Are the Join and Meet operators on complete lattices both continuous?

Question

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?

Answer 1

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 $\sup_n \inf P_n = \sup_n -1 - 1/n = -1 \neq 0$.

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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.

Answer 2

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.