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Consider the set $X = \{1,2\} \times \mathbb{Z}_+$ in the dictionary order. Then this will be an ordered set with smallest element.

Denote $a_n = (1,n)$ and $b_n = (2,n)$.

Then elements of $X$ will be $a_1, a_2, \dots, b_1, b_2, \dots$.

Singleton sets are open except $\{b_1\}$.

This is not clear to me. Please help.

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    $\begingroup$ How do you define the topology? $\endgroup$ – Rasmus Jul 12 '13 at 14:08
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The (sub)basic open sets of the order topology (which I assume is what interests us here) are of the form $(-\infty,a)=\{a\mid x<a\}$ and $(a,\infty)=\{x\mid x>a\}$. The singleton $a_k$ is open because it is $(-\infty,a_k+1)\cap (a_{k-1},\infty)$ for $k>1$ and $(-\infty,a_2)$ for $k=1$. Similarly, $\{b_k\}=(-\infty,b_{k+1})\cap(b_{k-1},\infty)$ if $k>1$. However, this does not work for $b_1$, neither with the axact trick as above nor any other way: If we progress from our subbasis to a basis of the topology, we have to consider finite intersections. Those are (as far as neighbourhoods of $b_1$ are concerned) of the form $(a_k,b_l)$ with $l>1$, or at least half-inifite. Since none o fthese is the singleton $\{b_1\}$, a union of such sets connot be that singleton either.

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The basic open sets of an ordered set $X$ (with order relation $\prec$) in the order topology are:

  • $(a,b):=\{x\in X:a\prec x\prec b\}$, where $a,b\in X$ with $a\prec b$
  • $(\leftarrow,b):=\{x\in X:x\prec b\}$, where $b\in X$
  • $(a,\rightarrow):=\{x\in X:a\prec x\}$, where $a\in X$

Let's denote the dictionary order on $\{1,2\}\times\Bbb Z_+$ by $\prec$. For $n>1,$ can you find $a,b$ with $a\prec b$ and such that $(a,b)=\{a_n\}$? What about $\{b_n\}$? We can't do such a thing for $\{a_1\},$ since there are no elements less than $a_1,$ but we can show that $\{a_1\}=(\leftarrow,b)$ for some $b$ (which?).

Now, the problem with trying to do this for $\{b_1\}$ is that no matter what $a\prec b_1\prec b$ we choose, there will be infinitely-many $a_n$ in $(a,b)$. (Why?) Likewise, with any $(a,\rightarrow)$ with $a\prec b_1$ or $(\leftarrow,b)$ with $b_1\prec b$. Thus, $b_1$ is the only non-isolated point in the space (the only limit point of the space).

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  • $\begingroup$ Thank you very much for hints. $(a,b) = (a_{n-1},a_{n+1})$. Similarly for $b_n$. Any b can be considered. Any open set containing $b_1$ will contain infinitely many $a_n$, in its left side. $\endgroup$ – Dutta Jul 12 '13 at 15:11
  • $\begingroup$ Almost perfect! It isn't enough to consider any $b$. There is exactly one $b$ such that $(\leftarrow,b)=\{a_1\}.$ $\endgroup$ – Cameron Buie Jul 12 '13 at 15:23
  • $\begingroup$ As $\{a_1\} = (\leftarrow,b)$, we shall take $b = a_1$. $\endgroup$ – Dutta Jul 12 '13 at 17:05
  • $\begingroup$ $b=a_2,$ actually. $\endgroup$ – Cameron Buie Jul 12 '13 at 22:17

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