I am self-learning Real Analysis from Understanding Analysis by Stephen Abbott.
In the introduction to the topology of $\mathbf{R}$, the author proves that countably finite unions and intersections of open sets results in an open set.
Exercise 3.2.1 asks for the following rather interesting questions.
After giving these answers a thought, I would like to ask for some help; verify if my intuition is correct. I could be entirely wrong, in which case, please share some hints. (Please feel free to add any asides/interesting remarks.)
(a) Where in the proof of the theorem part (ii) does the assumption that the collection of open sets be finite get used?
(b) Give an example of a countable collection of open sets $\{O_1,O_2,O_3,\ldots\}$ whose intersection $\cap_{n=1}^{\infty}O_n$ is closed, not empty and not all of $\mathbf{R}$.
(Story) Proof.
(a) In the proof of the theorem, "The intersection of an arbitrary (finite) collection of open sets is open", we argue that if $x \in \cap_{k=1}^{N}O_k$, then for all $1 \le k \le N$, because $O_k$ is open, there is an $\epsilon_k$ neighbourhood of $x$, $V_{\epsilon_k}(x) \subseteq O_k$. If we choose $\epsilon = \min \{\epsilon_1,\epsilon_2,\ldots,\epsilon_N\}$, then $V_\epsilon(x) \subseteq V_{\epsilon_k}$.
But, instead, if we had a countably infinite sequence of sets, such as $A_n = (0,\frac{1}{n})$, then $\cap_{n=1}^{\infty}A_n = \{0\}$. Every $\epsilon$-neighborhood around this point will not by contained in $\cap_{n=1}^{\infty}A_n$, so the countably infinite intersection is not open.
I know that, $\lim_{n \to\infty}\frac{1}{n} = 0$. We can make this set as small as we like. I am not very confident, about $\cap_{n=1}^{\infty}A_n$, if its a isolated point or $\emptyset$.
(b) I am tempted to think that the Cantor set $C$ is an open set. Basically, its Cantor dust. So, I wonder if $\left[\left(\frac{1}{3},\frac{2}{3}\right) \cup \left(\frac{1}{9},\frac{2}{9}\right) \cup \left(\frac{7}{9},\frac{8}{9}\right) \ldots \right] $ formed by the intersection of the open middle one-thirds of the sets $C_1,C_2,\ldots$, that is $[0,1] – C$ is a closed set.
Best Answer
If there were say countably infinite number of open sets, so that the sequence $(\epsilon_1,\epsilon_2,\cdots)$ went on forever with no end, then the minimum of the $\epsilon_k$ might be $0$ [for example if $\epsilon_k=1/k$ for each $k$].