Order topology on the set $X = \{ 1,2 \} \times \mathbb{Z}_{+}$ I am learning topology with the book "Topology" by James R. Munkres and have some difficulty in understanding the Example 4 of section 14 titled "The Order Topology" (page 85 of the second edition).

Example 4: The set $X = \{ 1,2 \} \times \mathbb{Z}_{+}$ in the dictionary order is another example of an ordered set with a smallest element. Denoting $1 \times n$ by $a_n$ and $2 \times n$ by $b_n$, we can represent $X$ by $$a_1, a_2, \ldots; b_1, b_2, \ldots.$$ The order topology on $X$ is not the discrete topology. Most one-point sets are open, but there is an exception --- the one-point set $\{ b_1 \}$. Any open set containing $b_1$ must contain a basis element about $b_1$ (by definition), and any basis element containing $b_1$ contains points of the $a_i$ sequence.

I am confused about the last statement: any basis element containing $b_1$ contains points of the $a_i$ sequence. What is the reason? And why is $\{ b_2 \}$ (or $\{ b_3 \}, \ldots$) open? What is key distinction between $b_1$ and $b_2$?   

They key distinction is that $b_1$ does not have an immediate predecessor. Given any interval $(x,y)$ containing $b_1$, $x$ is necessarily one of the $a_j$, and so $(x,y)$ contains $a_{j+1},a_{j+2},\ldots $. 
The set $\{b_1\}$ is not open, because any neighbourhood of $b_1$ contains points other than $b_1$ (namely some of the $a_j$). 
The set $\{b_2\}$, on the other hand, equals the interval $(b_1,b_3)$ and is thus open. More generally $\{b_j\}=(b_{j-1},b_{j+1})$, $\{a_j\}=(a_{j-1},a_{j+1})$ if $j\geq2$, and $\{a_1\}=(-\infty,a_2)$. So all other singletons are open. 
