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Let $X$ be a topological space

Let $A$ be the set of all closed, irreducible subsets of $X$ equipped with a topology that contains all sets of the form $V(U)=\{a\in A| a\cap U\neq\emptyset, \text{where $U\in$ (the topology of $X$, $T_X$)}\}$. So the topology of $A$ is $\{V(U)|U\in T_X\}$

What then is a closed, irreducible subset of $A$?

Sets of sets confuse me. If anyone has any good way of thinking about them do please divulge!

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I would be tempted to look at the complements in $A$ of the $V(X\setminus a)$ where $a\in A$. What have you tried? Also, what are you really asking? Are you looking for a caracterisation of all irreducible closed subset of $A$ with this topology, or just some examples of irreducible ones? –  Olivier Bégassat May 8 '12 at 14:02
    
Also, I am not entirely convinced the topology of $A$ you describe consists solely of sets of the form $V(U)$ with $U$ open in $X$... Did you prove this? –  Olivier Bégassat May 8 '12 at 14:05
    
Perhaps it's worth looking up "sober space" and "soberification". $A$ is the soberification of $X$. –  Zhen Lin May 8 '12 at 17:44
    
@ZhenLin: Thank you, I have done as you suggested, but the Wikipedia definition of "sober space" seems to require the condition that there exist a "unique generic point" for every irreducible set. Perhaps it is true here and perhaps you have seen that it is true here! But unfortunately I can't see it, would you mind explaining? Thanks! –  Fancourt May 8 '12 at 18:03
    
@ZhenLin: I am starting to suspect some correlation in structure between $A$ and $X$, but I am not sure how strong that correlation is and what I can say from that. –  Fancourt May 8 '12 at 18:05

1 Answer 1

up vote 1 down vote accepted

I'm going to change the notation a bit to something with which I'm more comfortable. Let $\langle X,\tau\rangle$ be a topological space, and let $\mathscr{A}$ be the set of irreducible closed subsets of $X$. For $U\in\tau$ let $$\mathscr{V}(U)=\{A\in\mathscr{A}:A\cap U\ne\varnothing\}\;.$$

It's easy to check that $\mathscr{V}(U)\cap\mathscr{V}(W)=\mathscr{V}(U\cap W)$ for any $U,W\in\tau$ and that $$\mathscr{V}\left(\bigcup\mathscr{U}\right)=\bigcup\{\mathscr{V}(U):U\in\mathscr{U}\}$$ for any $\mathscr{U}\subseteq\tau$, so $\mathfrak{T}=\{\mathscr{V}(U):U\in\tau\}$ is a topology on $\mathscr{A}$.

Now let $\mathscr{F}\subseteq\mathscr{A}$, and suppose that $\mathscr{F}\cap\mathscr{V}(U)\ne\varnothing\ne\mathscr{F}\cap\mathscr{V}(W)$ for some $U,W\in\tau$. If $\mathscr{F}$ is irreducible, there must be some $A\in\mathscr{F}\cap\mathscr{V}(U)\cap\mathscr{V}(W)\ne\varnothing$; then $U\cap A\ne\varnothing\ne W\cap A$, so $A\cap U\cap W\ne\varnothing$, and in particular $U\cap W\ne\varnothing$ and $A\in\mathscr{F}\cap\mathscr{V}(U\cap W)$. Conversely, if $U,W\in\tau$ and $\mathscr{F}\cap\mathscr{V}(U\cap W)\ne\varnothing$, then clearly $\mathscr{F}\cap\mathscr{V}(U)\ne\varnothing\ne\mathscr{F}\cap\mathscr{V}(W)$. Thus, for irreducible $\mathscr{F}\subseteq\mathscr{A}$ and $\mathscr{V}(U),\mathscr{V}(W)\in\mathfrak{T}$ we have $$\mathscr{F}\cap\mathscr{V}(U)\ne\varnothing\ne\mathscr{F}\cap\mathscr{V}(W)\;\text{ iff }\;\mathscr{F}\cap\mathscr{V}(U\cap W)\ne\varnothing\;.$$

Now $\mathscr{F}\cap\mathscr{V}(U)\ne\varnothing\ne\mathscr{F}\cap\mathscr{V}(W)$ iff there are $A,B\in\mathscr{F}$ such that $A\cap U\ne\varnothing\ne B\cap W$, so for irreducible $\mathscr{F}$ we must have $$\exists A,B\in\mathscr{F}(A\cap U\ne\varnothing\ne B\cap W)\;\text{ iff }\;\exists A\in\mathscr{F}(A\cap U\cap W\ne\varnothing)\;.$$ This can be stated more simply: $$U\cap\bigcup\mathscr{F}\ne\varnothing\ne W\cap\bigcup\mathscr{F}\;\text{ iff }\;\exists A\in\mathscr{F}(A\cap U\cap W\ne\varnothing)\;.\tag{1}$$

Let $F=\bigcup\mathscr{F}$. Then $\exists A\in\mathscr{F}(A\cap U\cap W\ne\varnothing)$ iff $F\cap U\cap W\ne\varnothing$ (since every $A\in\mathscr{F}$ is irreducible in $X$), so $(1)$ can be simplified further to $$U\cap F\ne\varnothing\ne W\cap F\;\text{ iff }\;(U\cap W)\cap F\ne\varnothing\;,$$ which simply says that $F$ is irreducible in $X$.

In other words, $\mathscr{F}\subseteq\mathscr{A}$ is irreducible in $\mathscr{A}$ iff $\bigcup\mathscr{F}$ is irreducible in $X$.

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Wow, thank you very much! Your thought processes are so clear! –  Fancourt May 8 '12 at 21:44
    
@Fancourt: Thank you! I try; I don't always succeed. –  Brian M. Scott May 8 '12 at 21:50
    
Slightly vague question: How do you approach questions like this? Is there a good way of breaking up such problems into more manageable bits to tackle? The definition given was enough to confuse me, not to mention doing any problems about it. (It is alright if you decide not to answer this question, it is really quite vague...) –  Fancourt May 8 '12 at 21:55
    
@Fancourt: My first step here was to convince myself that we really did have a topology on $\mathscr{A}$; that gave me a little bit of a feel for what was going on. Then I asked myself what it would mean for $\mathscr{F}\subseteq\mathscr{A}$ to be irreducible and tried to use the various definitions progressively to translate that into something more immediately comprehensible. In this case my answer actually follows my train of thought fairly closely; it just omits a couple of blind alleys and cleans up the rough spots, because in this case my thinking went pretty smoothly. It doesn't always. –  Brian M. Scott May 8 '12 at 22:15
    
Thanks again, Brian!! –  Fancourt May 9 '12 at 7:47

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