Re: non-separable linearly ordered space

From: thomas@xxxxxxxxxxx

> If I have a linearly ordered space with its order topology, and
> there is no countable dense subset, is there necessarily (or under
> reasonable additional conditions) an uncountable nowhere-dense
> subset?

No, additional conditions are necessary. Have you any in mind?
Let S be any uncountable ordinal such as omega_1, with each point
replaced by Z. S is an uncountable discrete linear order. As such,

S is not separable and every nonnul subset has a nonnul interior.
Hence, though there is no countable dense subset, there is also
no uncountable nowhere-dense subset.

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