Midterm 1: Review

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. It turns out 3 problems of the following kinds maybe either too easy or too hard. The number of problems may vary, but the difficulties will include some easy ones and some hard ones.

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