Exercise 1.4.8

Question

Give an example of each or state that the request is impossible. When a request is impossible, provide a compelling argument for why this is the case.
  \begin{enumerate}[label=(\alph*)] 
  \item Two sets $A$ and $B$ with $A \cap B=\emptyset, \sup A=\sup B, \sup A 
otin A$ and $\sup B 
otin B$.
  \item A sequence of nested open intervals $J_{1} \supseteq J_{2} \supseteq J_{3} \supseteq \cdots$ with $\bigcap_{n=1}^{\infty} J_{n}$ nonempty but containing only a finite number of elements.
  \item A sequence of nested unbounded closed intervals $L_{1} \supseteq L_{2} \supseteq L_{3} \supseteq \cdots$ with $\bigcap_{n=1}^{\infty} L_{n}=\emptyset$. (An unbounded closed interval has the form $[a, \infty)=$ $\{x \in R: x \geq a\} .)$
  \item A sequence of closed bounded (not necessarily nested) intervals $I_{1}, I_{2}$, $I_{3}, \ldots$ with the property that $\bigcap_{n=1}^{N} I_{n} 
eq \emptyset$ for all $N \in \mathbf{N}$, but $\bigcap_{n=1}^{\infty} I_{n}=\emptyset$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $A = \mathbf{Q} \cap (0, 1)$, $B = \mathbf{I} \cap (0, 1)$. $A \cap B = \emptyset$, $\sup A = \sup B = 1$ and $1 
otin A$, $1 
otin B$.
  \item Defining $J_i = (a_i, b_i)$, $A = \{a_n : n \in \mathbf{N}\}$, $B = \{b_n : n \in \mathbf{N}\}$, $\bigcap_{n=1}^\infty J_n$ will at least contain $(\sup A, \inf B)$. Thus, a necessary condition to meet the request is $\sup A = \inf B$.

$J_i = (-1/n, 1/n)$ satisfies this condition ($\sup A = \inf B = 0$) and by inspection, $\bigcap_{n=1}^\infty J_n = \{0\}$, which meets the request.

\item $L_n = [n, \infty)$ has $\bigcap_{n=1}^\infty L_n = \emptyset$
  \item Impossible. Let $J_n = \bigcap_{k=1}^n I_k$ and observe the following
    \begin{enumerate}[label=(oman*)] 
      \item Since $\bigcap_{n=1}^N I_n 
e \emptyset$ we have $J_n 
e \emptyset$.
      \item $J_n$ being the intersection of closed intervals makes it a closed interval.
      \item $J_{n+1} \subseteq J_n$ since $I_{n+1} \cap J_n \subseteq J_n$
      \item $\bigcap_{n=1}^\infty J_n = \bigcap_{n=1}^\infty\left(\bigcap_{k=1}^n I_kight) = \bigcap_{n=1}^\infty I_n$
     \end{enumerate}

By (i), (ii) and (iii) the Nested Interval Property tells us $\bigcap_{n=1}^\infty J_n 
e \emptyset$. Therefore by (iv) $\bigcap_{n=1}^\infty I_n 
e \emptyset$.
   \end{enumerate}

Exercise 1.4.8

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Exercise 1.4.8

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