A question on linear ordered space A space $X$ is called left-separated if it can be well-ordered in such a way that every initial segment is closed in $X$. And we know every space contains a dense left-separated subspace.
My question is this:

Is such left-separated space as a subspace linear ordered space?

Thanks for any help and the references on the left-separated space are also welcome.

Without adding extra conditions on the original topology, you can come up with easy examples of topological spaces such that no dense left-separated subspace is a LOTS.
Consider $\omega_1$ topologised by making all final segments open (i.e., $[ \alpha , + \infty ) = \{ \xi \in \omega_1 : \alpha \leq \xi \}$ is open for all $\alpha < \omega_1$).  This is clearly a left-separated topological space (under the usual order), and every dense left-separated subspace will be homeomorphic to the original space.  As the space is not even T$_1$ it cannot be a LOTS.
