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Suppose that $(A,\le)$ is a complete lattice, that means $(A,\wedge,\vee)$ is a lattice which satisfies $$\forall B \subseteq A[\bigwedge B\text{ and }\bigvee B\text{ exist}].$$ And of course $(\wp(A),\subseteq)$, in which $\wp(A)$ is the powerset of $A$, is a complete lattice too (let $\bigcap \emptyset=A$). Furthermore, Let $(D,\sqsubseteq)$ be a directed set, and $P \colon D \to \wp(A)$ s.t. $\forall \alpha,\beta \in D[\alpha \le \beta \Rightarrow P_{\alpha} \supseteq P_{\beta}]$. Then if $\bigcap_{\alpha \in D}P_{\alpha} \ne \emptyset$, do

  • $\bigvee \bigcap_{\alpha \in D}P_{\alpha}=\bigwedge_{\alpha \in D}\bigvee P_{\alpha}$?

  • $\bigwedge \bigcap_{\alpha \in D}P_{\alpha}=\bigvee_{\alpha \in D}\bigwedge P_{\alpha}$?

That is, in discrete topology, are the limit superior and limit inferior of a directed net exactly the supremum and infimum of this net's limit set respectively?

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I think it could be puzzled out, but to help readers, could you please indicate which lattice has which operations? I presume you want $\vee$ and $\wedge$ for $A$ and $\cup$ and $\cap$ for $\wp(A)$, which I am guessing is the powerset of $A$. Please include this information in the statement of the question, thank you :) –  rschwieb Jun 5 '12 at 16:05
    
OK. No problem. –  Popopo Jun 6 '12 at 1:49
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I am just correcting the statement of Question 4, not answering it: the meet of the emptyset in P(A) is A (the top element of P(A)), not P(A) itself! –  user48292 Nov 5 '12 at 16:56
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2 Answers

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I don't think so. Let $A$ be the extended real line with the usual ordering. Let $D$ be the naturals. Let $P_n = (-1/n - 1,-1) \cup \{0\} \cup (1,1+1/n)$. Then $\cap P_n = 1$ and hence its sup and inf are both just 1. But $\inf_n \sup P_n = \inf_n 1 + 1/n = 1 \neq 0$, and $\sup_n \inf P_n = \sup_n -1 - 1/n = -1 \neq 0$.


Edit: The above example takes advantage of the fact that the directed set $(\mathbb{N},\leq)$ has no upper bound. If you require $D$ to be a dcpo instead, then $\cap_{\alpha \in D} P_\alpha = P_{\sup D}$, which using the condition that $$\alpha \leq \beta \implies \inf P_\alpha \leq \inf P_\beta, \sup P_\alpha \geq \sup P_\beta$$ you get the formulae you want.

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By the definition it seems $\cap P_n=[-1,1)$? So these two equations seem also hold in this case. –  Popopo Jun 7 '12 at 1:11
    
@Popopo: typo on my part. The domain should be symmetric about the origin. So the lower interval is $(-1 - 1/n,-1)$ instead of $(-1 -1/n,1)$ as I mistyped previously. –  Willie Wong Jun 7 '12 at 7:39
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For another counter example, let $S$ be any countably infinite set and take $A = S\cup \{0,1\}$ with the order that $0 \leq s \leq 1$ for any $s\in S$ and no other comparison possible. Enumerate $S$ by $s_n$ and let $P_{n} = A \setminus \left( \{0,s_1,s_2,\ldots,s_n\}\right)$. You have $\sup_n \inf P_n = \sup_n 0 = 0$ while $\inf \cap_n P_n = \inf 1 = 1$. Note that in this lattice every element is compact and hence the lattice is algebraic. (Together with completeness it is almost as nice as it can be...) –  Willie Wong Jun 7 '12 at 7:49
    
You see easily that if $A$ is a finite set your condition must hold. Do you have an example of $(A,\leq)$ where your equation holds for all $P_\alpha$ and the set $A$ is infinite? I've been trying to come up with one and not succeeding. –  Willie Wong Jun 7 '12 at 7:52
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@Willie_Wong:I found these two are exactly infallible in every finite case but at least one be fallible in every infinite case. But my proof is too long to be written in comment, so let me post it in a new reply. –  Popopo Jun 8 '12 at 9:10
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Let $\mathbf{A}$ be infinite, then obviously we can find some strictly increasing chains from $0$ to $1$. Let $C$ be one of the longest ones. Then switch weather $C$ is finite or infinite.

(1)Case finite: then it must has an atom, let $a$ denote it. Beside, there are infinite many elements b which satisfies $a\vee b>a$. Hence let us choose countable ones $\{b_1,b_2,...\}$. Let $P_n=\{a\} \cup \{b_j|j>n\}$. Then it is easy to see that every $\bigvee P_n>a$ and $\bigwedge \bigvee P_n>a=\bigvee \cap_{n<\omega}P_n$ since $(\bigvee P_n)_{n=0}^{<\omega}$ is a non-increasing sequence and has a minimum.

(2)Case infinite: then at least one of ACC and DCC is fallible, suppose ACC is fallible, then there is a infinite increasing chain $0<b_1<b_2<\dots$. Let $P_n=\{0\} \cup \{b_j|j>n\}$, similarly $\bigwedge_{n<\omega} \bigvee P_n=\bigvee P_0>0=\bigvee \cap_{n<\omega}P_n$. Therefore the first equation failed.

In a word, Meet and Join cannot be both continuous where $\mathbf{A}$ is infinite but surely be both continuous where $\mathbf{A}$ is finite.

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Looks good to me. Nice job. –  Willie Wong Jun 11 '12 at 13:58
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