Subspace topology and order topology (2)

Yesterday I asked a very similar question but now I am confused again...

Let $I=[0,1]$. The dictionary order on $I\times I$ is just the restriction to $I\times I$ of the dictionary order on the plane $\mathbb{R}\times\mathbb{R}$. now, how can I show that the set $\{1/2\}\times (1/2,1]$ is not open in the order topology? The maximum element of $I$ is 1. So the order topology on $I\times I$ has intervals of the form $(a,1]\times (b,1]$ as basis elements. Isn't $\{1/2\}\times (1/2,1]$ itself a basis element of the order topology?

Solution 1:

You may have somewhat misunderstood the lexicographic order, since the lexicographic order topology on $I\times I$ does not have sets of the form $(a,1]\times(b,1]$ as basis elements. To avoid confusion with the usual order $\le$ on $I$, I’ll use $\preceq$ for the lexicographic order on $I\times I$. Note also that I use angle brackets for ordered pairs, e.g., $\langle x,y\rangle$, and parentheses for open intervals.

If $p=\langle x_0,y_0\rangle\in I\times I$ and $q=\langle x_1,y_1\rangle\in I\times I$, then $p\preceq q$ iff either $x_0<x_1$, or $x_0=x_1$ and $y_0\le y_1$. Suppose that $p\prec q$. If $x_0=x_1$, so that $p$ and $q$ are both on the vertical segment $\{x_0\}\times I$, then the open interval $(p,q)$ in $\langle I\times I,\preceq\rangle$ is the vertical open interval

$$(p,q)=\{x_0\}\times(y_0,y_1)=\{\langle x,y\rangle\in I\times I:x=x_0\text{ and }y_0<y<y_1\}\;.$$

If $x_0<x_1$, it’s a little more complicated:

$$\begin{align*} (p,q)&=\Big(\{x_0\}\times(y_0,1]\Big)\cup\Big((x_0,x_1)\times I\Big)\cup\Big(\{x_1\}\times[0,y_1)\Big)\\ &=\{\langle x,y\rangle\in I\times I:x=x_0\text{ and }y_0<y\le 1\}\\ &\qquad\cup\{\langle x,y\rangle\in I\times I:x_0<x<x_1\text{ and }y\in I\}\\ &\qquad\cup\{\langle x,y\rangle\in I\times I:x=x_1\text{ and }0\le y<y_1\}\;. \end{align*}$$

One base for the lexicographic order topology on $I\times I$ is the collection of all of these open intervals together with all of the intervals $[\langle 0,0\rangle,p)$ and $(p,\langle 1,1\rangle]$ for $p\in(I\times I)\setminus\{\langle 0,0\rangle,\langle 1,1\rangle\}$, since these latter form local bases at the endpoints $\langle 0,0\rangle$ and $\langle 1,1\rangle$, respectively.

Let $A=\left\{\frac12\right\}\times\left(\frac12,1\right]$, the specific set about which you asked. (Note that it is not of the form $(a,1]\times(b,1]$ in the first place.) Let $p=\left\langle\frac12,\frac12\right\rangle$ and $q=\left\langle\frac12,1\right\rangle$. Then $A$ is the half-open, half-closed interval $(p,q]$, which is not open, because $q$ is not in its interior. To see this, suppose that $U$ is an open nbhd of $q$. Then there are $u,v\in I\times I$ such that $u\prec q\prec v$ and $(u,v)\subseteq U$. Say $v=\langle x_0,y_0\rangle$; then $q\prec v$ implies that $x_0>\frac12$, so there is a real number $x\in\left(\frac12,x_0\right)$, and $q\prec\langle x,0\rangle\prec v$ and hence $\langle x,0\rangle\in(u,v)\subseteq U$. That is, every open nbhd of $q$ in $I\times I$ contains points with first coordinate larger than $\frac12$, so $(p,q]$, which does not contain any such points, cannot contain any open nbhd of $q$ and therefore cannot be open.