Morning reading a book that deals with metric spaces noticed this fact: Tell that $$\overline {(a,b)}=[a,b],$$ $$\overline{\{\frac{1}{n}}\}=\{\frac{1}{n}\}\cup \{0\}.$$
I do not know much about metric spaces and so I started to read about them, because the discipline looks interesting, please help me, if someone solves Previously, the thank
It is clear that $(a,b) \subset \overline{(a,b)}$, so we only need to show that $a,b \in \overline{(a,b)}$. For every sufficiently small $\varepsilon>0$ (say $\varepsilon<b-a$) we have $$ (a-\varepsilon,a+\varepsilon)\cap(a,b)=(a, a+\varepsilon)\ne \varnothing, $$ i.e. $a \in \overline{(a,b)}$. Similarly we have $$ (b-\varepsilon,b+\varepsilon)\cap(a,b)=(b-\varepsilon,b)\ne \varnothing, $$ i.e. $b \in \overline{(a,b)}$.
In the same manner we have $$ A:=\left\{\frac{1}{n}:\ n \in \mathbb{N}\right\} \subset \overline{A}=\overline{\left\{\frac{1}{n}:\ n \in \mathbb{N}\right\}}, $$ and only need to show that $0 \in \overline{A}$.
Since $\lim_{n\to \infty}\frac{1}{n}=0$, for every $\varepsilon>0$ there exists some integer $N=N(\varepsilon) \ge 1$ such that $$ \left|\frac{1}{n}\right| < \varepsilon \quad \forall n \ge N, $$ i.e. $$ (-\varepsilon,\varepsilon)\cap A =\left\{\frac{1}{n}:\ n \ge N\right\} \ne \varnothing, $$ and we conclude that $0 \in \overline{A}$.
The closure of the set is the set with all its boundary points. Look up the definition of the boundary, it is introduced using the metric. In your case, the interval $(a, b)$ doesn't contain its boundary points, hence it is open, and when you take its closure, you add boundary points which are $a$ and $b$, and you get $[a, b]$.
As for the second part, by definition of the boundary point, $0$ is the only boundary point of your set (any neighbourhood of $0$ contains points from your set). So, the closure will contain your set and its only boundary point $0$.
The closure of a set $A$ is defined as the set that contains the elements of $A$ and all the limit points of $A$. A limit point can be defined multiple ways, the most common way being that: a point $c$ is a limit point of $A$ iff every neighborhood surrounding $c$ contains elements of $A$. Here we understand neighborhood to be the set of points within a certain distance around $c$. Of course since $A$ is a metric space, the notion of distance is already defined.
Take the Real interval $(a,b)$. $a$ is a limit point of the interval, because every neighborhood surrounding $a$ contains points inside $(a,b)$, and the same is true for $b$. Since $a$ and $b$ are limit points, the closure must contain those points. Thus the closure $\overline{(a,b)}$ is the interval $[a,b]$.
For the set $\left\{\frac{1}{n}:n\in\mathbb{N}\right\}$, the point $0$ is a limit point. To see why, imagine any neighborhood $(-\varepsilon,\varepsilon)$ surrounding $0$. There must be some point $\frac{1}{N}$ that exists inside this neighborhood (if you pick a sufficiently large $N$), so every neighborhood surrounding $0$ contains an element of the set $\left\{\frac{1}{n}:n\in\mathbb{N}\right\}$. Thus $0$ is a limit point, and the closure $\overline{\left\{\frac{1}{n}:n\in\mathbb{N}\right\}}$ must also contain $0$.
