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Topology by Munkres says,

The set $X= \{1,2\}\times \mathbb{Z}_+$ in dictionary order is another example of ordered set with smallest element. Denoting $1\times n$ by $a_n$ and $2\times n$ by $b_n$ we can represent $X$ by $$a_1, a_2,\dots ; b_1, b_2, \dots$$

The order topology on $X$ is not 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 the basis element about $b_1$ (by definition), and any basis element containing $b_1$ contains point of the $a_i$ sequence.

I did not understand why "any basis element containing $b_1$ contains point of the $a_i$ sequence".

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A basis element containing $b_1$ is an open interval $(c,d)$ where $c<b_1<d$. We thus must have $c=a_n$ for some $n$. In that case, we have $a_{n+1}>a_n=c$ and $a_{n+1}<b_1<d$ so $a_{n+1}$ is also an element of $(c,d)$. Thus the interval $(c,d)$ contains a point of the $a_i$ sequence, namely $a_{n+1}$.

(Actually, there is one more possibility, which is that $c$ might be $-\infty$. I'll let you figure out how to handle that case.)

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    $\begingroup$ @ThomasAndrews: It does need it. If you don't include $-\infty$ as a possible endpoint, for instance, there would be no basic sets containing $a_1$, so it's not actually a basis. $\endgroup$ – Eric Wofsey May 26 '17 at 14:55
  • $\begingroup$ Ah, of course, missed that. $\endgroup$ – Thomas Andrews May 26 '17 at 17:58

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