Exercise 3.2.7

Question

Given $A \subseteq \mathbf{R}$, let $L$ be the set of all limit points of $A$.
  \begin{enumerate}[label=(\alph*)] 
  \item Show that the set $L$ is closed.
  \item Argue that if $x$ is a limit point of $A \cup L$, then $x$ is a limit point of $A$. Use this observation to furnish a proof for Theorem 3.2.12.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item Every $x_n \in L$ is $x_n = \lim_{m\to\infty} a_{mn}$ for $a_{mn} \in A$. Meaning if $\lim x_n = x$ then for $n > N$ and $m > M$ we have
    $$
    |a_{mn} - x| \le |a_{mn} - x_n| + |x_n - x| < \epsilon/2 + \epsilon/2 = \epsilon
    $$
    and thus $x$ is a limit point of $A$, so $x \in L$.
  \item Let $x_n \in A \cup L$ and $x = \lim x_n$. Since $x_n$ is infinite there must be at least one subsequence $(x_{n_k}) \to x$ which is either all in $A$ or all in $L$.
    If every $x_{n_k} \in L$ then we know $x \in L$ from (a), and if every $x_{n_k} \in A$ then $x \in L$ aswell.
   \end{enumerate}

Exercise 3.2.7

Answers

Comments

Exercise 3.2.7

Answers

Comments

Add answer