Let $X$ be a well-ordered set in the order topology. Let $x_0$ be the smallest element of $X$.

Then, the claim is that $\{x_0\}$ is both open and closed in $X$.

It is clear to me why $\{x_0\}$ is closed, but I do not see why that is open.

If it were open, then there must be a basis element containing only $x_0$, but every basis element of $X$ is either an interval $(a,b)$ or an interval $[x_0,b)$, neither of which contains only $x_0$.

Am I missing something?

  • $\begingroup$ Your definition of the order topology is... unusual. Typically, it's defined as the topology generated by the half-rays: all sets $\{x\mid a < x\}, \{x\mid a > x\}$ form a subbasis. See e.g. en.wikipedia.org/wiki/Order_topology $\endgroup$ – BrianO Apr 16 '16 at 2:34
  • $\begingroup$ @BrianO I took the definition from Munkres' book "Topology" Section 14. Using the subbasis, how do still show that the singleton set $\{x_0\}$ is open? $\endgroup$ – Jennifer Apr 16 '16 at 2:36
  • $\begingroup$ With the subbasis I described, if $X = \{x_0\}$ then we're done, so suppose $X$ isn't a singleton. In that case, $x_0$ has a successor, $x_1$, and then $\{x_0\} = \{x\in X\mid x < x_1\}$ is open. [Are you sure Munkres has "or an interval $(x_0, b)$?" not $[x_0, b)$?] $\endgroup$ – BrianO Apr 16 '16 at 2:45
  • $\begingroup$ @BrianO Oops, my apologies. It should be $[x_0,b)$; I just edited my question. Ok. I understand now. But, if you were to use a basis, instead of subbasis, how do you show that $\{x_0\}$ is open? $\endgroup$ – Jennifer Apr 16 '16 at 2:51
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    $\begingroup$ Same deal: if $x_0$ has a successor $x_1$, then $\{x_0\} = [x_0, x_1)$ is open; and if it doesn't have one, then $X=\{x_0\}$ is open. $\endgroup$ – BrianO Apr 16 '16 at 2:52

The result is trivial if $|X|=1$, so assume otherwise. Let $x_1$ be the smallest element of the set $\{x\in X: x \not= x_0\}$, which must exist as $X$ is well ordered. Then $\{x_0\}=(\leftarrow,x_1)$ is open.

  • 1
    $\begingroup$ To see the last identity in detail: $x_0 < x_1$ as $x_1 \neq x_0$ and $x_0 = \min(X)$, so $x_0 \in (\leftarrow, x_1)$, and if $p \neq x_0$, $x_1 \le p$ by definition of $x_0$, so $p \notin (\leftarrow, x_1)$. Hence equality. $\endgroup$ – Henno Brandsma Apr 16 '16 at 10:36

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