Is the interval $[-\frac{1}{n}, \frac{1}{n}]$ equal to $0$ as $n$ goes to $\infty$ Sorry if this is a dumb question, but does the interval $[-\frac{1}{n}, \frac{1}{n}]$ become $0$ as $n$ goes to $\infty$ or does it not quite get there...
In other words, does $[-\frac{1}{n}, \frac{1}{n}]$ as $n$ goes to $\infty$  leave a hole in the real line or does it become 0?
 A: To say it goes to $x$ you must have some kind of limit definition. So this questions does not make sense. If you take the intersection then indeed it is the singleton $\{0\}$.
A: By definition, $\left[-\tfrac{1}{n},\tfrac{1}{n}\right]=\left\{x\in\mathbb{R};\;-\tfrac{1}{n}\leq x\leq\tfrac{1}{n}\right\}$.
Consequence 1: $0\in \left[-\tfrac{1}{n},\tfrac{1}{n}\right]$ for all $n$.
Consequence 2: Given any $a\in \mathbb{R}$, $a\notin\left[-\tfrac{1}{n},\tfrac{1}{n}\right]$ for $n$ large enough (for example, for $n>1/|a|$).
Conclusion: The only number that belongs to all intervals is the number zero. In this sense, we can say that it become $0$.
A: This defines a sequence of sets
$$
A_n = [-1/n, 1/n]
$$
with
\begin{align}
\inf_{k \ge n} A_k &= \bigcap_{k=n}^\infty A_k= [0, 0] = \{ 0 \} \\
\sup_{k \ge n} A_k &= \bigcup_{k=n}^\infty A_k= [-1/n, 1/n] \\
\DeclareMathOperator*{\liminflim}{lim inf}
\DeclareMathOperator*{\limsuplim}{lim sup}
\liminflim_{n \to \infty} A_n &= \bigcup_{n \in \mathbb{N}} \inf_{k \ge n} A_k = 
\bigcup_{n \in \mathbb{N}} \{ 0 \} = \{ 0 \} \\
\limsuplim_{n \to \infty} A_n &= \bigcap_{n \in \mathbb{N}} \sup_{k \ge n} A_k 
= \bigcap_{n \in \mathbb{N}} [-1/n, 1/n] = \{ 0 \} \\
\end{align}
and finally
$$
\lim_{n\to\infty} A_n = \liminflim_{n \to \infty} A_n = \limsuplim_{n \to \infty} A_n = \{ 0 \}
$$
A: It leaves a hole in the real line.  Since $0$ is in each interval $[-\frac{1}{n}, \frac{1}{n}]$, it remains in the intersection of all of them.
