# Deductions about openness and closedness of sets

I would like someone to verify my solution attempt of this exercise on open and closed sets, from Stephen Abbott's Understanding Analysis. $$\newcommand{\absval}{\left\lvert #1 \right\rvert}$$ 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.

• Do you mean (b): since $\mathbf{N}$ is unbounded and has no limit point and, (e): since the harmonic series $\sum 1/n$ is divergent and has no limit point. – Quasar Jan 5 at 18:34