Consider $X=(1,2018]\cup\{\frac{1}{n}, n \in \mathbb{N}\}$. Let $ (X,d)$ be the usual metric space. Which of the following option are true?
(1) $ (1,2018] $ is an open set.
(2) $(1,2018]$ is not an open set
(3) $\{\frac{1}{n}, n \in \mathbb{N}\}$ is an open set.
(4) $0$ is not a limit point of the. $\{\frac{1}{n}, n \in \mathbb{N}\}$
My efforts.
(1) True. Take intersection of X with (1,2019)
(2) False
(3) False as if I take a open ball of radius $0.01$ around say $0.5$ it is not properly contained in the set.
(4) $0$ is a limit point of X. It juts shows that space is not complete. So this is also False.
(1) in fact $(1,2018]$ is a clopen (closed and open) set because $$\forall x\in (1,2018],\exists\epsilon>0,\{y\in X|d(x,y)<\epsilon\}\subseteq (1,2018]$$(2) answered as before
(3) it is not neither open nor closed. No neighborhood of $1$ is included by $\{\dfrac{1}{n}|n\in\Bbb N\}$ (while for any other point the exists some neighborhood contained by the set). Also take the sequence $a_n=\dfrac{1}{n}$ which is convergent to $0$ but $0\notin \{\dfrac{1}{n}|n\in\Bbb N\}$.
(4) $0$ is not a limit point; adding it to the set makes it closed, however it doesn't belong to $X$. You can observe this fact graphically over real line.