# Show that Every Point in a Topological Space is Contained in a Connected Set.

For a topological space $$X, \mathscr T$$ define the connectedness of points in $$X$$ as $$x$$ is connected to $$y$$ (denoted as $$x \sim y$$) if there is a connected set that contains them both. Then according to Munkres Topology p.157 $$\sim$$ is an equivalence relation on the points in $$X$$ and symmetry and reflexivity are obvious (he then goes on to prove transitivity).

Well, symmetry is pretty clear to me but reflexivity ($$x \sim x$$) doesn't seem obvious at all. How to show that every point is contained in a connected set when $$X$$ itself may not be connected ? My attempt follows and I'd appreciate feedback whether it's correct, and whether there is a more obvious proof of this.

By definition, $$X$$ itself must be an element of $$\mathscr T$$, and can be regarded as a closed (or open) set. So there is at least one closed set that contains $$x$$.
Let $$C$$ be the intersection of all the (possibly infinite number of) closed sets that contain $$x$$, so $$C$$ is a closed set in the topology.
Claim that $$C$$ is connected.
Suppose not. Then there are disjoint open sets $$A, B$$ with $$C \subset A \cup B$$ and $$A \cap C, B \cap C \not = \emptyset$$
Then either $$x \in A \cap C$$ or $$x \in B \cap C$$ but not both (as they are disjoint because $$A, B$$ are disjoint). Take $$x \in A \cap C$$.
Then $$x \in \overline{A \cap C}$$.
And since $$\overline{A \cap C}$$ is closed then by definition of $$C$$ we have $$C \subset \overline{A \cap C} \subset \overline A$$
And $$A \cap B = \emptyset \implies \overline A \cap B = \emptyset$$ (see e.g. https://math.stackexchange.com/q/444679)
Then $$C \cap B \subset \overline A \cap B = \emptyset$$ which contradicts the conditions assumed for $$C$$ to be disconnected. Therefore $$C$$ is connected.

By definition of connectedness $$\{x\}$$ is always connected since you cannot express this as the union of two non-emtpy disjoint open sets. Hence $$x \sim x$$.