically, we chose a number $x$ between $p$ and the smaller of $b$ and $d$. In either case, we have $p<x<b$ (hence $x\in M$) and $p<x<d$ (hence $x\in I$). Since $p<x$, we know $p\neq x$. This produces the needed value of $x$ to finish the proof that $p$ is a limit point of $M$.
Solution without infimum or supremum of $(a,b)$
Pick $p\in M = [a,b] $. There are two cases to consider, namely $p\in [a,b) $ and $p=b$. We first let $p\in [a,b)$. Let $I=(c,d)$ be an open interval that contains $p$. All we need to do is pick a value for $x$ that is larger than $p$ but less than both $b$ and $d$. How do we do this? We use the fact that between any two real numbers, there is another real number. Since both $d$ and $b$ are greater than $p$, we pick a number $x$ that is greater than $p$ and less than the smaller of $d$ and $b$. Since $c<p<x<d$, we know $x\in I$. Since $a\leq p<x<b$, we know $x\in M$. Since $p<x$, we know $p\neq x$. This completes that proof that $p\in [a,b)$ is a limit point of $M$.
To prove that $p=b$ is a limit point of $M$, we use similar reasoning as above, but this time pick a point left of $p=b$ rather than above it. Given an interval $I=(c,d)$ that contains $b$, we pick a number $x$ that is less than $b$ and greater than the larger of $a$ and $c$. Since $c<x<b$, we know $x\in I$. Since $a<x<b$, we know $x\in M$. Since $x<b$, we know $b\neq x$. This proves $p=b$ is a limit point of $M$.
Problem 23: (Points Not In A Closed Interval Are Not Limit Points)
Let $a,b\in \mathbb{R}$, with $a<b$. Let $M=[a,b]$. Prove that if $p\notin M$, then $p$ is not a limit point of $M$.
Problem 24: (Limit Points Of Subsets Are Limit Points Of The Larger Set)
Suppose that $A$ and $B$ are subsets of the real numbers. Prove that if $A\subseteq B$ and $p$ is a limit point of $A$, then $p$ is a limit point of $B$.
For more problems, see AllProblems