I think that the title of is pretty self-explanatory. I was trying to figure out whether or not the following holds:
Given $X$ a Noetherian space that decomposes into $X=X_1\cup\ldots\cup X_N$ and $U$ open in $X$, is it true that $U$ is dense in $X$ iff $U$ is dense in each of the $X_i'$s?
Uh, I think I figured it out.
First, suppose $U$ open dense in $X$ such that $U\cap X_N=\emptyset$. We will show that this is not possible. Take $V=X\setminus (X_1\cup\ldots X_{N-1})$. This is open in $X$ and $V\subseteq X_N$. Since $U$ is dense we have $U\cap V\neq \emptyset$ thus, $X_N\cap U\neq \emptyset$.
A similar proof holds for $G$ open in $X_N$, i.e $G= G_0\cap X_N$, $G_0$ open in $X$. Just take $V= (X\setminus(X_1\cup\ldots\cup X_{N-1}))\cap G_0\subseteq G$ is open in $X$ thus, $U\cap V\neq \emptyset$, in particalar $U\cap G\neq \emptyset$.
Conversely, if $U$ open in $X$ such that $U\cap X_i$ open dense in $X_i$ and $V$ open in $X$, then there is $j$ such that $V\cap X_j$ is non-empty. So, $(V\cap X_j)\cap (U\cap X_j)\neq \emptyset$, because $U\cap X_j$ dense in $X_j$, and hence $U\cap V\neq \emptyset$. Since $V$ was arbitrary we conclude that $U$ is dense in $X$.