Let $A_n:=\{\frac{1}{n}\}\times[0,1]$ and $X = \{(0,0),(0,1)\}\cup\bigcup_{n \in \mathbb{N}} A_n$. I'm being told that the connected component $C_{(0,0)}$ of $(0,0)$ is $\{(0,0)\}$, but there is something I'm not understading.
As far as I know, the connected component of a point, is, in particular, a connected space. So if I'm not mistaken, it should be at the same time, an open and a closed set. Clearly, $\{(0,0)\}$ is closed. But I think it's not open, so this would make $C_{(0,0)}$ larger than the point. What I'm thinking is, if $\{(0,0)\}$ is open, then exists an $\epsilon \gt 0$ such that $B((0,0), \epsilon) \subseteq \{(0,0)\}$. But I can find a point of $X$ for every $\epsilon$ in the following way.
There exists an $n_0 \in \mathbb{N}$ such that $\frac{1}{n_0} \lt \epsilon$. So the point $(\frac{1}{2n_0},\frac{1}{2n_0})$ is in $X \cap B((0,0), \epsilon)$, so $\{(0,0)\}$ is not open, so it's not a connected component.
Am I saying something wrong?
Thanks
My mistake was to think that a connected component it should be an open set in the larger set. For example, if I take $X = \mathbb{Q}$, each point is a connected component, but no point is and open set of $X$. But, the point is an open set of the set that only contains itself.
So, $\{(0,0)\}$ doesn't have to be an open set, and the thing is, that if I add any other point of $X$ is not connected.