Coincidence of classical notion of covering in classical topology and covering sieve in grothendieck topology

Question

Let $X$ a topological space, $\mathcal{O}(X)$ the usual category of the open sets of $X$ and $(\mathcal{O}(X),J)$ any site.

Do we necessarly have for each open set U and for each sieve in $J(U)$, $(U_i \rightarrow U)_i$, the union $U \subseteq \bigcup U_i $ or can we have for a certain grothendieck topology the reverse in some cases (strict subset $\bigcup U_i \subsetneqq U) $ ? (and why of course if yes or no)

Thanks

(extremely basic question i know)

Answer 1

No, there is nothing forcing the $U_i$ to cover $U$ as sets. For instance, if you define $J(U)$ to consist of all sieves on $U$ for each $U$, this is a Grothendieck topology.

Answer 2

For any partial order viewed as a category, the sieves are just downward closed subsets. You can call any sieve you want a covering sieve. The only thing that happens is that if you call one sieve a covering sieve, others may need to be considered covering sieves as well to meet the closure conditions of a Grothendieck topology.

By definition, you always have $\bigcup_i U_i \subseteq U$ where $U_i\to U$, i.e. $U_i \subseteq U$, but you can certainly choose a sieve such that $\bigcup_i U_i \neq U$. That said, any Grothendieck topology (on a topological space) containing such a covering sieve would fail to be subcanonical.