In the first part of this course, we covered the construction of real number, and some results about limit. Here is a list of key concepts

• The numbers $\N, \Z, \Q$.
• The axioms of field, an example of finite field $\F_5$.
• The order relation, and ordered set. upper bound, lower bound. The least upper bound property.
• $\R$ as equivalence classes of Cauchy sequences in $\Q$. Prove many familiar operations and properties of $\R$.
• $\R$ has least upper bound property. (hence $\sup$ and $\inf$ of bounded subset in $\R$ exists in $\R$)
• Sequences in $\R$, notion of convergence
• Monotone bounded sequences are convergent (for increasing sequence, the $\lim a_n = \sup\{a_n: n \in \N \}$; for decreasing one, $\lim a_n = \inf \{a_n: n \in \N\}$.
• lim-sup and lim-inf. The “epsilon of room” philosophy.
• Thm: Cauchy sequences are convergent.
• Limit Points, 3 equivalent definitions

Midterm will have 3 of such following questions.

1. Let $(a_n)$ be a Cauchy sequence of irrational numbers, then its limit has to be an irrational number.
2. Let $S$ be an ordered set, then any non-empty finite subset $E \In S$ has a least upper bound.
3. If $x$ is a limit point of sequence $(a_n)$, then there exists one $n \in \N$, such that $a_n = x$.
4. If $(a_n)$ is a sequence bounded above, and $L = \limsup(a_n)$, then for any $\epsilon > 0$, there exists an integer $N > 0$, such that $a_n < L + \epsilon$.
5. Let $(a_n)$ be a bounded sequence in $\R$. Let $A_n = \sup \{a_m : 0 \leq m \leq n \}$, then $\lim A_n = \limsup a_n$
6. Let $(a_n)$ and $(b_n)$ be convergent sequences with the same limit $x$. And choose any function $f: \N \to \{0,1\}$. We define a new sequence, by mixing $a_n$ and $b_n$ $$c_n = \begin{cases} a_n & \text{if } f(n) = 0 \cr b_n & \text{ if } f(n) = 1 \end{cases} $$ Then, $c_n$ converges to $x$.
7. If $(a_n)$ and $(b_n)$ are Cauchy sequence in $\R$, and they satisfy that $\lim(a_n b_n) = 1$, then $\lim a_n \neq 0$.
8. If $(a_n)$ is a sequence of positive real numbers, for $n \geq 1$, and $A_n = (a_1 + \cdots + a_n) / n$, show that if $a_n$ is convergent then $A_n$ is convergent. Give an example where $A_n$ is convergent, but $a_n$ is not convergent.
9. Is there a sequence $(a_n)$, where $|a_n - a_{n-1}|$ is monotone decreasing, but $(a_n)$ is not convergent?

Let $(a_n)$ be a Cauchy sequence of irrational numbers, then its limit has to be an irrational number.

False. Say $a_n = \pi / n$.

Let $S$ be an ordered set, then any non-empty finite subset $E \In S$ has a least upper bound.

True. One can prove this by induction on the size of $E$.

If $x$ is a limit point of sequence $(a_n)$, then there exists one $n \in \N$, such that $a_n = x$.

False. $0 = \lim_n 1/n$.

If $(a_n)$ is a sequence bounded above, and $L = \limsup(a_n)$, then for any $\epsilon > 0$, there exists an integer $N > 0$, such that $a_n < L + \epsilon$.

True. explained in class.

Let $(a_n)$ be a bounded sequence in $\R$. Let $A_n = \sup \{a_m : 0 \leq m \leq n \}$, then $\lim A_n = \limsup a_n$.

False. That's not the definition of $\limsup$. Say $a_n = 1/(n+1)$, for $n \in \N$, then $\lim A_n = 1$, and $\limsup a_n=0$.

Let $(a_n)$ and $(b_n)$ be convergent sequences with the same limit $x$. And choose any function $f: \N \to \{0,1\}$. We define a new sequence, by mixing $a_n$ and $b_n$ $$c_n = \begin{cases} a_n & \text{if } f(n) = 0 \cr b_n & \text{ if } f(n) = 1 \end{cases} $$ Then, $c_n$ converges to $x$.

True. By convergence of $a_n$ and $b_n$, for any $\epsilon>0$, exists $N_1 > 0$ and $N_2$, such that if $n > N_1$, then $|a_n - x| < \epsilon$, and if $n>N_2$, then $|b_n - x| < \epsilon$. Since $c_n$ is either $a_n$ or $b_n$, hence for $n > N=\max(N_1, N_2)$, we have $|c_n - x| < \epsilon$.

If $(a_n)$ and $(b_n)$ are Cauchy sequence in $\R$, and they satisfy that $\lim(a_n b_n) = 1$, then $\lim a_n \neq 0$.

True. Since if $a_n, b_n$ are convergent, then $1 = \lim(a_n b_n) = (\lim a_n ) (\lim b_n)$, hence $\lim a_n \neq 0$.

If $(a_n)$ is a sequence of positive real numbers, for $n \geq 1$, and $A_n = (a_1 + \cdots + a_n) / n$, show that if $a_n$ is convergent then $A_n$ is convergent. Give an example where $A_n$ is convergent, but $a_n$ is not convergent.

Proof: Let $x$ be the limit of $a_n$. Given any $\epsilon > 0$, we claim there exists an $N>0$, such that $|A_n - x| < \epsilon$ for all $n > N$. Let $\epsilon_1 = \epsilon/3$, and let $N_1>0$ be such that for all $n \geq N_1$, we have $|a_n - x| < \epsilon_1$. Since $a_n$ is convergent, we also have an $M>0$, such that $|a_n| < M$ for all $n$. Then, for any $n > N_1$, we have $$ A_n - x = (1/n) \sum_{m=1}^{N_1} a_m + (1/n) \sum_{m=N_1+1}^n (a_m-x) + (N_1/n) x $$ Thus, we have $$ |A_n - x| \leq (1/n) N_1 M + (1 - N_1/n) \epsilon_1 + (N_1/n) x $$ we may choose $N$ large enough, such that $N_1 / N x < \epsilon/3$, $N_1/N M < \epsilon / 3$, thus, for any $n > N$, we have $$ |A_n - x| \leq \epsilon/3 + \epsilon / 3 + \epsilon / 3 = \epsilon. $$ This finishes the proof of the first claim.

Here is an example where $A_n$ is convergent, but $a_n$ is not convergent: $a_n = (-1)^n$.

Is there a sequence $(a_n)$, where $|a_n - a_{n-1}|$ is monotone decreasing, but $(a_n)$ is not convergent?

Yes, for $n \geq 1$, $a_n = \sum_{m=1}^n 1/n$.