# Intersection of open sets

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.

• The intersection is a singleton as you have correctly deduced and not empty. Another way to see this is to choose $A_n=(-1/n, 1/n)$ which also has intersection $\{0\}$ and more obviously covers $0$. Commented Jan 4, 2021 at 21:04
• If $A_n=(0,1/n)$ then $\cap_n A_n=\emptyset$ which is open. Note that $0\not\in A_n, \forall n.$
– mfl
Commented Jan 4, 2021 at 21:04
• @CyclotomicField, Out of curiosity, is the Cantor set $C$ open, and therefore $[0,1] - C$ closed? Commented Jan 4, 2021 at 21:12
• Quasar--- The Cantor set is closed... it is the complement in $[0,1]$ of a union of the removed "middle third" open intervals. That union is open since any union of opens is open. Commented Jan 4, 2021 at 21:25
• In order to answer (b) you might want to go for $O_n = (-1-1/n, 1 + 1/n)$. Commented Jan 4, 2021 at 21:27

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 $$k$$].