Is $\left\{ \frac{1}{n}: n \in \mathbb{N} \right\} \cup \left\{ 0\right\}$ closed set? Is it true that $\left\{  \frac{1}{n}: n \in \mathbb{N} \right\}  \cup \left\{ 0\right\}$ is closed set? I suppose that yes, but I have no idea how can I prove it.
 A: Hint : It would be a bit easy if you can realize :
$\big(\{ \frac{1}{n} : n\in \mathbb{N}\}\cup \{0\}\big)^C=\bigcup_{n\in \mathbb{N}}\big(\frac{1}{n+1},\frac{1}{n}\big)\text{with} (-\infty,0)\cup (1,\infty)$
A: I have any other idea:
We have $\mathbb{R}  \setminus \left\{ 0,1,  \frac{1}{2}, \frac{1}{3} ,  \frac{1}{4} , \frac{1}{5},...  \right\} = (- \infty, 0)  \cup (1,+ \infty )  \cup  \bigcup_{n=1}^{\infty} \left(  \frac{1}{n+1} , \frac{1}{n} \right)$  so of course it must be open.
A: Showing that the set is closed.
Say we have a sequence $x_n$ contained in the set $X = \left\{  \frac{1}{n}: n \in \mathbb{N} \right\}  \cup \left\{ 0\right\}$ that converges as a sequence in $\Bbb R$. Then there are two options:


*

*It converges toward $0$. In this case, it converges to a point in $X$.

*It converges to some real number $a \neq 0$. Then $x_n$ has to be constantly equal to $a$ from some $n$ on. We can see this because there is some $\epsilon'$ such that $|x_n - 0| \geq \epsilon'$ from some $n$ on. That means that there are only a finite number of points it can vary between, and they all have non-zero distance between them. Choose an $\epsilon$ smaller than the smallest of the distances, and we see that $x_n$ has to be eventually constant. Thus $a\in X$ and we are done.
So any sequence of points in $X$ that converges in $\Bbb R$, converges in $X$, and therefore $X$ is closed.
Showing that the complement is open.
Let $a \in \Bbb R \setminus X$. Then I claim that $\inf_{x\in X} |a - x| > 0$. First of all, let's define $\epsilon' = |a - 0|/3$. Then there are only finitely many points $x\in X$ so that $|x - 0| \geq \epsilon'$, so the value
$$
\epsilon'' = \min_{x\in X, |x| \geq \epsilon'}|a - x|
$$
is defined and greater than $0$ (since $a \not \in X$). Any $x$ not considered here has distance at least $3\epsilon'/2$ from $a$, by the triangle inequality and definition of $\epsilon'$.
Let $\epsilon = \min(\epsilon', \epsilon'')/2$. Then the $\epsilon$-ball around $a$ will contain no points of $X$. Since $a$ was arbitrary in the set $\Bbb R \setminus X$, this set must be open.
A: Hint: Put $\{ \frac{1}{n} \} \cup \{0\} = A $. Maybe if you can show that the closure $\overline{A} = A $, then $A$ must be closed.
