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.


1 Answer 1


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

  • 2
    $\begingroup$ Thanks - surprisingly easy to miss the obvious ! $\endgroup$ Aug 14, 2019 at 11:16

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