I am working on the Osserman's book and encountered this exercise.
Problem 2.1.18.c
Show that if $Y \subseteq X$ is dense and $X$ is irreducible, then $Y$ is also irreducible.
Here is my attempt of a proof:
First, pick any $Z$ nonempty open in $Y$. If this is dense in $Y$, we are done (because if every nonempty open subset is dense, then the set is irreducible). Because $Y \subseteq X$, then this $Z$ is also open in $X$. Because $X$ is irreducible, then $Z$ is dense in $X$. This means that for any open $W$ in $X$, $Z \cap W \neq \emptyset$. But this includes all open sets in $Y$, hence we conclude that $Z$ is dense in $Y$. Hence, $Y$ is irreducible.
However, I have not used $Y \subseteq X$ is dense anywhere. I might have made a mistake, but I cannot find it now.
Can anyone give comments/corrections/hints on this proof? Thanks!
A topological space is irreducible iff every nonempty open subset is dense or equivalently, any two nonempty openset have nonempty intersection. Open sets of $Y$ are of the form $Y \cap U$ with $U$ open in $X$. It is nonempty iff it is nonempty so at least $U$ is nonempty. Intersection of two such is $Y \cap (U \cap V)$ which is nonempty as long as $U \cap V$ is nonempty because $Y$ is dense (being dense means it intersects every nonempty open set). Since $X$ is irreducible, it forces $U \cap V$ to be nonempty.