Irreducibility and Subspace Topology Is the following statement true? "Let $X$ be a topological space, $Y$ a subspace and $S$ a closed and irreducible subset of $X$. Then $Y \cap S$ is not necessarily irreducible in $Y$." Counterexamples from commutative algebra are particularly welcome.
 A: First note that there is really no reason to say "irreducible in $Y$," because irreducibility is an intrinsic property of topological spaces, and does not depend on any ambient space.
With this is in mind, your question is: "if I take an irreducible closed subspace of some space $X$, is the intersection of this space with any subspace of $X$ irreducible?" One trivial way this fails is if $Y\cap S=\emptyset$, i.e., if $X=\mathrm{Spec}(k)\coprod\mathrm{Spec}(k)$, and $S$ and $Y$ are the two copies of $\mathrm{Spec}(k)$. The empty set is not irreducible (by definition).
But even if $Y\cap S$ is non-empty, this can still fail. For example, if $Y$ has two irreducible components $Z_1$ and $Z_2$, so $Y=Z_1\cup Z_2$, then $Y\cap S=(S\cap Z_1)\cup(S\cap Z_2)$, both closed subsets of $S\cap Y$. So $Y\cap S$ will fail to be irreducible if these sets are proper and non-empty in $S\cap Y$.
For an example, let $X$ be the affine plane over $\mathbf{C}$, $Y$ the union of the two coordinate axes, i.e., $Y=\mathrm{Spec}(\mathbf{C}[X,Y]/(XY))$, and $S$ the parabola $\mathrm{Spec}(\mathbf{C}[X,Y]/(Y-X^2-1))$. The underlying topological space of the scheme-theoretic intersection is the intersection of the underlying topological spaces, and since $S$ meets both coordinate axes but $S\cap Y$ is not contained in either coordinate axis (because, e.g., there is a point on the intersection of the $X$-axis and $S$ which is not on the $Y$-axis, etc.), $S\cap Y$ is not irreducible.
If, however, $Y$ is open and $Y\cap S\neq\emptyset$, then $Y\cap S$ is a non-empty, open subset of the irreducible space $S$, hence irreducible.
EDIT: As Brian M. Scott's answer shows, my answer is overly complicated, as a particular case of this question is: "is every subspace of an irreducible space irreducible?" An example from algebraic geometry showing the answer is no would be the cross $Y\subseteq X$ above. 
A: Let $X=\{0,1,2\}$; $U\subseteq X$ is open iff $U=\varnothing$ or $0\in U$. $X$ is a closed, irreducible subset of itself, but $Y=\{1,2\}$ is a discrete space with more than one point and therefore not irreducible.
