What can we say about a locally compact Hausdorff space whose every open subset is sigma compact?
My question is induced from these problem and from Henno Brandsma's answer.
I want to show in double arrow space the subspace $(0,1)\times\{1\}$ is homeomorphic to $(0,1)$ in lower limit topology or the other name is Sorgenfrey line generated by $[a,b)$ type intervals.
Take double arrow space $[0,1]\times \{0,1\}$
My Attempt to show homeomorphism as follows:
Define $$\phi:(0,1)\times\{1\}\to (0,1)\\ (x,1)\mapsto x$$
from double arrow space to lower limit topology.
First of all it is clearly a bijection.
And continuous because take any open set in lower limit topology on $(0,1)$
as $[a,b)$
then $\phi^{-1}([a,b))=(a,\{0\}, b,\{1\})\cap (0,1)\times \{1\}$
it is open in subspace topology.
Is this way correct? And I cannot show $\phi $ is an open map because I cannot estimate general type of basic open sets in subspace of double arrow space.
Best Answer
I'll call $(0,1) \times \{1\}$ in the subspace topology $\Bbb S_1$ for short.
A basic open subset of a point $(x,1)$ in $\Bbb S_1$ is of the form $\langle A,B\rangle \cap \Bbb S_1$, where $\langle A,B\rangle$ is a basic open interval of Double Arrow, where $A < (x,1) < B$ (the definition of the order topology, and the definition of subspace topology are applied here).
If $A = (a,0)$, with $a \le x$, and $B=(b,0)$ or $B=(b,1)$ with $x < b$ we get that $\langle A,B\rangle \cap \Bbb S_1 = [a,b) \times \{1\}$
If $A =(a,1)$ with $a < x$ and $B=(b,0)$ or $(b,1)$ with $x < b$ we get that $\langle A,B\rangle \cap \Bbb S_1 = (a,b) \times \{1\}$
In all cases the map $\phi: ((0,1), \mathcal{T}_s) \to \Bbb S_1, x \to (x,1)$ sends basic elements of $\mathcal{T}_s$ (the Sorgenfrey topology on $(0,1)$, which is homeomorphic to the full Sorgenfrey line by any order homeomorphism from $(0,1) $ to $\Bbb R$) to open subsets of $\Bbb S_1$, so $\phi$ is bijective (obvious) and open. As $\phi^{-1}[B]$ is also open for every basic open set of $\Bbb S_1$, as we saw, $\phi$ is also continous. We can check openness and continuity by base elements alone; this is standard.
It follows that $\Bbb S_1$ is homeomorphic to the Sorgenfrey line.