# How to tell $\overline {(a,b)}=[a,b]$, $\overline{\{\frac{1}{n}:n=1,2,3,\ldots}\}=\{\frac{1}{n}:n=1,2,3,\ldots\}\cup \{0\}$

Morning reading a book that deals with metric spaces noticed this fact: Tell that $$\overline {(a,b)}=[a,b],$$ $$\overline{\{\frac{1}{n}}\}=\{\frac{1}{n}\}\cup \{0\}.$$

• It is the closure of a set. What do you want to know? Commented Nov 2, 2013 at 18:36
• if it comes to closure the set, I want to know sir solve, if possible Commented Nov 2, 2013 at 18:37
• I just tell you, if you started to read this today, you have to learn a lot of things to solve this. Learn about limit points, interior points, and so on. This is the starting point. Commented Nov 2, 2013 at 18:39
• yes definitely, I'm reading, but it seemed interesting, and I am convinced that there are many people who can help me Commented Nov 2, 2013 at 18:41
• The closure of a set is defined, at least one way, as the intersection of all closed sets containing the set, so the closure of a set is the smallest closed set that still contains the set. The closure of a set is also the set and then add in its limit points.
– user43138
Commented Nov 2, 2013 at 18:42

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 $0 \in \overline{A}$.

• it is very clear answer for me thanks sir Commented Nov 2, 2013 at 18:58

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 $0$.

• @MadritZhaku You are very welcome. If you like my answer, don't forget to upvote. Thanks!! Commented Nov 2, 2013 at 22:30