Counterexample of $\overline{\bigcup_i A_i }\subset \bigcup_i\overline{ A_i }$ with co-finite topology

Question

Let $A_i$, $i \in I$ subsets of a space X given topology $\tau$. Show that $\overline{\bigcup_i A_i }\subset \bigcup_i\overline{ A_i }$ does not hold necessarily :

My counterexample is in $\Bbb N$ with the co-finite topology and $A_k=\{2k\}=\overline{ A_k }$ because singletons are closed. Hence, $\bigcup_i\overline{ A_i } = 2\Bbb N$. However, $\overline{\bigcup_i A_i }$ is the smallest closed set containing $2\Bbb N$, since closed sets are finite or equal to $\Bbb N$, then $\overline{\bigcup_i A_i }=\Bbb N$ so $\overline{\bigcup_i A_i }\nsubseteq \bigcup_i\overline{ A_i }$.

But here, since $I$ is countable, it seems to disagree with the fact that $\overline{A \cup B}=\overline{A} \cup \overline{B}$. Is my counterexample wrong ? Should I find a case with $I$ uncountable ?

Answer 1

Your example is fine. The equality $\overline{A\cup B}=\overline A\cup\overline B$ holds for pairs of subsets of $X$ and, more generally, for finite sets of subsets of $X$. But you have provided an example with a countable set of subsets of $X$.

Another example would consist in taking any non-closed subset $A$ of a $T_1$ topological space, and then noticing that $\bigcup_{a\in A}\overline{\{a\}}=\bigcup_{a\in A}\{a\}=A$, but $\overline{\bigcup_{a\in A}\{a\}}=\overline A\supsetneq A$.

Answer 2

Your example is fine.

The equality $\overline{A\cup B}=\bar A\cup \bar B$ only tells you about binary unions (and, by induction, finite unions), not arbitrary ones. This is related to the fact that a finite union of closed sets is closed (but not an arbitrary one).

Answer 3

Consider, $(\Bbb{R}, \tau_{std}) $

Then, $\overline{\cup_{r\in \Bbb{Q}}\{r\}}=\overline{\Bbb{Q}}=\Bbb{R}$

And, $\cup_{r\in \Bbb{Q}}\ \overline{\{r\}}= \cup_{r\in \Bbb{Q}}\ \{r\} =\Bbb{Q}$