Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 2 down vote accepted

OK. Your question is :

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 $X_0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.