Countable product of $\mathbb{R}$ is not connected with respect to box topology

From the book Topology, by James Munkres, I am trying to understand that $\mathbb{R}^\omega$ is not connected with respect to box topology. It is written that, We can write $\mathbb{R}^\omega$ as the union of the set $A$ consisting of all bounded sequences of real numbers, and the set $B$ of all unbounded sequences. These sets are disjoint, and each is open in the box topology. For if $a$ is a point of $\mathbb{R}^\omega,$ the open set $$U=(a_1-1,a_1+1)\times (a_2-1,a_2+1)\times...$$ consists entirely of bounded sequences if $a$ is bounded, and of unbounded sequences if $a$ is unbounded....

I do not realise that, how the open set $$U=(a_1-1,a_1+1)\times (a_2-1,a_2+1)\times...$$ can contain all the bounded sequences of $\mathbb{R}?$ If I choose $a=(1,\frac{1}{2},\frac{1}{3},...) \in \mathbb{R}^\omega,$ then $a$ is a bounded sequence of $\mathbb{R}$. Therefore, $$U=(1-1,1+1)\times (\frac{1}{2}-1,\frac{1}{2}+1)\times...,$$ which implies that, $$U=(0,2)\times (-\frac{1}{2},\frac{3}{2})\times...,$$ Now, if I consider the bounded sequence $b=(5,5,...)\in \mathbb{R}^\omega,$ then how $b$ can be contained in $U?$
Surely I am missing something, please help me to figure out the concept.
Thanks in advance.

The set $U$ only contains bounded sequences if $(a_n)$ is bounded, so is a subset of $A$, and that's all you need because it shows that any bounded sequence is an interior point of $A$. We don't need it to contain all bounded sequences: just an open set around $(a_n)$ inside $A$.

The same holds for $B$: if $(a_n)$ is unbounded, the defined $U$ also contains only unbounded sequences as well, so $U \subseteq B$ and the sequence is an interior point of $B$.

So $A$ and $B$ only have interior points so both are open, and as $A \cup B = \Bbb R^\omega$, both are closed too, and we have a disconnection.