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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!

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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 $u$.

The one-point compactification of an uncountable discrete space has countable tightness yet it does not even have countable pseudocharacter.

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    $\begingroup$ @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. $\endgroup$
    – Henno Brandsma
    Commented Oct 19, 2018 at 22:45

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