Cartesian Product of Linearly ordered space and an example The base of the cartesian product of linearly ordered spaces $A$ and $B$ is the form $\{U\times V : U\text{ open in }A, B\text{ open in }B\}$.
By using this, we condiser the Long Line Topology which is the form cartesian product of the set of all countable ordinal numbers and the half open unit interval with dictionary order.
But I could not see the differences between the cartesian product and the cartesian  dictionary product of these space. Also some authors say that the long line is constructed by pasting the  half-open unit interval end-to-end. What is the reason of this? Thanks.
 A: Consider the sets $U = [ 0 , \omega+1 ) \subseteq \omega_1$ and $V = [ 0 , \frac{1}{2} )$.  Clearly $U$ and $V$ are open in $\omega_1$, $[0,1)$, respectively, and so $U \times V$ is open in $\omega_1 \times [0,1)$ with the product topology.
However this set is not open in the dictionary order topology.  Note that $\langle \omega , 0 \rangle \in U \times V$.  If $U \times V$ were open in the dictionary order, then there would be pairs $\langle \alpha , x \rangle , \langle \beta , y \rangle \in \omega_1 \times [0,1)$ such that


*

*$\langle \alpha , x \rangle <_\text{lex} \langle \omega , 0 \rangle <_\text{lex} \langle \beta , y \rangle$; and

*whenever $\langle \gamma , z \rangle \in \omega_1 \times [0,1)$ is such that $\langle \alpha , x \rangle <_\text{lex} \langle \gamma , z \rangle <_\text{lex} \langle \beta , y \rangle$, then $\langle \gamma , z \rangle \in U \times V$.


Clearly we must have $\alpha < \omega$ and so $\alpha < \alpha + 1 < \omega$, meaning that $\langle \alpha , x \rangle <_\text{lex} \langle \alpha+1 , \frac{3}{4} \rangle <_\text{lex} \langle \beta , y \rangle$, however $\langle \alpha+1 , \frac{3}{4} \rangle \notin U \times V$ (since $\frac{3}{4} \notin V$).

In the opposite direction, note that the set $W = \{ \omega \} \times (0,1)$ is open in the  dictionary order topology (it is the set of all points srictly between $\langle \omega , 0 \rangle$ and $\langle \omega , 1 \rangle$).  However this set is not open in the product topology, since if $U \subseteq \omega_1$ is open and contains $\omega$ then there is an $n < \omega$ such that $\{ m \in \omega : n < m \} \subseteq U$.

As for the intuitive picture of the long line as resulting in pasting together $\omega_1$ copies of $[0,1)$ end-to-end, this just comes up from making a picture of $\omega_1 \times [0,1)$ with the dictionary order.  First we have $\{ 0 \} \times [0,1)$, and immediately following this is $\{ 1 \} \times [ 0 , 1 )$ and then etc.
