Intuition for interior points and non-open sets I'm currently reading a book on real analysis. There is a statement right in the introductory section which I'm struggling to intuitively understand. It goes as follows:

Assume a sequence $(O_n)$ of open sets with $O_n := ]-1/n, 1/n[$. Their countably infinite intersection (infimum) $$M := \bigcap_{n=1}^\infty O_n =\{0\}$$ is not an open set because $0$ is not an interior point of $M$.

With respect to general topology, axiomatically it's clear.
Why is it the case from the point of view of the "basic notions" of real analysis, though?
I mean, $0$ is not an inner point of $M \subseteq \mathbb{R}$ if there does not exist a neighborhood of $0$ itself contained in $M$, i.e., for all $δ > 0$ we have $|x - 0| < δ$ and $x \notin M$ for some $x \in \mathbb{R}$. Which for $M = \{0\}$ is obviously true, since the only $x \in \mathbb{R}$ not fulfilling that negated implication is $0$, so, e.g., $x := δ/2$ would do.
This feels to me (subjectively) rather technical. By which I mean I'd need to think a little to come up with this, which may be due to only recently begun study. But still, is there a more conceptual or "intuitive", not necessarily purely geometric, interpretation of that result? The strong analogy to interval bisection is obvious. I think I'm interested in various perspectives, including a set-theoretic approach.
 A: Let's play a little game, "Who can choose the smaller positive real number." Because I'm generous, I'll let you choose first.... :)
Now, what does this droll prank have to do with openness of the intersection $\bigcap_{n} O_{n} = \{0\}$? In a sense, everything.
A set $U$ of real numbers is open if, for every $x$ in $U$, there exists a positive real number $\delta$ such that $(x - \delta, x + \delta) \subset U$.
Each set $O_{n}$ is open: If $x \in O_{n}$, then $|x| < \frac{1}{n}$ by definiton, so we may take $\delta = \frac{1}{n} - |x|$. In particular, if $x = 0$, we may choose $\delta = \frac{1}{n}$ (or any smaller positive number).
You can imagine an adversarial game, "Is $U$ open". The first, Player $x$ picks an $x$ in $U$. The second, Player $\delta$, tries to pick a real number $\delta > 0$ such that $(x - \delta, x + \delta) \subset U$. If Player $\delta$ succeeds, they "win the round"; otherwise Player $x$ wins. To say "$U$ is open" is to say "Player $\delta$ has a winning strategy against a perfect opponent."
Let's play the "Is $M$ open" game. Since $M = \{0\}$ is a single point, Player $x$ has no freedom of choice. Since $M$ is an intersection of open sets, Player $\delta$ can divide their strategy into an infinite family, one for each positive integer $n$. Great, $O_{n}$ is open; faced with the "challenge" $x = 0$, Player $\delta$ (successfully) chooses any real number $0 < \delta \leq \frac{1}{n}$. They win because they were playing the "Who can choose the smaller positive real number" game, and their opponent had to choose $\frac{1}{n}$ first.
But when faced with the infinite sequence of games implied by "Is $M = \bigcap_{n} O_{n}$ open", the tables are turned; Player $\delta$ has to pick one single number $\delta$ so that
$$
(-\delta, \delta) \subset M = \bigcap_{n=1}^{\infty} O_{n},
$$
i.e., a single $\delta > 0$ that "wins simultaneously" against every number $\frac{1}{n}$. But there is no such number: For every real number $\delta > 0$, there exists a positive integer $N$ such that $\frac{1}{N} < \delta$. Consequently, Player $\delta$ loses; $M$ is not open.
A: An arbitrary subset $A\subset{\mathbb R}$ is open if each point $x\in A$ has a (maybe small) neighborhood $U:=\ ]x-\delta, x+\delta[\ $ such that $U\subset A$. I think this is pretty intuitive: Standing at any point $x\in A$ you have some room to maneuvre without leaving $A$.
Now we come to your one-element set $M=\{0\}$. In order to check whether this set is open we have to test its single point $x=0$, and clearly this point fails the test. Therefore $M$ is not open.
It so happens that $M$ can be written as an intersection of infinitely many open sets. This shows that such intersections need not be open.
A: Well, informally speaking an interior point of a set is one which has other points of the same set around, however close you are to the interior point. If your set is just {0}, then such close-by points cannot exist.
