We say that a topological space $X$ is reducible if $X$ can be written as a union of two proper non-empty closed subsets $X_1$ and $X_2$ i.e.
$$X=X_1\cup X_2.$$
Is there an equivalent definition of an irreducible set $X$ where we take $X_1$ and $X_2$ to be open?
I think that I found one good example which suggests "No" as an answer. Let's say that we want to prove a proposition that every Noetherian space $X$ can be written as a finite union of irreducible component $X_i$ where $X_i$ are open for all $i$. Then we will struggle to show the existence since we use the fact that $X_i$ are closed for all $i$ in the original prove to construct a proper chain of closed sets which will stabilize. If $X_i$ are open, then we cannot do it anymore.
So, are there other reasons why $X_i$ should be closed?
Also, the original definition of $X$ being irreducible can be reformulated that any open proper subset $Y$ of $X$ is dense in $X$. So, if we are going to mirror the definition, then $Y$ cannot be dense in $X$ as $Y$ is closed i.e. $\overline{Y}=Y$.
Best Answer
Actually, the "dual" open set definition is equally useful. We can say that a topological space $X$ is irreducible if every pair of nonempty open sets $U_1$ and $U_2$ has $U_1\cap U_2\ne \varnothing$. This is equivalent to your definition because every closed set is the complement of an open set and conversely: so for $F_i^c=U_i$, $F_i$ closed, we have $$ F_1\cup F_2\ne X \iff U_1\cap U_2\ne \varnothing.$$ This is the same as saying that every open set is dense. In the context of algebraic geometry, this says that an algebraic variety with its Zariski topology has "big" open sets.