Finding the Intersection of Sequences of Open Intervals on the Real Line

We are asked to determine the nature of the intersection of sequences of open intervals given by $]-1/n, 1/n[$ for $n = 1, 2, 3, \dots$ on the real line $\mathbb{R}$ with the general metric.

What is the Intersection of the Given Open Intervals?

The sequence of intervals is $I_n = ]-1/n, 1/n[$. We are interested in the intersection of all these intervals as $n$ goes to infinity:

$$ \bigcap_{n=1}^\infty ]-1/n, 1/n[ $$

Let $x$ be an element in this intersection of intervals. This means $x \in ]-1/n, 1/n[$ for every positive integer $n$.

So, for all $n \ge 1$, we have $-1/n < x < 1/n$. This inequality can be written compactly as $|x| < 1/n$ for all $n \ge 1$.

Now, let's think about which real numbers $x$ satisfy this condition:

• Suppose $x \ne 0$. Then $|x|$ is a positive number. By the Archimedean property of real numbers, for any positive number like $|x|$, we can always find a natural number $N$ such that $1/N < |x|$. If we find such an $N$, then the condition $|x| < 1/n$ for all $n \ge 1$ is not met for $n=N$, because $|x| > 1/N$. This is a contradiction. Therefore, a non-zero $x$ cannot be in the intersection of sequences.
• Suppose $x = 0$. Then $|x| = |0| = 0$. For any positive integer $n$, we know that $0 < 1/n$. So, $|0| < 1/n$ for all $n \ge 1$. This condition is satisfied for $x=0$.

Thus, the only number that is present in every interval $]-1/n, 1/n[$ is $0$. The resulting intersection set is $\{0\}$.

$$ \bigcap_{n=1}^\infty ]-1/n, 1/n[ = \{0\} $$

Analyzing the Nature of the Intersection Set (Closed Set or Open Set)

We now need to determine if the set $\{0\}$ is open, closed, or neither on the real line $\mathbb{R}$ with the standard general metric.

Why the Resulting Set is a Closed Set

Let's recall the definitions of open and closed sets in $\mathbb{R}$:

• Open Set: A set $S$ is open if for every point $x \in S$, there exists an open interval $(a, b)$ such that $x \in (a, b) \subseteq S$. For the set $\{0\}$, the only point is $0$. If $\{0\}$ were open, there would have to be an open interval $(a, b)$ containing $0$ such that $(a, b) \subseteq \{0\}$. An open interval containing $0$ must include points other than $0$, no matter how small the interval is (e.g., $r/2$ for $r>0$ such that $]-r, r[ \subseteq (a,b)$). Since $(a,b)$ would contain points other than $0$, it cannot be a subset of $\{0\}$. Hence, $\{0\}$ is not open.
• Closed Set: A set $S$ is closed if its complement $\mathbb{R} \setminus S$ is open. The complement of $\{0\}$ in $\mathbb{R}$ is $\mathbb{R} \setminus \{0\} = ]-\infty, 0[ \cup ]0, \infty[$. Both $]-\infty, 0[$ and $]0, \infty[$ are open intervals. The union of any collection of open sets is always open. Therefore, $\mathbb{R} \setminus \{0\}$ is an open set. Since the complement of $\{0\}$ is open, the set $\{0\}$ itself is a closed set.

Alternatively, a set is closed if it contains all its limit points. The only possible limit point of the set $\{0\}$ is $0$. Since $0$ is an element of $\{0\}$, the set is closed.

Thus, the intersection of the sequences of open intervals is the set $\{0\}$, which is a closed set on the real line with the general metric.