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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.

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I am a bit confused as to what your question is actually asking. Are you asking whether every left-separated space is a subspace of a LOTS? or whether every left-separated dense subspace of a space is a LOTS? or something else? –  Arthur Fischer Jan 15 '13 at 9:17
    
Sorry. Yes. My question is this: every left-separated dense subspace of a space is a LOTS? –  Paul Jan 15 '13 at 9:31
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up vote 2 down vote accepted

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.

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@Brian: Wow! Bad mistake on my part. Thanks a lot, for the $(n+1)$st time! –  Arthur Fischer Jan 16 '13 at 13:59
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