Disjoint union topology

Remarks

It is the final topology on $X$ for the family of maps $\iota_i$. Each $\iota_i$ is an open topological embedding, and the image of each $X_i$ is open and closed in $X$. Universal property: a map $f \colon X \to Y$ to a topological space $Y$ is continuous if and only if $f \circ \iota_i$ is continuous for every $i$.

Example: if each $X_i$ is a one-point space, then every subset of $X$ is open, so $X$ is the discrete space on $I$. Indeed, the preimage of any subset under each $\iota_i$ is empty or a point, and both are open.

Non-example: let $X_0$ and $X_1$ be two copies of a two-point space with the trivial topology, and let $X = X_0 \sqcup X_1$. The trivial topology on the four-point set $X$ is not the disjoint union topology. The set $\iota_0(X_0)$ has open preimages in both pieces, but it is not open in the trivial topology.

Notation: $\bigsqcup_i X_i$ or $\coprod_i X_i$; for two spaces, $X_1 \sqcup X_2$.