# Deductions for the sets $A = \left\{(-1)^n + \frac{2}{n}:n=1,2,3,\ldots\right\}$ and $B = \{x \in \mathbf{Q}:0<x<1\}$

I would like to ask, if my proof and deductions for the below sets with respect to openness, closure checks out and is technically correct.

Exercise 3.2.2 from Understanding Analysis by Stephen Abbot.

Let \begin{align*} A = \left\{(-1)^n + \frac{2}{n}:n=1,2,3,\ldots\right\} \end{align*} and \begin{align*} B = \{x:\mathbf{Q}:0

Answer the following questions for each set. (a) What are the limit points? (b) Is the set open? Closed? (c) Does the set contain any isolated points? (d) Find the closure of the set.

Proof.

(a) Writing out a few elements of the set $$A$$, we have: \begin{align*} A = \left\{1,2,-\frac{1}{3},\frac{5}{4},-\frac{3}{5},\frac{4}{3},\ldots\right\} \end{align*}

The points $$-1$$ and $$1$$ are the limit points of $$A$$. There exists a subsequence $$a_{2n-1} = -1 + \frac{2}{2n-1}$$ in $$A$$ such that $$\lim_{n\to\infty} a_{2n-1} = -1$$ and $$a_{2n-1} \ne 1$$ for all $$n \in \mathbf{N}$$. Similarly, look at the subsequence $$(a_{2n})$$. $$\lim_{n\to\infty} a_{2n} = 1$$.

Every real number in $$[0,1]$$ is the limit of a sequence of rational numbers in $$B$$. So, the closed interval $$[0,1]$$ is the set of all limit points of $$B$$.

(b) The set $$A$$ is open. For all $$a \in A$$, there exists an $$\epsilon$$-neighbourhood of $$a$$, $$V_\epsilon(a)$$ that is contained in $$A$$, $$V_\epsilon(a) \subseteq A$$.

The limit points of $$A$$ are not elements of the set $$A$$. $$A$$ is not closed.

The set $$B$$ is open. $$B$$ has no largest or smallest element. For every rational number $$x \in B$$, there exists an $$\epsilon$$-neighbourhood of $$B$$, that intersects $$B$$ and contains points other than $$x$$.

The set $$B$$ is not closed, because $$B$$ does not contain its limit points.

(c) Pick any point $$a = (-1)^n + 2/n$$. Let $$\epsilon = 2(\frac{1}{n} - \frac{1}{n+1})$$. Then, $$V_\epsilon(a) \cap A = \{a\}\ \subseteq A$$. So, every point in $$A$$ is an isolated point.

The set $$B$$ does not contain any isolated points.

(d) \begin{align*} \bar{A} &= A \cup \{-1,1\}\\ \bar{B} &= [0,1] \end{align*}

Overall, if this were a test and I was grading your results, you'd get 6 out of 10 points.

For (a), you are correct.

For (b), not so much.

1. $$A$$ is not open. Take $$a=\frac54\in A$$. Then, clearly, $$a+\frac\epsilon2$$ is not a member of $$A$$, meaning that no neighborhood of $$a$$ can ever be fully included in $$A$$.
2. You claim that the limit points of $$A$$ are not elements of the set $$A$$. This is partially true. One of the limit points is an element of $$A$$. For the other, you should show that it is not (the proof is easy, but it should be present).
3. Similarly to 1, $$B$$ is also not open. For any rational number $$q$$ and any $$\epsilon > 0$$, there exists an irrational number $$x$$ such that $$|x-q|<\epsilon$$.

For (c), you are again sloppy. Not every point of $$A$$ is an isolated point. One of them is not.

For (d), you are correct.

• That's generous, I clearly overlooked the definition of an open set: $\forall a \in A$, $\exists V_\epsilon(a) \subseteq A$ (and not $V_\epsilon(a) \cap A \ne \emptyset$ !). For (b) 2. would it suffice to argue, $\lim (-1)^{2n-1} + 2/(2n-1) = -1$ and $a_n > -1$ for all $n \in \mathbf{N}$. That is, $a_n \ne -1$ for all $n$. So, $-1$ is limit point that does not belong to $A$. – Quasar Jan 5 at 7:44
• For (c), $1$ is a limit point of $A$, contained in $A$, so it is not an isolated point. – Quasar Jan 5 at 8:01
• @Quasar Yes, for (b) 2., what you wrote seems sufficient. And for (c), indeed, $1$ is not an isolated point. For all other points, your proof that they are isolated is a little short, but it is correct. – 5xum Jan 5 at 8:06