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According to Brris & Sankappanavar's "A course in universal algebra," the set $L_C$ of closed subsets of a set $A$ forms a complete lattice under $\subseteq$.

Here, a subset $X$ of $A$ is said to be closed if $C(X) = X$, where $C$ is a closure operator on $A$ in the sense that it satisfies C1 - C3 below:

(For any $X, Y \subseteq A$)
C1: $X \subseteq C(X)$
C2: $C^2(X) = C(X)$
C3: $X \subseteq Y \Rightarrow C(X) \subseteq C(Y)$.

They say that the supremum of a subset $\{C(A_i): i \in I\}$ of the lattice $\langle L_C, \subseteq \rangle$ is $C(\bigcup _{i \in I} A_i)$. If so, it must be that $$C(\bigcup _{i \in I} A_i) \subseteq \bigcup _{i \in I} C (A_i)$$ (since $\bigcup _{i \in I} C (A_i)$ is also an upper bound). But, I cannot so far show how this is so.

It was an error to think that the above inclusion had to hold if $C(\bigcup _{i \in I} A_i)$ is $sup \{C(A_i): i \in I\}$. This inclusion does not follow, and its converse follows, actually, as pointed out by Brian and Abel. Still, $C(\bigcup _{i \in I} A_i)$ is the supremum of the set since, among the closed subsets of $A$, it is the set's smallest upper bound, as explained by Brian and Alexei.

This question was very poorly and misleadingly stated. I will delete it if it's requested.

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I thank all three people who responded to this poorly & confusingly stated question. Specifically, I thank all of them for pointing out my error of having the inclusion backwards; thank Brian and Abel for diagnosing exactly where my reasoning went wrong in holding the wrong inclusion; thank Alexei for explaining why then C(U Ai) is the smallest upper bound in L_C. All three replies (with the comments) helped me to figure out the whole thing, and my "accepting" Alexei's reply simply means it was the last piece for me to end this (self-created) puzzle. My "acceptance" goes to all of you. – Seiichiro Apr 17 '13 at 23:12
I should have thanked Brian as well for helping me to see why C(U Ai) is the smallest upper bound in L_C. I will be more careful to study each reply more carefully before responding to them... – Seiichiro Apr 18 '13 at 1:42
up vote 3 down vote accepted

HINT: The result given by B&S follows easily from the following useful fact:

Proposition. For any $X\subseteq A$, $C(X)=\bigcap\{Y\subseteq A:X\subseteq Y\text{ and }C(Y)=Y\}$.

Proof. Let $\mathscr{C}=\{Y\subseteq A:X\subseteq Y\text{ and }C(Y)=Y\}$, and let $Z=\bigcap\mathscr{C}$. If $Y\in\mathscr{C}$, then $Z\subseteq Y$ and hence $Z\subseteq C(Z)\subseteq C(Y)=Y$, so $Z\subseteq C(Z)\subseteq\bigcap\mathscr{C}=Z$, and $Z$ is therefore closed. And since $X\subseteq Y$ for each $Y\in\mathscr{C}$, it’s also true that $X\subseteq Z$, so $C(X)\subseteq C(Z)=Z$. On the other hand, $X\subseteq C(X)$ and $C\big(C(X)\big)=C(X)$, so $C(Z)\in\mathscr{C}$, and therefore $Z\subseteq C(X)$. Thus, $C(X)=Z$. $\dashv$

Added: I should probably note that your displayed inclusion is actually backwards: C3 ensures that $C(A_k)\subseteq C\left(\bigcup_{i\in I}A_i\right)$ for each $k\in I$, so $\bigcup_{i\in I}C(A_i)\subseteq C\left(\bigcup_{i\in I}A_i\right)$, but the reverse inclusion fails whenever $\bigcup_{i\in I}C(A_i)$ is not closed.

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I think you misread the inclusion. – Abel Apr 17 '13 at 10:10
@Abel: No, I did not: I ignored it in favor of providing a tool to get the actual result. However, I probably should say something about the error. – Brian M. Scott Apr 17 '13 at 10:11
Indeed, I'm sorry. It was me who misread what you are proposing to prove. +1 for helping to aquire the actual result. – Abel Apr 17 '13 at 10:19
@Abel: That’s okay; your comment led me to improve my answer, so I reckon that it was a good thing. – Brian M. Scott Apr 17 '13 at 10:21
Thank you very much Brian, for realizing my error and giving this hint for the actual result. I really appreciated this. – Seiichiro Apr 17 '13 at 23:40

$\bigcup C(A_i)$ is not necessarily closed, and the smallest closed set containing it is $C[\bigcup C(A_i)]$. Now, $\bigcup A_i \subset \bigcup C(A_i)$, thus $C(\bigcup A_i) \subset C[\bigcup C(A_i)]$.

Conversely, $A_i \subset \bigcup A_i$, so $C(A_i) \subset C(\bigcup A_i)$. Therefore, $\bigcup C(A_i) \subset C(\bigcup A_i)$, so $C[\bigcup C(A_i)] \subset C(\bigcup A_i)$.

Thus, $C[\bigcup C(A_i)] = C(\bigcup A_i)$, Q.E.D.

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Thank you very much, Alexei. I really appreciated your help, as well as Brian's and Abel's! – Seiichiro Apr 17 '13 at 23:38
Alexi: If you realize that I once "accepted" your reply and changed it to Brian's, I'm sorry for this awkward behavior. Due to my slowness, I could not see Brian's direct relevance to my problem at first. Still, I needed both your help and Brian's, to finally see how to get B&S's correct result. Thank you very much for your help. – Seiichiro Apr 18 '13 at 1:36

We can prove the following:

$A_i\subseteq \cup_{i\in I} A_i$, thus by the third property $C(A_i)\subseteq C\left(\cup_{i\in I} A_i\right)$ for all $i\in I$. Thus $$\cup_{i\in I}C(A_i)\subseteq C\left(\cup_{i\in I}A_i\right).$$

The formula in you question is about the converse inclusion and is false in general. Consider $I=\mathbb{N}$ and $A_i = [i^{-1},1-i^{-1}]\subseteq\mathbb{R}$. Clearly $C(A_i) = A_i$ and hence $\cup_{i\in I} C(A_i) = \cup_{i\in \mathbb{N}}[i^{-1},1-i^{-1}] = (0,1)$.

On the other hand $C(\cup_{i\in I} A_i) = C((0,1)) = [0,1]$.

The problem in your reasoning is that while $\cup_{i\in I} C(A_i)$ is an upper bound it is not in general an element of your lattice.

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Thank you very much Abel, for pointing out exactly where I went wrong in making up the wrong question! Slow as usual, I finally realized my error thanks to this reply! – Seiichiro Apr 17 '13 at 23:42
@Seiichiro Glad I could help. – Abel Apr 18 '13 at 0:17

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