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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 $M$.

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.

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 $z>x+y$.
Definition (Maximum and Minimum)

Let $S\subseteq \mathbb{R}$.

  • We say $m$ is a minimum of $S$ if $m$ is a lower bound for $S$ and $m\in S$. We write $\min S$ for the minimum of $S$.
  • We say $m$ is a maximum of $S$ if $m$ is an upper bound for $S$ and $m\in S$. We write $\max S$ for the maximum of $S$.

Problem 28: (Minimums And Maximums Are Unique)

Prove that a minimum of set $S$, if it exists, is unique. In other words, prove that if $m_1$ and $m_2$ are both minimums of $S$, then we must have $m_1=m_2$. A similar proof will show that maximums are unique.

Problem 29: (Relation Between Minimums And Infimums)

Suppose $S$ is a set of real numbers.

  • If $m$ is the minimum of $S$, must $m$ be the infimum of $S$? Prove your claim.
  • If $m$ is the infimum of $S$, must $m$ be the minimum of $S$? Prove your claim.

(Similar facts hold true for the maximum and supremum of a set.)

The next theorem lists several facts about set unions and intersections that we would like to refer to from here on out.

Theorem (Union And Intersection Properties)

Let $A,B,C$ be sets. Then the following facts are true.

  1. $A\subseteq A$ (Every set is a subset of itself.)
  2. $A\subseteq A\cup B$ (A set is a subset of the union of itself and another set.)
  3. $A\cap B\subseteq A$ (The intersection of two sets is a subset of the first set.)
  4. $A\cup B = B\cup A$ (Set unions are commutative.)
  5. $A\cap B = B\cap A$ (Set intersections are commutative.)
  6. $A\cup (B\cup C)=(A\cup B)\cup C$ (Set unions are associative.)
  7. $A\cap (B\cap C)=(A\cap B)\cap C$ (Set intersections are associative.)
  8. $A\cup (B\cap C)=(A\cup B)\cap(A\cup C)$
  9. $A\cap (B\cup C)=(A\cap B)\cup(A\cap C)$

The last two we call the distributive laws for unions and intersections.

Take some time to prove that each of the statements above is true. The next exercise includes several of the proofs. I'll leave some for you to prepare for class.

Exercise (Union And Intersection Properties)

Prove the statements in the Union and Intersection Properties theorem.

Click to see a solution.

  1. To prove $A\subseteq A$, let $a\in A$. Then clearly $a\in A$ which means $A\subseteq A$ as desired.
  2. We now prove $A\subseteq A\cup B$. Let $a\in A$. Then clearly $a\in A$ or $a\in B$. This means that $a\in A\cup B$, which proves that $A\subseteq A\cup B$.
  3. We now prove $A\cap B\subseteq A$. Let $x\in A\cap B$. This means $x\in A$ and $x\in B$. In particular, notice that we know $x\in A$. This completes the proof that $A\cap B\subseteq A$.
  4. Let $x$ be a real number. Let $P$ be the statement $x\in A$, and let $Q$ be the statement $x\in B$. Note that the statements $P\vee Q$ and $Q\vee P$ are logically equivalent from a truth table. The rule $A\cup B=B\cup A$ immediately follows. Alternately, we can prove this fact using the standard technique. Let $y\in A\cup B$. Then we know $y\in A$ or $y\in B$. This is equivalent to $y\in B$ or $y\in A$, which means $y\in B\cup A$. Hence we've shown $A\cup B\subseteq B\cup A$. The proof that $B\cup A\subseteq A\cup B$ is similar. Thus $A\cup B= B\cup A$.
  5. The proof that $A\cap B=B\cap A$ is almost identical to the previous.
  6. This is the next problem.
  7. This is the next problem.
  8. We now prove that $A\cup(B\cap C) = (A\cup B)\cap (A\cup C)$. We first show $A\cup(B\cap C) \subseteq (A\cup B)\cap (A\cup C)$. Let $x\in A\cup (B\cap C)$. This means that $x\in A$ or $x\in B\cap C$. We must show that $x\in A\cup B$ and that $x\in A\cup C$. There are two cases, namely $x\in A$ or $x\notin A$. Suppose first that $x\in A$. Then clearly $x\in A\cup B$ and $x\in A\cup C$ as $x$ is a member of the first set in each union. This shows that $x\in (A\cup B)\cap(A\cup C)$. Now suppose $x\notin A$. This means $x\in B\cap C$ (see the third sentences in this proof). Hence we know that $x\in B$ and $x\in C$. Since $x\in B$, we now $x\in A\cup B$. Since $x\in C$, we know $x\in A\cup C$. This shows that $x\in (A\cup B)\cap(A\cup C)$ as desired. This completes the proof that $A\cup(B\cap C) \subseteq (A\cup B)\cap (A\cup C)$. To finish, we must prove $(A\cup B)\cap (A\cup C) \subseteq A\cup(B\cap C)$. Let $y\in (A\cup B)\cap (A\cup C)$. We again use two cases. Suppose $y\in A$. Then clearly $y\in A\cup (B\cap C)$ by definition of union. The only other option is $y\notin A$. Recall we assumed that $y\in (A\cup B)\cap (A\cup C)$, which means $y\in (A\cup B)$ and $y\in (A\cup C)$. This means $y\in A$ or $y\in B$, and it means $y\in A$ or $y\in C$. Since we have assumed that $y\notin A$, this means that $y\in B$, and it means that $y\in C$. Together, this gives $y\in B\cap C$, which shows that $y\in A\cup (B\cap C)$.
  9. Your proof should be very similar to the previous.

Problem 30: (Associative Laws For Set Unions And Intersections)

Let $A$, $B$, and $C$ be sets.

  1. Prove that $A\cup (B\cup C) = (A\cup B)\cup C$.
  2. Prove that $A\cap (B\cap C) = (A\cap B)\cap C$.
Here is another problem to help you with limit points.

Problem 31: (The Integers Have No Limit Points)

There are three parts to this problem.

  • Start by writing the definition of a limit point $p$ of a set $S$ using the quantifiers $\forall$ and $\exists$. Feel free to use set operations $\cup$ and/or $\cap$ in your definitions.
  • Then write, using these quantifiers, what it means to not be a limit point.
  • Finish by proving that if $p\in\mathbb{R}$, then $p$ is not a limit point of $\mathbb{Z}$. In other words, prove that $\mathbb{Z}$ has no limit points.

For more problems, see AllProblems