Let X is a irreducible noetherian topological space, and $U \subseteq X$ is a nonempty open subset. If $B \subseteq U$ is dense, then so is $B \subseteq X$. Is this true or false, and why?
Here noetherian means that the closed subsets satisfy the descending chain condition and X irreducible iff $X=Y_1 \cup Y_2$ implies that $X= Y_1$ or $X=Y_2$.
I am rather stuck on this question because I do not see how DCC and irreducibility relate to density (I have little topological intuition).
The definition of irreducibility is $X=Y_1 \cup Y_2$ with $Y_1,Y_2$ closed implies that $X=Y_1$ or $X=Y_2$. By taking the complements, this corresponds to say that two non-empty open subsets always intersect. Equivalently, that every non-empty open subset is dense.
In your case, $U$ is open and nonempty in $X$ so is dense. As $B$ is dense in $U$ and $U$ is dense in $X$, $B$ is dense in $X$.
The fact that $X$ is noetherian is not important.