Is there a sequence of unbounded closed intervals such that its intersection is empty? Is there a sequence of unbounded closed intervals $J_1\supseteq J_2 \supseteq J_3 \supseteq \cdots$ such that $\bigcap _{n=1} ^{\infty} J_n = \emptyset$. By an unbounded closed interval I mean an interval of the form $[a,\infty) = \{x \in \mathbb{R} \mid x \ge a\}$.
My answer to the question is yes taking the sequence $[1,\infty),[2,\infty),[3,  \infty), ...$ because suppose $b \in  \bigcap _{n=1} ^{\infty} J_n$. Then there exists an interval in the sequence such the $b\lt a_i$ whereby $b \notin [a_i, \infty) $ hence $b $ cannot be in $\bigcap _{n=1} ^{\infty} J_n$. Therefore the intersection is empty.
Is my answer correct and if so is my reasoning correct? If not can you please provide the correct justification for the correct answer. Thanks in advance!
 A: Yup, that's right (although see my comment in response to Rob Arthan's answer - it could be written more clearly). In particular, note that even though the notation "$[1,\infty)$" (for example) may suggest that the set is not closed, it is in fact closed.
A: Your answer is correct but your proof isn't phrased quite right: if $b \in \bigcap_{n=1}^{\infty}J_n$, then $b \in [n, \infty)$ for every natural number $n$, which is impossible, because if $b$ is any real number there is a natural number $n$ such that $b < n$, so that $b \not\in [n, \infty)$. (The way you have phrased it, it's not very clear what $a_i$ is.)
A: Yes, that is essentially a "proof by contradiction".  You start by assuming the negation of what you want to prove ("$b\in \cap_{n=1}^\infty J_n$") and arrive at a contradiction ("b is not in $\cap_{n=1}^\infty J_n$").
A "direct proof" would be "If $p\in R$ then, by the Archimedian property, there exist an integer, n, n> p.  Then p is not $J_n$ so p in not in $\cap_{n=1}^\infty J_n$".
