MATH 447: Real Variables - Homework #3
Jerich Lee
December 22, 2024
Problem.
[11.5] Let $(q_n)$ be an enumeration of all the rationals in the interval $(0,1]$.
[11.5] Let $(q_n)$ be an enumeration of all the rationals in the interval $(0,1]$.
- Give the set of subsequential limits for $(q_n)$.
- Give the values of $\lim \sup q_n$ and $\lim \inf q_n$.
Solution.
- The set of subsequential limits for $(q_{n} )$ is ${q \in Q : 0 < q \leq 1}$.
- $\lim \sup q_{n} =1$
- $\lim \inf q_{n} = 0$
Problem.
[11.11] Let $S$ be a bounded set. Prove there is an increasing sequence $(s_n)$ of points in $S$ such that $\lim_{n \to \infty} s_n=\sup{} S$. Compare Exercise 10.7. Note: If $\sup{} S$ is in $S$, it's sufficient to define $s_n=\sup{} S$ for all $n$.
[11.11] Let $S$ be a bounded set. Prove there is an increasing sequence $(s_n)$ of points in $S$ such that $\lim_{n \to \infty} s_n=\sup{} S$. Compare Exercise 10.7. Note: If $\sup{} S$ is in $S$, it's sufficient to define $s_n=\sup{} S$ for all $n$.
Solution.
$u_{n} =\sup{} s_{n} $, and $s_{n} \leq{} \sup{} s_{n} $. If $S$ is bounded, then there exists a real number $a$ such that $s_{n} \leq{} a$. By Theorem 11.4 in Ross, there exists a monotonic subsequence for bounded set $S$. We shall prove that this sequence is increasing.
Proof.
Because $S$ is bounded, we know that there exists $\sup{} S$ by the Axiom of Completeness. We also know that monotone bounded sequences converge by Theorem 10.2 in Ross. Therefore, we can say:
$$\begin{align}
\sup S-s_{n} <1
\end{align}$$
Where 1 is chosen arbitrarily. We can then choose a subsequence of $S$ such that it satisfies the following conditions:
$$\begin{align}
\sup S - s_n = \min \left( \sup S - s_{n-1}, \frac{1}{n} \right)
\end{align}$$
Choosing such $s_{n} $ enforces that the sequence is increasing, and is getting closer and closer to the supremum. With the the values selected, we now have a subsequence of $S$ such that $n_1<n_{2}<\ldots{} n_k $, $s_{n_1}<s_{n_2}<\ldots{} s_{n_k}$, and $\lim{} S_{n_k} = \sup{} S$.
∎
Problem.
[12.3 (d,e)] Let $(s_n)$ and $(t_n)$ be the following sequences that repeat in cycles of four:
[12.3 (d,e)] Let $(s_n)$ and $(t_n)$ be the following sequences that repeat in cycles of four:
- $(s_n)=(0,1,2,1,0,1,2,1,0,1,2,1,0,\dots)$
- $(t_n)=(2,1,1,0,2,1,1,0,2,1,1,0,\dots)$
- $\lim \sup (s_n+t_n)$
- $\lim \sup s_n + \lim \sup t_n$
Solution.
- $\lim \sup (s_{n} + t_{n}) = 3$
- $\lim \sup (s_{n} )+\lim \sup (t_{n} )=4$
Problem.
[12.4] Show $\lim_{N \to \infty} \sup_{n>N}(s_{n} +t_n)\leq{} \lim_{N \to \infty} \sup_{n>N}s_n+\lim_{N \to \infty} \sup_{n>N} t_n$ for bounded sequences $(s_n)$ and $(t_n)$. Hint: First show:
[12.4] Show $\lim_{N \to \infty} \sup_{n>N}(s_{n} +t_n)\leq{} \lim_{N \to \infty} \sup_{n>N}s_n+\lim_{N \to \infty} \sup_{n>N} t_n$ for bounded sequences $(s_n)$ and $(t_n)$. Hint: First show:
- $\sup\{s_{n} +t_{n} : n> N\}\leq \sup\{s_{n} :n>N\}+\sup \{t_{n} :n>N\}$.
- Prove that if $\lim s_{n} $ and $\lim t_{n} $ exist, then $\lim s_n\leq \lim t_n$.
Solution.
Proof.
We want to find the following:
$$\begin{align}
\limsup_{N \to \infty} s_{N}+t_{N} \leq (\limsup_{N \to \infty} s_{n} )+(\limsup_{N \to \infty} t_{n} )
\end{align}$$
We know that the following holds true, as this is the definition of $\lim{}\sup{} $.
$$\begin{align}
\limsup_{N \to \infty} s_{N} = \limsup_{M \to \infty}\left\{ s_{N} | N\geq M \right\}
\end{align}$$
We also know that any element of a sequence $s_{n} $ is always less than or equal to the supremum of $s_{n} $ due to the Axiom of Completeness. Then, we can state the following.
$$\begin{align}
N>M\implies s_{N} \leq \sup \left\{ s_{N} | N\geq M \right\}
\end{align}$$
Therefore, the following also holds true from above:
$$\begin{align}
s_{N} +t_{N} \leq \sup \left\{ s_{N} |N>M \right\} + \sup \left\{ t_{N} |N>M \right\}
\end{align}$$
The RHS of the above is a constant, so the following is true:
$$\begin{align}
(s_{N} +t_{N} \leq C)
\end{align}$$
If any value of $s_{n} +t_{n} $ is always less than the RHS of above, then we can say the the supremum of $s_{n} +t_{n} $ is also less than or equal to the RHS of above: $$\begin{align}
\sup \left\{ s_{N} +t_{N} |N>M \right\} \leq \sup \left\{ s_{N} |N>M \right\} +\sup \left\{ t_{N} |N>M \right\}
\end{align}$$
We have now established $\sup{} (s_{n} +t_{n})\leq{} \sup{} s_{n} +\sup{} t_{n} $. Now, we can use a familiar limit theorem to prove our original hypothesis.
$$\begin{align}
\limsup_{N \to \infty} s_{N} +t_{N}
\end{align}$$
We know that the above is equal to the following from the definition of $\limsup{}$. Then, we can take the limit of both sides. From Exercise 9.9c, we know that if if given $N_{0}\ s.t. \ s_{n} \leq{} t_{n} $ for all $n>N_0$ $\lim_{n \to \infty} s_{n} $ and $\lim_{n \to \infty} t_{n} $ exist, then $\lim_{n \to \infty}\leq{} \lim_{n \to \infty} t_{n}$
$$\begin{align}
=\limsup_{M \to \infty} \left\{ s_{N} +t_{N} |N>M \right\} \leq \lim_{M \to \infty} (\sup \left\{ s_{N} |N>M \right\} +\sup \left\{ t_{N} |N>M \right\} )
\end{align}$$
From a known limit theorem, we know that $\lim_{n \to \infty} s_{n} +t_{n} =\lim_{n \to \infty} s_{n} +\lim_{n \to \infty} t_{n} $:
$$\begin{align}
=\limsup_{M \to \infty} \left\{ s_{N} |N>M \right\} +\limsup_{M \to \infty} \left\{ t_{N} |N>M \right\}
\end{align}$$
$$\begin{align}
\limsup_{N \to \infty} (s_{N} +t_{N} )\leq \limsup_{N \to \infty} s_{N} + \limsup_{N \to \infty} t_{N}
\end{align}$$
∎
Problem.
[12.12] Let $(s_{n} )$ be a sequence of nonnegative numbers, and for each $n$ define $\sigma_n{}=\frac{1}{n}(s_1 +s_2 +\dots{} +s_n )$.
[12.12] Let $(s_{n} )$ be a sequence of nonnegative numbers, and for each $n$ define $\sigma_n{}=\frac{1}{n}(s_1 +s_2 +\dots{} +s_n )$.
- Show: $$\begin{align} \lim_{N \to \infty} \inf_{n>N}s_n \leq \lim_{N \to \infty} \inf_{n>N} \sigma_n \leq \lim_{N \to \infty} \sup_{n>N} \sigma_n \leq \lim_{N \to \infty} \sup_{n>N} s_n \end{align}$$\\ Hint: for the last inequality, show first that $M>N$ implies: $$\begin{align} \sup \left\{ \sigma_n:n>M \right\} \leq \frac{1}{M}(s_1 +s_2 + \dots +s_n)+\sup \left\{ s_n:n>N \right\} \end{align}$$
- Show that if $\lim_{n \to \infty} s_n$ exists, then $\lim_{n \to \infty} \sigma_n$ exists and $\lim_{n \to \infty} \sigma_n=\lim_{n \to \infty} s_n$.
- Give an example where $\lim_{n \to \infty} \sigma_n$ exists, but $\lim_{n \to \infty} s_{n} $ does not exist.
Solution.
- Proof. $$\begin{align} \liminf_{N \to \infty} s_{N} \leq \liminf_{N \to \infty} \sigma_{N} \leq \limsup_{N \to \infty} \sigma_{N} \leq \limsup_{N \to \infty} s_{N} \end{align}$$ $$\begin{align} \lim_{n \to \infty} s_{N} = s \implies s=\liminf_{N \to \infty} s_{N} \leq \liminf_{N \to \infty} \sigma_{N} \leq \limsup_{N \to \infty} \sigma_{N} \leq \limsup_{N \to \infty} s_{N} = s \end{align}$$ By the Squeeze Theorem, $$\begin{align} \implies \liminf_{N \to \infty} \sigma_{N} =\limsup_{n \to \infty} \sigma_{N} =s \end{align}$$ $$\begin{align} \implies \lim_{N \to \infty} \sigma_{N} =s \end{align}$$ We wish to find the third inequality as shown below, as the second inequality is obvious, and the first inequality can be proved similarly to the third inequality. $$\begin{align} \limsup_{N \to \infty} \sigma_{N} \leq \limsup_{N \to \infty} s_{N} \end{align}$$ We will fix $L$. Then, if $M>L$, we can consider $\sup{} \{ \sigma_{N} |N>M \} $ $$\begin{align} \sigma_{N} &= \frac{s_{1}+\ldots s_{N}}{N}\\[10pt] &= \frac{s_{1} +\ldots +s_{N} }{N}+\frac{s_{L+1}+\ldots +s_{N} }{N} \end{align}$$ We know that there are $N-(L+1)+1=N-L$ total terms in the denominator of the second term of the RHS above. Therefore, we can state the following: $$\begin{align} \sigma_{N} \leq \frac{s_{1}+\ldots s_{L}}{N} + \underbrace{\frac{N-L}{N}}_{\leq 1}\sup \left\{ s_{N} |N>L \right\} \end{align}$$ $$\begin{align} \leq \frac{s_{1} +\ldots +s_{L} }{M}+\sup \left\{ s_{N} |N>L \right\} \end{align}$$ $$\begin{align} \sup \left\{ \sigma_{N} |N>M \right\} \leq \frac{s_{1} +\ldots +s_{L} }{M}+\sup \left\{ s_{N} |N>L \right\} \end{align}$$ We know that the RHS above is independent of $N$, so we take the limit of both sides above. Doing so results in the following: $$\begin{align} \limsup_{N \to \infty} \sup \sigma_{N} =\limsup_{M \to \infty} \sigma_{N} \left\{ N|N>M \right\}\\[10pt] \leq \lim_{M \to \infty} (\underbrace{\frac{s_{1} +\ldots +s_{L}}{M}}_{\text{goes to $0$ as $M\to infinity$ } } +\sup \left\{ s_{N} |N>L \right\} ) \end{align}$$ $$\begin{align} \limsup_{N \to \infty} \sigma_{N} \leq \sup \left\{ s_{N} |N>L \right\} \end{align}$$∎
- Proof. $$\begin{align} \left\vert s_{n} -L \right\vert < \varepsilon \\[10pt] \sigma_{n} = \frac{1}{n} \Sigma_{i=1}^{n}s_{n} \\[10pt] \lim \frac{1}{n}\Sigma_{i=1}^{n}s_{n}\\[10pt] n>N \implies \Sigma_{i=1}^{n}s_{n} \\[10pt] \implies \left\vert ns_{n} -nL \right\vert \\[10pt] \frac{1}{n}\left\vert n(s_{n} -L) \right\vert <\varepsilon \\[10pt] =\left\vert (s_{n} -L) \right\vert <\varepsilon \\[10pt] = \left\vert s_{n} -L \right\vert <\varepsilon \\[10pt] = \lim \sigma_{n} = L=\lim s_{n} \end{align}$$∎
Proof. \item{} Suppose $s_{n} = (-1)^n$. Then $\lim{} s_{n} = \text{DNE} $. But we know: $$\begin{align} \frac{\text{$-1$ or $0$ } }{N} \underbrace{\to }_{N\to \infty } 0 \end{align}$$ Therefore, the $\lim_{n \to \infty} \sigma_{n}$ exists if $\lim_{n \to \infty} s_{n}=\text{DNE} $.∎
Problem.
[12.13] Let $(s_{n} )$ be a bounded sequence in $\mathbb{R}$. Let $A$ be the set of $a\in{}\mathbb{R} $ such that $\{n \in{} \mathbb{N} :s_{n} <a \} $ is finite, i.e., all but finitely many $s_{n} $ are $\geq{} a$. Let $B$ be the set of $b\in{}\mathbb{R} $ such that $\{ n\in{} \mathbb{N}: s_{n} >b \} $ is finite. Prove $\sup{} A=\lim_{N \to \infty} \inf_{n>N}s_{n} $ and $\inf{} B=\lim_{N \to \infty} \sup_{n>N}s_{n} $.
[12.13] Let $(s_{n} )$ be a bounded sequence in $\mathbb{R}$. Let $A$ be the set of $a\in{}\mathbb{R} $ such that $\{n \in{} \mathbb{N} :s_{n} <a \} $ is finite, i.e., all but finitely many $s_{n} $ are $\geq{} a$. Let $B$ be the set of $b\in{}\mathbb{R} $ such that $\{ n\in{} \mathbb{N}: s_{n} >b \} $ is finite. Prove $\sup{} A=\lim_{N \to \infty} \inf_{n>N}s_{n} $ and $\inf{} B=\lim_{N \to \infty} \sup_{n>N}s_{n} $.
Solution.
From Theorem 11.7, there exists a monotonic subsequence whose limit is $\lim{} \sup{} s_{n} $ and $\lim{} \inf{} s_{n} $. So $\lim{} \sup{} s_{n} $ and $\lim{} \inf{} s_{n} $ exist. From Theorem 10.1, we know that these limits converge because $s_{n} $ is bounded. Our goal is to show that $\lim{} \sup{} s_{n}=\inf{} B $, and $\lim{} \inf{} s_{n} =\sup{} A$, for sets $A=\{ s_{n} :s_{n} <a \} $, and $B = \{ s_{n} :s_{n} >b \} $, where $A,B$ both have finite cardinality.
Proof.
Choose $N_{1}\ s.t. \ n>N_{1}\implies{} \{ n:\vert s_{n} -a \vert < \varepsilon{} \} $ is infinite. By Theorem 11.2, $a$ is a subsequential limit. Choose $N_{2} \ s.t. \ n>N_{2}\implies{} \{ n:\vert s_{n} -b \vert <\varepsilon{} \} $ is infinite. By 11.2, $b$ is also a subsequential limit. To prove that $a=\sup{} A$, choose $N=\mathop{\min} \{ d(s_{N}, a) \} $. Then $N=n+1\implies{} s_{n+1} >a$, so $s_{n} $ is not the least upper bound of A. So $a=\sup{} A$. A similar argument follows for $\inf{} B=b$. We know that $a<b$, or else $B$ would be infinite, contradictory to $B$ being finite in the given. Let $S=\{ a,b \} $. Then, by Theorem 11.8, $\inf{} S=\lim{} \inf{} s_{n}, \sup{} S=\lim{} \sup{} s_{n} $. This implies $\sup{} A=\lim{} \inf{} s_{n}, \inf{} B=\lim{} \sup{} s_{n} $.
∎