I would like someone to verify my solution attempt of this exercise on open and closed sets, from Stephen Abbott's Understanding Analysis.
$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
Exercise 3.2.3
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 $\epsilon$-neighbourhood contained in the set. If a set is not closed, find a limit point that is not contained in the set.
(a) $\mathbf{Q}$
(b) $\mathbf{N}$
(c) $\{x \in \mathbf{R}: x \ne 0\}$
(d) $\{1 + 1/4 + 1/9 + \ldots + 1/n^2 : n \in \mathbf{N}\}$
(e) $\{1 + 1/2 + 1/3 + \ldots + 1/n : n \in \mathbf{N}\}$
Solution.
(a) Given any rational number $q$, and an $\epsilon > 0$, there exists an irrational number $x$, such that $\absval{x - q} < \epsilon$, so $V_\epsilon(x) \not\subseteq \mathbf{Q}$. Therefore $\mathbf{Q}$ is not open. For instance, if $q = \frac{m}{n}$, $\epsilon=\frac{1}{n}$ where $m,n \in \mathbf{Z}$ and if $x = \frac{m + 1}{\sqrt{2}n}$, $\absval{x - q} < \epsilon$. And $V_\epsilon(q) \not\subseteq Q$.
The set of all limit points of $\mathbf{Q}$ is $\mathbf{R}$. The irrational numbers $\mathbf{I}$ are not members of the set of rational numbers $\mathbf{Q}$. So, $\mathbf{Q}$ is not closed. For example, $x = \sqrt{2}$ is a limit of point of $\mathbf{Q}$, since every $\epsilon$-neighbourhood of $\sqrt{2}$, $(\sqrt{2} - \epsilon,\sqrt{2} + \epsilon)$ intersects $\mathbf{Q}$ at some point other than $\sqrt{2}$.
(b) Given any natural number $n$, and $\epsilon > 0$, there exists a rational number $p/q$, $q \notin \{0,1\}$ such that $\absval{\frac{p}{q} - n} < \epsilon$. Thus, $V_\epsilon(n) \not\subseteq \mathbf{N}$. Therefore, $\mathbf{N}$ is not open. For instance, if $n = 0, \epsilon = 1/2$, then $\absval{1/4 - n} < \epsilon$.
The members of $\mathbf{N}$, $\{0,1,2,3,\ldots\}$ are all isolated points, because if $\epsilon < 1$, then $V_\epsilon(0) \cap \mathbf{N} = \{0\}, V_\epsilon(1) \cap \mathbf{N} = \{1\}, V_\epsilon(2) \cap \mathbf{N} = \{2\}, \ldots$. Thus, there are no limit points in $\mathbf{N}$. Consequently, $\mathbf{N}$ is not closed.
(c) $\{x \in \mathbf{R}:x \ne 0\}$. For all $x \in \mathbf{R} - \{0\}$, there exists an $\epsilon$-neighbourhood $V_\epsilon(x)$, such that $V_\epsilon(x) \subseteq \mathbf{R} - \{0\}$. Thus, $\mathbf{R} - \{0\}$ is an open set.
The point $x = 0$ is a limit point of the set $\mathbf{R} - \{0\}$, since the sequence $a_n = \frac{1}{n}$ converges to $0$ as $n \to\infty$ and $a_n \ne x$. Moreover, $0$ is not element of the set $\mathbf{R} - \{0\}$, so $\mathbf{R} - \{0\}$ is not a closed set. This is the only defect in $\mathbf{R} - \{0\}$.
(d) This set contains the partial sums of the infinite series $\sum \frac{1}{n^2}$. Writing out the first few terms, $S = \{s_1,s_2,s_3,s_4,\ldots\} = \{1,5/4,49/36,\ldots,\}$. Given $s_n$, let $\epsilon = s_n - s_{n-1}$, $V_\epsilon(s_{n}) \not\subseteq S$. It is not an open set.
We know, that $(s_n)$ is a monotonic increasing sequence with an upper bound $2$, and therefore convergent. Since $S$ contains the terms of the sequence of the partial sums alone and $s_n \ne \lim s_n$ for all $n$, the limit point of $S$ is not an element of $S$. Consequently, it is not a closed set.
(e) This set contains the partial sums of the harmonic series $\sum \frac{1}{n}$. Writing out the first few terms, $S = \{s_1,s_2,s_3,s_4,\ldots\} = \{1,3/2,11/6,\ldots,\}$. Again, as before $\exists x$ such that $V_\epsilon(x) \not\subseteq S$, $\forall \epsilon > 0$. So, $S$ is not an open set.
The partial sums $s_n$ are isolated points of the set $S$. Given $s_n$, there exists an $V_\epsilon(s_n)$, such that $V_\epsilon(s_n) \cap S = \{s_n\}$. Since $(s_n)$ is an unbounded sequence; it is divergent and does not have any limit point. So, $S$ is not a closed set.
