Show that the largest connected exists (with respect to the inclusion of sets) $C_{x} \subseteq X$ containing $x$. If $X \subseteq \mathbb{R^n}$ and $x \in X$, show that the largest connected exists (with respect to the inclusion of sets) $C_{x} \subseteq X$ containing $x$.
The connected set $C_{x}$ is called the connected component of $x$ with respect to $X$.
I already know that the largest connected subset $C_{x}$ of $X$ exists as just the union of all the connected subsets of $X$ that contain $x$, but now I don't know how to use that to prove the above. Can someone help me?
 A: It's trivial if you know how to construct $C_x$.
So you define $$C_x = \bigcup\{C\subseteq X\mid x \in C, C \text{ connected }\}$$
This is non-empty, as we at least have $C=\{x\}$ as one of the sets in the union, contains $x$ for that same reason and is connected as a union of connected sets that all intersect at $x$ (this is a standard theorem on connectedness).
Now if $C$ is any connected set that contains $x$, it's by definition one of the sets we are taking the union of, so trivially $C \subseteq C_x$, whch is exactly the maximality under inclusion that you were looking for. It's no more than that. The fact that we're in $\Bbb R^n$ has no special relevance; this works in any topological space.
Also note that as $C_x$ is connected and the closure of a connected set is connected (another standard theorem) $\overline{C_x}$ is in particular such a connected set containing $x$ and so by that maximality $\overline{C_x} \subseteq C_x$ which implies $C_x$ is closed. So components are closed in $X$.
