Find a counterexample of:
the union of the closure of A_i = the closure of the union of A_i
I hope its understandable
How is closure defined here?
This is my understanding.
Definition: Given a set $\displaystyle S\subseteq \mathbb{R}^2$ we define $\displaystyle \bar S$ the "closure of $\displaystyle S$" to be the set of all points such that $\displaystyle \forall s\in S \ \exists \delta>0 \ \ N(s,\delta)\cap S\not = \{ \}$.
Note: The notation $\displaystyle N(s, \delta)$ is the open disk cented at $\displaystyle s$ with radius $\displaystyle \delta$.
Consider the set $\displaystyle A_n = \left( {\frac{1}{{1 + n}},1} \right]$ the closure of which is $\displaystyle \overline {A_n } = \left[ {\frac{1}{{1 + n}},1} \right]$.
But consider this: $\displaystyle \overline {\bigcup\limits_{n = 1}^\infty {A_n } } = \left[ {0,1} \right]\quad \mbox{but}\quad \bigcup\limits_{n = 1}^\infty {\overline {A_n } } = \left( {0,1} \right]$.
Beautiful conterexample.
I was trying to come up with a conterexample in $\displaystyle \mathbb{C}$ using the definition I posed above (same idea) but I used finite number of sets. And I forgot the important rule of analysis (or topology here): what works for finite sets does not necessarily work for infinite sets.
So I am guessing if $\displaystyle S=\{A_i | i \in I\}$ was such that $\displaystyle |S|<\aleph_0$ then the above would actually be true. Correct? Maybe this is consequence of Heine-Borel theorem, no?
The nature of a standard metric topology makes it impossible to find such an example. However, I quite sure that one could construct a general (probability a finite) topological space where that property is true. I just not have thought about it in a very long time. Good Luck!