Is whether a set is closed or not a local property?

If I want to show a topological subspace is closed in an ambient space, does it suffice to know what happens on an open cover of the ambient space? More specifically,

Let $X$ is a topological space with a given open cover ${ U_i }$. Suppose that $Z \subset X$ is a set such that $Z \cap U_i$ is closed in $U_i$ for all $i$. Does it follow that $Z$ is closed in X?

This is clearly true if there are finitely many ${ U_i }$. At first thought, it seems unlikely to be true in the infinite case, but I'm having trouble coming up with a suitable counter-example.

Solution 1:

Yes, it's true: being closed is a local property local on the base. Indeed, suppose $x \notin Z$. Then $x$ is in a neighborhood $U_i$ which is one of the sets in the open cover. Now $x \in U_i - Z \cap U_i$, which is an open set, so there is a neighborhood of $x$, contained in $U_i$, which does not intersect $Z \cap U_i$. This means that this neighborhood of $x$ does not intersect $Z$.

This is not true for closed covers (i.e. with the $U_i$ closed) in general, but it is true when they form a locally finite cover.

Incidentally, if each point $x \in Z$ has a neighborhood $U$ such that $Z \cap U$ is closed in $U$, then $Z$ is called locally closed: this means it is the intersection of a closed subset and an open subset, but is not itself necessarily closed.

Solution 2:

$U_i \setminus (Z \cap U_i)$ is open in $U_i$, thus also open in $X$. Then $X \setminus Z = \cup_ {i \in I} U_i \setminus (Z \cap U_i)$ is open in $X$, i.e. $Z$ is closed in $X$.