General topology on two specific set $X=\left\{1-\frac{1}{n^2+1}\,: n\in\Bbb N\right\}$ and $Y=\,]1,+\infty[\,\cap \,\mathbb Z$ Question to help an universitary student. I have two sets:
$$X=\left\{1-\frac{1}{n^2+1}\,: n\in\Bbb N\right\}$$
$$Y=\,]1,+\infty[\,\cap \,\mathbb Z$$
between five options I must to find the true answer.
$\fbox{A}$: $DX\cap DY\neq \emptyset$;
$\fbox{B}$: $\overset{\circ}{X}\cup \overset{\circ}{Y}\neq 0$;
$\fbox{C}$: $X\cup\{1\}$ and $Y\cup\{1\}$ are separate and contiguous (or adjacent);
$\fbox{D}$: $\text{Fr}X\equiv \partial X=X$;
$\fbox{E}$: None of the other answers.
This is the original image in Italian language (and excuse me very much for bad traslation):


My solution:
$$X=\left\{0,\frac12,\frac45,\frac9{10},\ldots\right\}$$
where $1$ it is an accumulation point because $\lim_n 1-\frac{1}{n^2+1}=1$. Hence $DX=\{1\}$ and being $Y=\{2,3,4,5,6,7, \ldots\}$ (isolated points), $DY=\emptyset$. The A is false because $DX\cap DY= \emptyset$. An internal point in a set is a point for which there is at least one circle of radius $\delta>0$ entirely contained in the set; for my humble opinion this condition not exists in $X$ and $Y$.
Hence $\overset{\circ}{X}=\emptyset=\overset{\circ}{Y}$ and the B is false. $\text{Fr}X=\{0,1\}\neq X$ and the D is false.
Now,
$$X'=X\cup\{1\}=\left\{0,\frac12,\frac45,\frac9{10},\ldots,1\right\}$$
and
$$Y'=Y\cup\{1\}=\left\{1,2,3,4,5,\ldots\right\}$$
It is true that $\forall x'\in X'$ and $\forall y'\in Y'\to x'\leq y'$ (definition of separated sets).
While if $X'$ and $Y'$ are contiguous (or adjacent),
$$\sup X'=\inf Y' \vee \sup Y'=\inf X' \iff \forall \epsilon>0, |x'-y'|<\epsilon.$$
For me it is true the C because $\sup X'=\inf Y'=1$, but I not remember how to prove with the condition $\forall \epsilon>0, |x'-y'|<\epsilon$.
Please, I ask you if my reasoning is correct or there are errors. Thank you to all users.
 A: The correct answer depends on what exactly is meant by contiguous and separated.
Some things are clear: $\overline{X}=X\cup \{1\}$ and $Y$ is closed, and both only have isolated points and no interior points.
A. is false as $DX = X \cup \{1\}$ and $DY= \emptyset$ so $DX \cap DY=\emptyset$ trivially.
B. is false as $\overset{\circ}{X} = \emptyset = \overset{\circ}{Y}$, as said.
D. is false as $FX = X \cup \{1\} \neq X$.
C.  depends on the precise meaning of "separati" and "contigui". From the question text and comments we see that the former means that one set lies the left of the other, so any element of the "leftmost" set is $\le$ than any element of the rightmost set, and this is surely the case for $X \cup \{1\}$ and $Y \cup \{1\}$. The "contigui"  sets are those that the infimum of one set equals the sup of the other, and this is also formulated as $$\forall \varepsilon >0: \exists a \in A: \exists b \in B: |a - b| < \varepsilon\tag{1}$$
which I also found on this Italian page, where they treat the exist same notions.
There, though, they state (contrary the OP's statement) that "separati" means
$$\forall a \in A: \forall b \in B: a < b\tag{2}$$
which would be violated by $1$. So in definitions on the referred page, $X$ and $Y \cup \{1\}$ are "separati" and "contigui" but $X \cup \{1\}$ and $Y \cup \{1\}$ are not.
As $\sup(X \cup \{1\})=1 = \inf(Y \cup \{1\})$  as well, the sets are then indeed "contigui". So C holds and thus E does not, or the other way around depending on whether "separati" is strict $<$) or not ($\le$). So make your pick...
