The Munkres' topology book provides Example 30.5 (p.193, 2nd Ed) for a subspace of a Lindelöf space that need not be Lindelöf as follows:
The ordered square $I_0^2$ is compact; therefore it is Lindelöf
trivially. However, the subspace $A = I \times (0,1)$ is not Lindelöf.
For $A$ is the union of the disjoint sets $U_x = \{x\} \times (0,1)$,
each of which is open in $A$. This collection of sets is uncountable,
and no proper subcollection covers $A$.
I could largely understand this argument. But it is not clear to me how $U_x$ is open in $A$. Could someone please explain? Thanks.
RD
Best Answer
The set $U_x$ is open, since the topology on the ordered square is not the standard Euclidean topology. The definition of the order on the square is lexicographic, ie $(x_1,y_1) < (x_2, y_2)$ if either $x_1 < x_2$ or if $x_1 = x_2$ and $y_1 < y_2$. The corresponding order topology is generated by sets of the form $$ \{ (x, y) : (x_1, y_1) < (x,y) < (x_2, y_2)\} $$
In this example $U_x$ is obtained by taking $(x_1, y_1) = (x,0)$ and $(x_2, y_2) = (x,1)$.