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From Ben:

Here is who we'll have present. Please come on time, so we can have you start putting things up when class starts.

  • 63 Intersection -
  • 66 - Junseong
  • 67 -
  • 68 -
  • 69 -
  • 70 -
  • 71 -

The following students were absent

  • James
Exercise (Unions And Intersections Of Opens Sets)

Prove each of the following:

  1. The union of any collection of open sets is open.
  2. The intersection of finitely many open sets is open.

Click to see a solution.

Let's first prove that the union of any collection of open sets is open. Let $J$ be a set and for each $j\in J$ let $U_j$ be an open set. This gives us a way to talk about an arbitrary collection of open sets. We must prove that $\ds\bigcup_{j\in J}U_j$ is an open set. So pick $\ds x \in \bigcup_{j\in J}U_j$. Since $x$ is an element of this union, we know that $x\in U_j$ for some $j\in J$. Since $U_j$ is an open set, we know we can pick $\varepsilon>0$ such that $N_\varepsilon (x)\subseteq U_j$. Since we know $U_j\subseteq \ds\bigcup_{j\in J}U_j$, this means $N_\varepsilon(x)\subseteq \ds\bigcup_{j\in J}U_j$. Since this entire argument holds for any $x\in \ds\bigcup_{j\in J}U_j$, we have shown that $\ds\bigcup_{j\in J}U_j$ is an open set.

We now prove that the intersection of finitely many open sets is open. One way to prove this is to refer to a previous problem where we used induction to prove this is true. Here is another proof. Let $n\in\mathbb{N}$ and suppose $U_1, U_2, \ldots, U_n$ are open sets. Let $x\in \ds\bigcap_{i=1}^n U_i$. To complete this proof, we must produce a postive $\varepsilon$ and prove that $N_{\varepsilon}(x)\subseteq \ds\bigcap_{i=1}^n U_i$. Pick $i\in\{1,2,\ldots,n\}$. Because $x\in \ds\bigcap_{j=1}^n U_j$, we know that $x\in U_i$. We assumed that $U_i$ is open, which means we can pick $\varepsilon_i$ such that $N_{\varepsilon_i}(x)\subseteq U_i$. Since the argument above holds for each relevant $i$, we pick $\varepsilon_i$ for each relevant $i$ so that $N_{\varepsilon_i}(x)\subseteq U_i$. Now comes the key part, namely we let $\varepsilon$ be the smallest of these positive values, which gives $$\varepsilon = \min\{\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n \}.$$ Clearly $\varepsilon>0$ by construction. In addition, because of how we defined $\varepsilon$, we know $N_{\varepsilon}(x)\subseteq N_{\varepsilon_i}(x)$ for each relevant $i$. This fact, together with the fact that $N_{\varepsilon_i}(x)\subseteq U_i$ for each relevant $i$, means we know $N_{\varepsilon}(x)\subseteq U_i$ for each relevant $i$. This fact proves that $N_{\varepsilon}(x)\subseteq \ds\bigcap_{i=1}^n U_i$, as needed. Our proof is complete (and should look very similar to the proof for two open sets).

Notice that the min function in the proof above can fail to produce a positive $\varepsilon$ if there is an infinite number of open sets. There is no guarantee that a minimum will even exist when a set has infinitely many elements. The problem before this exercise clearly shows us that the intersection of infinitely many sets does not have to be open, as we proved $\ds\bigcap_{n=1}^\infty \left(0,1+\frac{1}{n}\right) =(0,1] $.

Problem 63: (The Union And Intersection Of Infinitely Many Closed Sets)

For each $x$ such that $3<x<4$, let $A_x = [2,x]$.

  1. State an interval that equals each of $\ds \bigcup_{3<x<4} A_x$ and $\ds \bigcap_{3<x<4} A_x$.
  2. Prove your claims.

Problem 66: (Triangle Inequality)

For any real numbers $u$ and $v$, let $d(u,v)$ be the distance between $u$ and $v$, which means $d(u,v)=|u-v|=|v-u|$.

  1. Let $a,b,c\in\mathbb{R}$. Prove that $d(a,b)\leq d(a,c)+d(c,b)$.
  2. Let $x,y\in\mathbb{R}$. Use the previous result to prove that $|x+y|\leq |x|+|y|$.
Both facts above we call the triangle inequality. Both facts basically state that the distance from point $A$ to point $B$ is less than or equal to the distance traveled if you take the shortest route from $A$ to $B$ that must also pass through a third point $C$. Equality holds if $C$ is already on the shortest path from $A$ to $B$, otherwise the distance must increase.

Problem 67: (Proving A Quotient Of Two Linear Sequences Converges)

Prove that $\left(\frac{2n+1}{3n+4}\right)$ converges to $\frac{2}{3}$.

Hint: Given $\varepsilon>0$, solve the equality $|a_M-L|=\varepsilon$ for $M$, which should help you find a value $M$ you can choose to satisfy the definition of converges. So start by solving $\left|\frac{2M+1}{3M+4} - \frac{2}{3}\right|=\varepsilon$ for $M$.

Problem 68: (Limit Of A Sum Equals Sum Of Limits)

Suppose $(a_n)$ converges to $A$ and $(b_n)$ converges to $B$. Prove that $(a_n+b_n)$ converges to $A+B$. The triangle inequality should help.

Definition (Invertible Function)

We say that a function $f:D\to C$ is invertible if there exists a function $g:C\to D$ such that $f(g(c))=c$ for every $c\in C$ and $g(f(d))=d$ for every $d\in D$. If such a function $g$ exists, then we use the notation $f^{-1}$ as the name for the function $g$.

Problem 69: (Functions Are Invertible Iff Bijective)

Prove that a function $f:D\to C$ is invertible if and only if $f$ is bijective.

Problem 70: (Proving A Quotient Of Two Quadratic Sequences Converges)

Prove that $\left(\frac{n^2}{n^2+1}\right)$ converges.

Hint: Given $\varepsilon>0$, solve the equality $|a_M-L|=\varepsilon$ to find a value $M$ you can choose to satisfy the definition of converges.

Definition (Image Of A Function)

Let $f:X\to Y$. The image of $f$ is the set $f(X)$.

Some people call the image of $f$ the "range of $f$", while other people use the word "range" to denote the codomain. The word "range" is unfortunately dependent on author. For this reason, we'll stick to the words "codomain" and "image" to separate these two different sets.
Exercise (A Function Is Surjective Iff Codomain And Image Are Equal)

Prove that a function is surjective if and only if the codomain of $f$ and the image of $f$ are equal.

Click to see a solution.

Let $f:X\to Y$. Suppose $f$ is surjective. Clearly the image of $f$ is a subset of the codomain by definition. Let $y$ be an element of the codomain. Since $f$ is surjective, this means we can pick $x$ in the domain such that $f(x)=y$. This shows that $y$ is in the image of $f$, which completes the proof that if $f$ is surjective, then the image of $f$ equals the codomain of $f$.

Now suppose that the image of $f$ equals the codomain of $f$. We need to show that $f$ is surjective. Pick $y$ in the codomain of $f$. Since the image equals the codomain, this means $y$ is in the image of $f$. This means that we can pick $x$ in the domain such that $f(x)=y$. This completes the proof that $f$ is surjective when the image equals the codomain.

Exercise (The Composition Of Surjective Functions Is Surjective Take 2)

Suppose both $f:A\to B$ and $g:B\to C$ are surjective. Prove that $g\circ f:A\to C$ is surjective.

Click to see a solution.

First note that a function is surjective if and only if the image of the domain equals the codomain. We will use both the "if" and "only if" parts of that statment. We now begin the proof. Because the two functions are surjective, we know $f(A)=B$ and $g(B)=C$ (we used the "only if", or implies, part). This means that $(g\circ f)(A)=g(f(A))=g(B)=C$. Since we know $(g\circ f)(A)=C$, this proves that $g\circ f$ is surjective (we used the "if" part).

Definition (The Image Of A Sequence Is A Set)

Given a sequence $(a_n)$ of real numbers, note that the image of the sequence, namely $a(\mathbb{N}) = \{a_n\mid n\in\mathbb{N}\}$, is a subset of the real numbers. Because the image of the sequence is a set of real numbers, we can use any of our previous words that we defined on sets of real numbers, and now apply them to a sequence. Here are some examples:

  • We say a sequence is bounded if the image of the sequence is a bounded set.
  • A lower bound for a sequence is a lower bound for the image of the sequence.
  • The supremum of a sequence is the supremum of the image of the sequence.
  • A limit point of a sequence is a limit point of the image of the sequence.

Problem 71: (Convergent Sequences Are Bounded)

Prove that if a sequence of real numbers converges, then the sequence is bounded.

Problem 72: (Limit Of Product Equals Product Of Limits)

Suppose $(a_n)$ converges to $A$ and $(b_n)$ converges to $B$. Prove that $(a_nb_n)$ converges to $AB$. The triangle inequality should help, along with the fact that convergent sequences are bounded.

Click for a hint.

The key here is to rewrite $|a_nb_n-AB|$ in a new way so that the quantities $a_n-A$ and $b_n-B$ appear. There are lots of ways to do this. One is to just force them to appear by replacing $a_n$ with $a_n-A+A$, giving us $$ |a_nb_n-AB|= |(a_n-A+A)b_n-AB|= |(a_n-A)b_n+Ab_n-AB|= |(a_n-A)b_n+A(b_n-B)|. $$ At this point, the triangle inequality should help to separate things. Then you'll need to pick $M$ large enough so that both $|(a_n-A)b_n|\leq \frac{\varepsilon}{2}$ and $|A(b_n-B)|<\frac{\varepsilon}{2}$. You'll need to guarantee $|a_n-A||b_n|\leq \frac{\varepsilon}{2}$ and $|b_n-B||A|<\frac{\varepsilon}{2}$. You will need an estimate for $|b_n|$ (did you see the bounded part), and use facts about $(a_n)$ and $(b_n)$ converging to get the needed $M$.

Problem 73: (Limits Of Sequences And Limit Points Of Images)

Is a limit point of the image of a sequence equal to the limit of that sequence?

  • Give an example of a sequence $(a_n)$ that converges to $L$, such that $L$ is a limit point of the image of the sequence.
  • Give an example of a sequence $(a_n)$ that converges to $L$, such that $L$ is not a limit point of the image of the sequence.
  • Make a conjecture about when limits of sequences and limit points of images of sequences are equal.
  • Give an example of a sequence $(a_n)$ that does not converge, yet the image of the sequence has one (or more) limit points.
We'll have a class discussion on this one, rather than have someone present it. Please come with examples and a conjecture. If you choose to type this one up, you will need to prove any claims you make (about a sequence converging, or not, to what you say it does (which means you probably want to focus on typing up a different problem, as doing so could be quite time consuming if your examples are non trivial.

Problem 74: (Converges To L Can Be Written As Converges To 0)

Let $(a_n)$ be a sequence of real numbers, and $A$ a real number. Prove that $(a_n)$ converges to $A$ if and only if the sequence $(a_n-A)$ converges to $0$. This will simplify proving that some sequences converge.

Problem 75: (Proving A Rational Sequence Converges)

Consider the sequence $\ds (s_n)=\left(\frac{4n^3-2n^2-7n}{5n^3-3n^2+2n-1}\right)$. In this problem, your job is to prove that $s_n\to 4/5$. You may assume that for all natural numbers, we have $5n^3-3n^2+2n-1>0$.

  1. Show that $\ds(s_n-4/5)= \left(\frac{2 n^2-43 n+4}{5 \left(5 n^3-3 n^2+2 n-1\right)}\right).$
  2. Find a $k_1>0$ and $M_1$ so that $|2 n^2-43 n+4|\leq k_1 n^2$ for all natural numbers $n>M_1$.
  3. Find a $k_2>0$ and $M_2$ so that $|5 \left(5 n^3-3 n^2+2 n-1\right)|\geq k_2 n^3$ for all natural numbers $n>M_2$.
  4. Prove that $(s_n)$ converges to $4/5$.

Problem 76: (Limit Of Quotient Equals Quotient Of Limits)

Suppose $(a_n)$ converges to $A$ and $(b_n)$ converges to $B\neq 0$, and also suppose $b_n\neq 0$ for every natural number $n$. Prove that $(a_n/b_n)$ converges to $A/B$.

Definition (Increasing Decreasing Monotonic Sequences)

Let $(a_n)$ be a sequence of real numbers.

  • We say that $(a_n)$ is (strictly) increasing if $a_n<a_{n+1}$ for every $n\in\mathbb{N}$.
  • We say that $(a_n)$ is (strictly) decreasing if $a_n>a_{n+1}$ for every $n\in\mathbb{N}$.
  • We say that $(a_n)$ is nonincreasing if $a_n\geq a_{n+1}$ for every $n\in\mathbb{N}$.
  • We say that $(a_n)$ is nondecreasing if $a_n\leq a_{n+1}$ for every $n\in\mathbb{N}$.
  • We say that $(a_n)$ is monotonic if $(a_n)$ is either nonincreasing or nondecreasing.

You should notice that a strictly decreasing sequence is nonincreasing, and a strictly increasing sequence is nondecreasing.

Problem 77: (Monotonic Sequences Converge If And Only If Bounded)

Let $(a_n)$ be a monotonic sequence. Prove that $(a_n)$ converges if and only if $(a_n)$ is bounded.

Definition (Diverges To Infinity)

Let $(a_n)$ be a sequence of real numbers. We say that $(a_n)$ diverges to infinity if for every $V$, there exists $H$ such that for every $n\in \mathbb{N}$ we know $n>H$ implies $a_n>V$. When $(a_n)$ diverges to infinity, we write $(a_n)\to \infty$.

Think of $V$ as vertical and $H$ as horizontal.

Problem 78: (A Sequence Diverges To Infinity)

Prove that the sequence $(n^2)$ diverges to $+\infty$.

Problem 79: (Diverges To Negative Infinity)

Construct a definition of what it means to diverge to $-\infty$, by appropriately modifying the definition of diverges to $\infty$. Then prove that the sequence $(-n^3+2n)$ diverges to $-\infty$.

For more problems, see AllProblems