Bounded sequences in the box topology

Question

Could anyone explain to me why the set of all bounded sequences of real numbers is open in the box topology? Can we say the same for the set of all unbounded sequences?

Answer 1

Let $U_n = \{(x_i)\mid |x_i| < n \text{ for all }i\in \Bbb N\}$, i.e. all sequences that never exceed $n$. This is an open set, because it's an element in the standard box topology basis. The set of all bounded sequences is the union of all these open $U_n$'s.

Similarily, an unbounded sequence is one that is not bounded, so the set of unbounded sequences is the complement of the set of bounded sequences. Therefore the set of unbounded sequences is closed in the box topology.

One would think we could do something similar with unbounded sequences, i.e. define the open $V_n = \{(x_i)\mid |x_i| > n \text{ for at least one } i\in \Bbb N\}$. This does not work because in this case we want the intersection of all the $V_n$, and an infinite intersection of open sets is not necessarily open. In fact, if you use $\geq$ in the definition of $V_n$ instead, they become closed (they're the complements of the $U_n$), and you get that the set of unbounded sequences is an intersection of closed sets, and therefore closed.

Answer 2

If a sequence $(x_n)$ is bounded, so is every sequence in its box open neighbourhood $U = \prod_n (x_n -1 ,x_n +1)$. So bounded sequences form an open subset.

If that sequence is unbounded, so is every sequence in $U$ as well, making the unbounded sequences open as well.

So the set of bounded sequences is both open and closed. This shows that the infinite box product of connected spaces need not be connected.