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<x<1\}
\end{align*}
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*}
Best Answer
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.
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.