# Hartshorne Chapter 2, Proposition 2.6 bijection between open sets in $X$ and open sets in $t(X)$

If $X$ is a topological space, let $t(X)$ be the set of nonempty irreducible closed subsets of $X$. We put a topology on $t(X)$ by saying that the closed sets are $t(Y)$, where $Y$ is closed in $X$. We have a continuous function $\alpha: X \rightarrow t(X)$ given by $\alpha(P) = \overline{\{P\}}$. It is clear that $\alpha$ is continuous, since if $E \subseteq t(X)$ is the collection of nonempty irreducible closed subsets of some closed set $Y \subseteq E$, then $$\alpha^{-1}E = \{ P \in X : \overline{\{P\}} \in E \} = \{ P \in X : \overline{\{P\}} \subseteq Y\} = Y$$ where the last equality follows because $Y$ is closed. Then Hartshorne says Note that $\alpha$ induces a bijection between open subsets of $X$ and open subsets of $t(X)$.

Why is this the case? It's just as good to give a bijection between closed sets. Is Hartshorne saying that $E \mapsto \alpha^{-1}E$ gives such a bijection? In that case, I am wondering why it is true that $Y_1 \subsetneq Y_2$ for closed sets $Y_1, Y_2$ implies that there is an irreducible closed subset of $Y_2$ which is not contained in $Y_1$.

• What does it mean for a closed set to be irreducible closed? – DanielWainfleet Dec 26 '15 at 23:12
• It means it's closed and irreducible? – D_S Dec 27 '15 at 1:33
• What does irreducible mean here? – DanielWainfleet Dec 27 '15 at 1:50
• Oh it means that the space cannot be written as the union of two proper closed sets. – D_S Dec 27 '15 at 2:20

Wait I'm dumb. If $y \in Y_2 \setminus Y_1$, then $\overline{\{y\}} \subseteq Y_2$ is such an irreducible closed set.