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

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.  
