# Atiyah-Macdonald Ex7.22: Equivalent condition for $E$ to be open in a Notherian space

Atiyah-Macdonald Ex7.22 Let $X$ be a Notherian topological space and let $E$ be a subset of $X$. Show that $E$ is open in $X$ if and only if, for each irreducible closed subset $X_0$ in $X$, either $E \cap X_0 = \emptyset$ or else $E \cap X_0$ contains a non-empty open subset of $X_0$.

$\Rightarrow$ direction is easy and I'm trying to solve $\Leftarrow$ direction by contrapositve. So assume that $E$ is not open in $X$. To use the Noetherian condition, I came up with the set of closed sets $X' \subseteq X$ such that $E\cap X'$ is not open in $X$. Then I showed that the minimal element $X_0$ of this set is irreducible, and $E\cap X_0 \neq \emptyset$. But there seems no way to deduce a contradiction if $E\cap X_0$ contains a non-empty open subset of $X_0$.

So I come up with the set of closed sets $X' \subseteq X$ such that $E\cap X'$ is not open in $X'$. Then the others can be proved, but now I can't show that $X_0$ is irreducible. If $X_0=C_1 \cup C_2$ with proper closed subsets of $X_0$ then $E\cap X_0 = (E \cap C_1) \cup (E \cap C_2)$. By the minimality of $X_0$, $E \cap C_i$ is open in $C_i$ so that $E \cap C_i=U_i \cap C_i$. But I think it needs not be true that $(U_1 \cap C_1) \cup (U_2 \cap C_2)$ is open in $X_0$, by sketching some sets in $\mathbb{R}^2$.

So how can I solve the problem? Or is there any way to correct my above trials?

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Suggestion: The Noetherian condition on $X$ means that all closed subsets of $X$ can be written as the finite union of irreducible closed subsets of $X$. – Thomas Andrews Jun 17 '11 at 5:19

If $F_1$ and $F_2$ are closed, $E\cap F_1$ is open in $F_1$ and $E\cap F_2$ is open in $F_2$, is $E\cap (F_1\cup F_2)$ open in $F_1\cup F_2$?
The answer is Yes. Since $F_i-E$ is closed in $F_i$, moreover $F_i-E$ is closed in $F_1\cup F_2$,for $i=1,2$ we get $F_1\cup F_2-E$ is closed in $F_1\cup F_2$, thus $E$ is open in $F_1\cup F_2$.
The following is a proof of this exercise: Let $\mathscr{F}=\{F\;\big|\;F \text{ is closed, but } F\cap E \text{ is not open in } F\,\}$. Assume $E$ is not open, then $X\cap E$ is not open in $X$.(Thus $\mathscr{F}$ is not empty) We can pick a minimal element $X_0$ of $\mathscr{F}$. Thus $X_0$ is closed and $X_0\cap E$ is not open in $X_0$. We have already shown that $X_0$ is irreducible. Thus $X_0\cap E$ contains a nonempty open set $U\cap X_0$ in $X_0$ (here $U$ is open in $X$). But notice that $U^{c}\cap X_0$ is closed and strictly contained in $X_0$, thus $E\cap U^{c}\cap X_0$ is open in $U^{c}\cap X_0$. There exists an open set $V$ such that $E\cap U^{c}\cap X_0=V\cap U^{c}\cap X_0$. Finally, we have $E\cap X_0=E\cap X_0\cap(U\cup U^c)=\cdots=(U\cup V)\cap X_0$ is open in $X_0$. This is impossible by our choice of $X_0$.