Proof by counter example to false analysis statements. (i)Disprove that the isolated points of a set form a closed set.
(ii)Disprove
every open set contains at least 2 points.
(iii) Disprove $\partial ( S \cup D) = \partial S \cup \partial D$
$S= \mathbb {Q}$
$D = \mathbb{R} \backslash \mathbb {Q}$
$\partial A = \phi$ $\partial B = \mathbb {R}$
$\partial A \neq \partial B$
(iv)Disprove $\partial ( S \cap D) = \partial S \cap \partial D$
$ S= \phi $
$D= ( 0,1)$
$\partial A = \mathbb {R}$
$\partial B = \{ 0 \} \cup \{ 1\}$
$\partial A \neq \partial B$
(v) Disprove: the supremum of a bounded nonempty set is the greatest of its limit points.
let $S= [0,1] \cup \{ 7 \}$  7 is the supremum but its clearly not a limit point.
(vi) Disprove: If S is any set, then $\partial (\partial S)= \partial S$
Let $S= \mathbb {Q}$
then $\partial (\partial S)=\mathbb {Q}$
and $\partial S= \mathbb {R} \backslash \mathbb {Q}$
thus $\partial (\partial S) \neq \partial S$
(vii) if S is any set, then $\partial S_{c} = \partial S$ where $S_{c}$= closure of S
let $S=\{ (1,2) \cup (2,3)$
then $\partial S_{c}= \{ 1\} \cup \{ 3\}$ $ \neq \partial S = (\{ 1 \} \cup \{ 2 \} \cup \{ 3 \}+$
(viii) this one is a little weird in that i can't find in my textbook what this notation means???
Disprove: if $S_{1}$ and $S_{2}$ are arbitary sets then  $( S_{1} \cup S_{2})^{0} = S^{0}_{1} \cup S_{2}^{0}$
Please check my arguments and if possible give me hints to better answer or to approach these simple proofs. HW so hints only please.
 A: I assume that throughout this problem, only subsets of the set $\mathbb R$ of real numbers (with its usual topology) are concerned. In more general contexts, e.g. question (ii) has a less trivial answer.
(i) Consider $S:=\{\frac1n\mid n\in\mathbb N\}$. Note that all points are isoloated and that $\overline S$ is strictly larger than $S$.
(ii) Recall that $U$ is open if for every $x\in U$ there is bla bla. There exists one special subset of $\mathbb R$ such that "For every $x\in U$" is a void condition.
(iii) Your suggestion to take $S=\mathbb Q$ and $D=\mathbb R\setminus \mathbb Q$ is fine.
However, $\overline S=\overline D=\mathbb R$ and $S^o=R^o=\emptyset$, hence $\partial S=\partial D=\mathbb R$. As $S\cup D=\mathbb R$ and $\partial R=\emptyset$, this is a valid counterexample. However, $S=[0,1]$, $D=[1,2]$ would have worked just as well (maybe less complicated). Go ahead and calculate the boundaries for these $S, D, S\cup D$!
(iv) Here I guess you over-think the problem statement (and then miscalculate the boundaries: $\partial\emptyset=\emptyset$). Inspired by my alternative suggestion for (iii), have a look at $S=[0,2]$, $D=[1,3]$.
(v) Your example is fine (the supremum - in fact maximum - is an isolated point and therefore not a limit point)
(vi) Your idea to take $S=\mathbb Q$ is fine. However, you miscalculate the boundaries.
As seen above, $\partial \mathbb Q=\mathbb R$ and $\partial \mathbb R=\emptyset$ (why?)
(vii) Absolutely correct (modulo typoes). However, why don't you write simply $\{1,2,3\}$ instead of $\{1\}\cup\{2\}\cup\{3\}$?
(viii) The notation stands for the interior, that is the biggest contained open set (this is dual to the closure, which is the smallest containing closed set). We have $\partial S=\overline S\setminus S^o$ (how else did you define the boundary?). You may want to try $S_1=\mathbb Q$, $S_2=\mathbb R\setminus\mathbb Q$ again: Both sets have no interior points, but their union is all of $\mathbb R$ (which is open, hence its own interior). Alternatively - and maybe less confusing - let $S_1=[0,1]$, $S_2=[1,2]$. What do you get here for the interiors?
A: $\newcommand{\bdry}{\operatorname{bdry}}$(i) Suppose that $\langle x_n:n\in\Bbb N\rangle$ is a sequence of distinct real numbers, and $\lim_{n\to\infty}x_n=a$. Let $A=\{x_n:n\in\Bbb N\}$. What are the isolated points of $A$? Is $A$ closed?
(ii) There’s only one open set in $\Bbb R$ that contains fewer than two points; it’s not a neighborhood of any point.
(iii) What are $A$ and $B$? If $A=S\cup D=\Bbb R$, then it’s true that $\bdry A=\varnothing$, and it’s also true that $\bdry S=\bdry D=\Bbb R$, so that $\bdry S\cup\bdry D=\Bbb R\ne\varnothing=\bdry(S\cup D)$. However, there is nothing that could be represented by $B$, since $\bdry S\cup\bdry D$ isn’t of the form $\bdry B$.
(iv) Here again you’ve failed to tell us what $A$ and $B$ are. I suspect that $A$ is intended to be $S\cap D=\varnothing$, but $\bdry\varnothing=\varnothing$, not $\Bbb R$. And I suspect that you’re misusing ‘$\partial B$’ as an abbreviation for $\bdry S\cap\bdry D$. If so, it should be $\varnothing$, since $\bdry S=\bdry\varnothing=\varnothing$. Thus, this is not actually a counterexample. You can get a counterexample by taking $S$ and $D$ to be disjoint sets that share a boundary point. If $S=[0,1)$, for instance, the boundary points of $S$ are $0$ and $1$; can you find a set $D$ disjoint from $S$ that also has $0$ or $1$ as a boundary point?
(v) $S=[0,1]\cup\{7\}$ isn’t ‘sort of bounded’: it is bounded. And yes, $7=\sup S$, and $7$ is not a limit point of $S$, so this is a fine example.
(vi) No, $\bdry\Bbb Q=\Bbb R$: every real number is a limit point of both $\Bbb Q$ and $\Bbb R\setminus\Bbb Q$. But even though your reasoning is wrong, $S=\Bbb Q$ does work as a counterexample here, because $\bdry\Bbb R=\;$?
(vii) This counterexample is correct, though you have some extra symbols that shouldn’t be there. $S=(1,2)\cup(2,3)$, without the {, and $\bdry S=\{1\}\cup\{2\}\cup\{3\}$, without the ( and the +. Note that you can write both boundaries more simply, the first as $\{1,3\}$ and the second as $\{1,2,3\}$.
(viii) Check the table of contents for Interior: $S^\circ$ is a common notation for the interior of the set $S$. For a subset $S$ of $\Bbb R$, $S^\circ=\{x\in S:(x-\epsilon,x+\epsilon)\subseteq S\text{ for some }\epsilon>0\}$. More general, $S^\circ$ is the set of all points of $S$ that have an open neighborhood contained in $S$. If $S=[0,1]$, for instance, $S^\circ=(0,1)$; this is also the case if $S$ is $[0,1)$, $(0,1]$, or $(0,1)$. On the other hand, $\Bbb Q^\circ=\varnothing$: every non-empty open interval contains an irrational number, so no non-empty open interval is a subset of $\Bbb Q$. $\Bbb R^\circ$, on the other hand, is $\Bbb R$: every real number is the centre of an open interval contained in $\Bbb R$.
