Exercise 3.2.3

Question

Decide whether the following sets are open, closed, or neither. If a set is not open, find a point in the set for which there is no $\epsilon$-neighborhood contained in the set. If a set is not closed, find a limit point that is not contained in the set.
  \begin{enumerate}[label=(\alph*)] 
  \item $\mathbf{Q}$.
  \item $\mathbf{N}$.
  \item $\{x \in \mathbf{R}: x 
eq 0\}$.
  \item $\left\{1+1 / 4+1 / 9+\cdots+1 / n^{2}: n \in \mathbf{N}ight\}$
  \item $\{1+1 / 2+1 / 3+\cdots+1 / n: n \in \mathbf{N}\}$
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item
    Neither, not open as $(a,b) \subseteq \mathbf{Q}$ is impossible since $\mathbf{Q}$ contains no irrationals but $(a,b)$ does.
    and not closed since every irrational can be reached as a limit of rationals ($\sqrt 2$ is a simple example).
  \item Clearly not open, but ironically closed since it has no limit points.
  \item Open since every $x \in \{x \in \mathbf{R} : x 
e 0\}$ has an $\epsilon$-neighborhood around it excluding zero. But closed since $(1/n) 	o 0$.
  \item
    Neither, not closed, as the limit $\sum_k^n 1/n^2 = \pi^2/6$ is irrational but every term is rational.
    and not open as it does not contain any irrationals.
  \item Closed as it has no limit points, every sequence diverges. Not open because it contains no irrationals.
   \end{enumerate}

Exercise 3.2.3

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Exercise 3.2.3

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