Given a chain of topological spaces, $X_1 \subset X_2 \subset \cdots$, where each $X_i$ is a closed subspace of $X_{i+1}$, we define the coherent/weak topology on $X = \bigcup_n X_n$ by declaring a set $U \in X$ open when $U \cap X_n$ is open in $X_n$ for all $n$. Why do we require each $X_i$ to be closed in $X_{i+1}$ if the topology is still well-defined without that assumption?
Definition of the Coherent/Weak Topology on a Union
general-topology
Best Answer
It's all a special case of a so-called final topology induced by the inclusions of subspaces. So it's well-defined also if $X_i$ is not closed in $X_{i+1}$.
But if you're worried, go back to definitions: it's clear that $\emptyset$ and $X$ are open as $\emptyset$ and $X_n$ are both open in $X_n$ for all $n$.
If $U,V$ are open (under this definition), then noting that $U \cap V \cap X_n = (U \cap X_n) \cap (V \cap X_n)$ for all $n$ should convince you that $U \cap V$ is open too, under this definition.
And if all $O_i, i \in I$ are all open, then note too that for $O=\bigcup_{i \in I} O_i$ we have
$$O \cap X_n = \bigcup_{i \in I} (O_i \cap X_n)$$ and the sets in that union are all open in $X_n$ by definition, so their union is too, and as $n$ is arbitrary, $O$ is open in $X$ too.
You see that nothing is used of $X_n$ except that it has a topology defined on it. But maybe the applications are mainly for closed subspaces (CW complexes being a prime example) and we might want the $X_n$ to be closed in $X$ too.