# A question on monotonically normal spaces

This question is related to one of previous questions.

For any generalized order space $$X$$, $$X$$ has countable tightness iff $$X$$ is first countable.

Since a generalized order space is monotonically normal, the following question is natural.

Is there a monotonically normal space $$X$$ with countable tightness which is not first countable?

thanks a lot!

There are many examples. Take one ultrafilter $$u$$ on $$\mathbb{N}$$ and consider $$\mathbb{N}\cup\{u\}$$ as a subspace of $$\beta\mathbb{N}$$ (every point of $$\mathbb{N}$$ is isolated and the basic open neighbourhhods for $$u$$ are the sets of the form $$U\cup\{u\}$$ with $$U\in u$$). The resulting space has countable tightness but it is not first-countable at $$u$$.
• @Joe For such spaces with a single non-isolated point we can just use a trivial monotone-normality operator $\mu(x,U)$. $\mu(x,U) = \{x\}$ for all isolated $x$ and $\mu(x,U) = U$ for the other point. – Henno Brandsma Oct 19 '18 at 22:45