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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 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!2012-11-05

2 Answers 2

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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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    @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.2012-06-08
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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. 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.2012-06-11