Let $X=\Bbb R^{\Bbb N}$ be equipped with the box topology. Let $A \subset X$ be the set of bounded sequences. Show that $A$ is clopen. 
Let $X=\Bbb R^{\Bbb N}$ be equipped with the box topology. Let $A \subset X$ be the set of bounded sequences. Show that $A$ is clopen.

To show that $A$ is open I think I don't need to consider $A^c$. If I pick $(x_n) \in A$, then $\forall n :\exists M$ for which $|x_n|\le M$.
I feel like I should now use the box topology to show that this set is open, but don't quite know how?
 A: If $(x_n)\in A$, then $\prod_{n\in\mathbb N}(x_n-1,x_n+1)$ is open in the box topology, contains $(x_n)$, and is included in $A$. (To show this product is included in $A$, notice that if $|x_n|\leq M$ for all $n$, then all elements of the intervals $(x_n-1,x_n+1)$ are bounded by $M+1$.)  Thus $A$ is open.
To show that $A$ is closed, i.e., that its complement is open, use the same product construction. If $(x_n)$ is unbounded, then so are all elements of $\prod_{n\in\mathbb N}(x_n-1,x_n+1)$.
A: HINT:
$A$ equals
$$\bigcup_{m=1}^{\infty} \prod_{n\in \mathbb{N}} (-m, m)$$
and so open. We can also use the closed intervals $[-m, m]$, but that would show $A$ a union of closed sets. However,  note that $\mathbb{R}^{\mathbb{N}}$ with the box topology is a topological group, and $A$ is an open subgroup.
A: As X is equipped with the box topology wich is generated by :${\displaystyle B = \left\{\prod_{i\in I} U_{i}\mid U_i{\text { open in }}X_{i}\right\}.}$
you have to prove tha each sequence is open in $\mathbb{R}$,
So A is open in $\mathbb{R}^n$
