A disjoint union is not connected.

Question

Consider the set $\coprod_{i \in I} X_i$ with the disjoint union topology. Prove that this space is not connected, even if all the components are connected.

Attempt: Consider the disjoint union $(\mathbb{R} \times \{1\}) \cup (\mathbb{R} \times \{2\})$

Then, $\mathbb{R} \times \{1\}$ is open, and also closed because $$\mathbb{R} \times \{1\} = (\mathbb{R} \times \{1\}) \cup (\mathbb{R} \times \{2\}) \setminus(\mathbb{R} \times \{2\})$$

Answer 1

You've got the right idea, but this doesn't count as a proof. What you have proved is that the following statement is false:

A disjoint union of (connected) spaces, $\coprod_{i \in I} X_i$, is always connected.

However, what you have been asked to prove is that the following statement is true:

A disjoint union of (connected) spaces, $\coprod_{i \in I} X_i$, is always disconnected.

(Pedantic remark: it actually isn't true as it stands. But if you exclude the cases when $|I| = 0$ or $|I| = 1$, then it becomes true. I presume this was the intention of the author of the question.)

Answer 2

If we have at least $|I| \ge 2$ and all $X_i \neq \emptyset$ (or at least two of them are non-empty) then $X \times \{i\}$ is open and closed and non-empty (as $X_i$ is) and not the whole space (if $j \neq i$, then $X_j \times \{j\}$ is a non-empty set in its complement). A disjoint union of at least $2$ non-empty open sets is never connected.