Exercise 3.2.6

Question

Decide whether the following statements are true or false. Provide counterexamples for those that are false, and supply proofs for those that are true.
  \begin{enumerate}[label=(\alph*)] 
  \item An open set that contains every rational number must necessarily be all of $\mathbf{R}$.
  \item The Nested Interval Property remains true if the term ``closed interval'' is replaced by ``closed set.''
  \item Every nonempty open set contains a rational number.
  \item Every bounded infinite closed set contains a rational number.
  \item The Cantor set is closed.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item False, $A = (-\infty, \sqrt 2) \cup (\sqrt 2, \infty)$ contains every rational number but not $\sqrt 2$.
  \item False, $C_n = [n, \infty)$ is closed, has $C_{n+1} \subseteq C_n$ and $C_n 
e \emptyset$ but $\bigcap_{n=1}^\infty C_n = \emptyset$.
  \item True, let $x \in A$ since $A$ is open we have $(a,b) \subseteq A$ with $x \in (a,b)$ the density theorem implies there exists an $r \in \mathbf{Q}$ with $r \in (a,b)$ and thus $r \in A$.
  \item False, $A = \{1/n + \sqrt 2 : n \in \mathbf{N}\} \cup \{\sqrt 2\}$ is closed and contains no rational numbers.
  \item True, as it is the intersection of countably many closed intervals.
   \end{enumerate}

Exercise 3.2.6

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Exercise 3.2.6

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