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Let $V$ be a topological space. Show that $V$ is Noetherian space if only if every open subset of $V$ is compact.

$(\Longrightarrow)$ If $W \subset V$ ($W$ open) and $C = \lbrace C_{\alpha} \rbrace$ a cover of $W$, let $F$ be a collection of all finite unions of elements of $C$. Since $V$ is Noetherian, $F$ has a maximal element $C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}}$. If $w \in W\setminus (C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}})$. $C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}} \subsetneq C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}}\cup C_{\beta}$ with $w \in C_{\beta}$, but $C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}}$ is a maximal. Therefore, $W \subset C_{\alpha_{1}}\cup...\cup C_{\alpha_{n}}$ so, $W$ is compact. (Is there an error? Or suggestion?)

$(\Longleftarrow)$ I couldn't prove. Any hint?

PatrickR
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Lucas
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2 Answers2

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If $X$ is Noetherian, let $O\subseteq X$ be open. Let $\mathcal{U}$ be an open cover of $O$. Note that its members (being open in $O$) are also open in $X$. If $\mathcal{U}$ were an infinite cover, pick $U_1, U_2, \ldots\in \mathcal{U}$ all distinct and non-empty, and then define $V_n = \cup_i^n U_i$ for all $n$ and note that this then would be an infinite chain of open subsets of $X$. This cannot exist, so $\mathcal{U}$ is a finite cover already, and so has itself as a finite subcover. So $O$ is compact.

For the reverse direction: if $U_1 \subseteq U_2 \subseteq U_3 \ldots$ is a chain of open subsets of $X$, then define $O = \cup_{i=1}^\infty U_i$ is open and $\{U_i: i=1,2, \ldots\}$ is by definition an open cover of $O$ which by assumption has a finite subcover $\{U_{i_1}, U_{1_2}, \ldots , U_{i_N}\}$. But then $O = U_j$ where $j = \max(i_1, i_2, \dots, i_N)$ by the $U_i$ being a chain, and so the chain stops at $i_N$.

Henno Brandsma
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  • Nice proof! Was my idea for the first part wrong? – Lucas Apr 20 '18 at 04:58
  • @LucasCorrêa your proof was also correct when you use the alternative definition that a family of open sets has a maximal element. This follows by Zorn from the ascending chain definition that I use. – Henno Brandsma Apr 20 '18 at 05:01
  • Yeah. I used it because I proved in a previous question. Thanks for the help! – Lucas Apr 20 '18 at 05:04
  • $\cup_{a \in \mathbf{C}}D(x - a)$ is an infinite cover of distinct non-empty opens of the complex affine line, so it is not true that any cover of an open in a noetherian topological space is already finite. – Sam Cassidy Jan 20 '20 at 11:44
  • @SamCassidy then this is not a Noetherian space. From these sets you can build an infinite chain. – Henno Brandsma Jan 20 '20 at 22:18
  • @HennoBrandsma It is a Noetherian space. The covering I gave is an infinite cover of distinct non-empty opens, but any ascending chain constructed from this cover will stabilize, since any two distinct opens in the cover form a subcover. Noetherianness means that infinite chains eventually stabilize, not that they don't exist. – Sam Cassidy Jan 22 '20 at 08:36
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Suppose one has an ascending chain $(U_n)$ of open subsets. The union $V$ of the $U_n$ is open. As $V$ is compact, and the $U_n$ form an open covering, then finitely many of the $U_n$ cover $V$. But $(U_n)$ is a chain, so $V=U_n$ for some $n$.

Angina Seng
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