The set $A=\{0\} \cup \{\frac 1n \mid n \in \mathbb N\}$

Question

For the set $A=\{0\} \cup \{\frac 1n \mid n \in \mathbb N\}$, I understand that $\{\frac 1n \mid n \in \mathbb N\}$ is open and closed in $A$ because it is a union of all the connected components $\{\frac 1n\}$ in $A$ for all $n \in \mathbb N$. Even though $\{0\}$ is also a connected component of $A$, why is $\{0\}$ closed but not open? I thought $\{0\}$ is closed and open in $A$ as well just like each $\{\frac 1n\}$.

Answer 1

Viewing A as a subspace of R, since {0} is closed, within A, B = A - {0} is open. B is not closed within A because 0 is an adherance point of B that is not in B.

Using the clumbsy definition of closed, B is not closed
within A because 0 is a limit point of B that is not in B.

Answer 2

Let $X$ be an uncountable set with the cocountable topology. As all convergent sequences in $X$ are eventually constant and then have that constant as its limit, all subsets are sequentially closed. But only $X$ and at most countable subsets are closed.

Answer 3

If $\{0\}$ is open then there must exist some $\epsilon>0$ such that $$\{x\in A:|x|<\epsilon\}\subseteq\{0\}$$if such an $\epsilon$ exists we must have $$\{\dfrac{1}{n}:n>\dfrac{1}{\epsilon}\}\subseteq\{0\}$$which doesn't hold at all then the set is not open