2
$\begingroup$

Recall that a space $X$ is called basically disconnected [1] if every cozero-set has an open closure. According to Tkačuk [2], a space $X$ said to be $b$-discrete if every countable subset of $X$ is closed (equivalently, closed and discrete) and $C^∗$-embedded in $X$. Note that if $X$ is basically disconnected then $X$ is $b$-discrete [1].

Is every $b$-discrete space $X$ with countable injective weight basically disconnected?

[1] L. Gillman, M. Jerison, Rings of continuous functions, The University Series in Higher Mathematics. Princeton, New Jersey: D. Van Nostrand Co., Inc., 1960. 300 p

[2] V.V. Tkachuk, The spaces $C_p(X)$: decomposition into a countable union of bounded subspaces and completeness properties, Topology and its Applications, n 22, (1986), 241–253

Improve this question
$\endgroup$
2
  • $\begingroup$ countable injective weight means that $X$ maps 1-1 onto a subspace of $\mathbb{R}^\omega$? $\endgroup$
    – Henno Brandsma
    Commented Apr 4, 2018 at 22:10
  • $\begingroup$ continuously one-to-one in $R^{\omega}$ $\endgroup$
    – Alexander Osipov
    Commented Apr 6, 2018 at 15:56

1 Answer 1

Reset to default
-1
$\begingroup$

Just came across this. b-discrete is a subclass of the class of weak P-spaces aka $\omega$-discrete (that is, every countable set is closed). It is not the case that every basically disconnected space is b-discrete. E.g. $\beta \mathbb{N}$ is extremally disconnected and $\mathbb{N}$ is not closed.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .