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From Ben: Remember to complete the Prep Report in I-Learn. I'll list, below, the people who are presenting each problem.
  • 20 - Done
  • 21 -
  • 22
  • 23
  • 24
  • 25
  • 26

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 $B$.

We have seen that the truth value of many sentences cannot be determined without an appropriate context. This requires that we quantify any possible variable in the sentence. Do we consider all possible values of the variable over some range, or should we consider just one possible instance of the variable. Consider the open sentence "$x^2+5x+6=0$." Two ways we can quantify what the variable $x$ represents are given below.

  • For all real numbers $x$, we have $x^2+5x+6=0$.
  • There exists a real number $x$ such that $x^2+5x+6=0$.

The first sentence is false (since when $x=0$ we do not have $6= 0$), and the second is true (let $x=-2$). We'll see the phrases "for every" and "there exists" quite often in mathematical writing. Because they occur so often, mathematicians have agreed upon some shorthand symbols for writing these phrases.

Exercise

Are the statements "It is not true that P implies Q" and "P does not imply Q" logically equivalent? Prove that they are, or provide a counter example.

Definition (The Quantifiers $\forall$ and $\exists$)
  • We'll use $\forall$ as shorthand in place of the phrases "for every," "for all," "for each," or any equivalent expression that suggest for every possible case. We call $\forall$ the universal quantifier.
  • We'll use $\exists$ as shorthand in place of the phrases "there exists," "there is at least one," or any equivalent expression that suggest there is at least one possible case. We call $\exists$ the existential quantifier.

These symbols are used often in open discussions, presentations, and informal work. However, when publishing formal papers, it is common practice to avoid using these symbols and instead just write the words.

Problem 25: (The Order Of Quantifiers Matters)

Translate each of the following into an English sentence. Then determine the truth value of each statement.

  1. $\forall x \in \mathbb{R}, \exists y\in\mathbb{R}$ such that $y+1>x$.
  2. $\exists y \in \mathbb{R}$ such that $\forall x \in \mathbb{R}$, we have $y+1>x$.

Problem 26: (Negating Quantifiers)

Rewrite each statement below using the quantifiers $\forall$ and/or $\exists$. Then write the negation of each statement.

  1. For each $x\in\mathbb{N}$ we have $x>4$.
  2. There exists $y\in \mathbb{R}$ such that $y\in (-3,4)$.
  3. For every $\varepsilon>0$, there exists a $\delta>0$ such that $0<|x-c|<\delta$ implies $\left|f(x)-L\right|<\varepsilon$.

Problem 27: (Practice Finding Truth Values With Universal Quantifiers)

Determine the truth value of each statement below. Be prepared to justify your claim.

  1. $\forall x\in \mathbb{R}$ and $\forall y\in \mathbb{R}$, $\exists z\in \mathbb{R}$ such that $x+y=z$.
  2. $\forall x\in \mathbb{R}$ and $\forall y\in \mathbb{R}$, $\exists z\in \mathbb{R}$ such that $xz=y$.
  3. $\forall x\in \mathbb{R}$, $\exists y\in \mathbb{R}$ such that $\forall z\in \mathbb{R}$, $z>y$ implies $z>x+y$.
  4. $\exists x\in \mathbb{R}$ such that $\forall y\in \mathbb{R}$, $\exists z\in \mathbb{R}$ such that $z>y$ implies $z>x+y$.

The following problem gets at the heart of what it means to be a limit point.

Problem 27.5: (Limit Points Of A Singleton Set)

Let $a\in \mathbb{R}$. Suppose $S=\{a\}$, so a set with a single number. What are the limit points of $S$? Prove your claim.

For more problems, see AllProblems