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From Munkres p.90-91

Definition Given an ordered set $X$, a subset $Y$ of $X$ is convex in $X$ if for each pair of points $a<b$ in $Y$, the entire interval $(a,b)$ of points of $X$ lies in $Y$.

Theorem Let $X$ be an ordered set in the order topology; let $Y$ be a subset of $X$ that is convex in $X$. Then the order topology on $Y$ is the same as the topology $Y$ inherits as a subspace of $X$.

Aren't $[0,1]$ and $[0,1]\times [0,1]$ convex? But according to the book the dictionary order topology on $[0,1]\times [0,1]$ is not the same as the subspace topology on $[0,1]\times [0,1]$ obtained from the dictionary order topology on $\mathbb{R}\times \mathbb{R}$. Doesn't that contradict the theorem?

Ludolila
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Xena
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1 Answers1

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No, it doesn’t contradict the theorem, because $[0,1]\times[0,1]$ isn’t a convex subset of $\Bbb R\times\Bbb R$ in the lexicographic order topology. For instance, $\langle 0,0\rangle\in[0,1]\times[0,1]$ and $\langle 1,1\rangle\in[0,1]\times[0,1]$, and $\langle 0,0\rangle<\langle 0,2\rangle<\langle 1,1\rangle$ in the lexicographic order on $\Bbb R\times\Bbb R$, but $\langle 0,2\rangle\notin[0,1]\times[0,1]$.

The smallest convex subset of $\Bbb R\times\Bbb R$ containing $\langle 0,0\rangle$ and $\langle 1,1\rangle$, i.e., the closed interval in the lexicographic order with $\langle 0,0\rangle$ and $\langle 1,1\rangle$ as endpoints, is

$$\Big(\{0\}\times[0,\to)\Big)\cup\Big((0,1)\times\Bbb R\Big)\cup\Big(\{1\}\times(\leftarrow,1]\Big)\;.$$

(You’re probably more familiar with the notations $\infty$ and $-\infty$ for my $\to$ and $\leftarrow$, respectively.)

Brian M. Scott
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