How to show that $C(\bigcup _{i \in I} A_i)$ is a supremum of a subset $\{C(A_i): i \in I \}$ of the lattice $L_C$ of closed subsets?

Question

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.

---

Postscript
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.

Answer 1

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.

Answer 2

$\bigcup C(A_i)$ 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.

Answer 3

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.