Let for \(n \in \mathbb{N}\), \(I_n = \left(0, \frac{1}{n}\right)\), \(J_n = \left[0, \frac{1}{n}\right]\), \(K_n = (n, \infty)\) and \(L_n = [n, \infty)\). Which of the following sets is non-empty ?
\(\bigcap_{n=1}^{\infty} J_n\)
\(\bigcap I_n\): if \(x\) belonged to every \(I_n\), then \(x \gt 0\); by the Archimedean property choose \(n\) with \(\frac{1}{n} \lt x\), so \(x \notin I_n\). Hence \(\bigcap I_n = \emptyset\).
\(\bigcap J_n\): the point \(0\) lies in every \(\left[0, \frac{1}{n}\right]\), so \(0 \in \bigcap J_n\). Any \(x \gt 0\) is excluded as above, so \(\bigcap J_n = \{0\}\), which is non-empty (nested closed bounded intervals have non-empty intersection).
\(\bigcap K_n\) and \(\bigcap L_n\): for any real \(x\), choose \(n \gt x\); then \(x \notin (n, \infty)\) and \(x \notin [n, \infty)\). So both intersections are empty (closed but unbounded sets need not have a common point).
Hence, the answer is \(\bigcap_{n=1}^{\infty} J_n\).
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