Which are the connected components of $K=\{1/n\mid n\in\mathbb{N}\}\cup\{0\}$? Which are the connected components of the topological subspace $K=\left\{\left.\dfrac{1}{n}\ \right|\ n\in\mathbb{N}\right\}\cup\{0\}$ of $\mathbb{R}$?
I think they are every single point, but I can't find open set of $\mathbb{R}$ separating the point $0$ from the others, thanks.
 A: 1 -- Topology on $K$ is subspace topology.
This means $U \subset K$ is open if $U = V\cap K$ for some open set $V \subset \mathbb{R}$.
2 -- Definition of connected
A space $K$ is connected if and only if for $U,V \subset K$ satisfying $U\cap V = \emptyset$ and $U \cup V = K$, then $U = K$ and $V = \emptyset$ or vice versa. A subset is connected if it is connected in the subspace topology.
3 -- each set $\{1/n\}$ is open.
Observe $\{1/n\} = K \cap \{x \in \mathbb{R} : |x-1/n| < \frac{1}{n(n+1)}\}$.
4 -- The set $\{0\}$ is not open.
Any open set in $\mathbb{R} \ni 0$ contains an element of the form $1/n$. If $U$ is an open set in $\mathbb{R}$ that contains 0, for some $\epsilon$ then $(-\epsilon,\epsilon) \subset U$ then
 $$
1/n \in \{x \in \mathbb{R} : |x| < \epsilon\}\cap K \subset U \cap K,
$$
So $\{0\}$ is not open. An open set containing $0$ contains a set of the form $(-\epsilon,\epsilon) \cap K$.
5 -- The connected sets are single element sets of the form $\{1/n\}$ and $\{0\}$.
Clearly the only open subsets of $\{1/n\}$ are the $\emptyset$ and $\{1/n\}$. So these sets are connected. Also in the subspace topology on $\{0\}$ from $K$ $\{0\}$ is open and thus connected.
6 -- Other Sets are not connected
If $U$ doesn't contain $0$. We can disconnect it into a possibly infinite union of sets of the form $\{1/n_i\}$. Pick any two subsets and they will be open. So $U$ is no connected.
Suppose that $U$ is a set containing $0$. Then
$$ 
U = \{0\} \cup \bigcup_{i=1}^{\infty}\{1/n_i\}.
$$
This can't be a finite union by 4. So we can disconnect $U$ into 
$$ 
\bigcup_{i=1}^{N}\{1/n_i\} \text{ and } \{0\} \cup \bigcup_{i=N}^{\infty}\{1/n_i\}
$$
for any finite $N$.
A: You cannot find such an open set. What you need to check is that all points of the form $\frac{1}{n}$ are isolated points.
Any subset $C$ with more than 1 point and at least one isolated point $p$ is disconnected (use $\{p\}$ and its complement in $C$, both of which are open and closed in $C$).
A: We don't have to separate $0$ from the rest, instead, we have to show that if any subset of $K $ contains $0$ and another point, then that set is not connected. This shows that no point besides $0$ itself is in the same connected component at $0$.
