# 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. $$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$$ 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.

## 1 Answer

Your solution is almost perfect.

In b) and e) however, you arrive that the set has no limit points. But then the conclusion is that it's closed, since the set of limit points is the empty set which is a subset of every 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
• @BrianM.Scott Yes, thanks, edited. – Berci Jan 5 at 20:41
• @Quasar: it's because they contain all the limit points (which is immediate if there isn't any). As the exercise says, you should be able to show up a limit point of the set which is outside of the set in case it weren't closed. – Berci Jan 5 at 20:43
• Frankly these questions, as posed, make no sense because whether a set is open or closed (or neither or both) depends upon the topology of the super set they are subsets of. Are these all considered as subsets of the real numbers, with the usual topology? If so you need to say that! – user247327 Jan 5 at 20:50