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

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

• countable injective weight means that $X$ maps 1-1 onto a subspace of $\mathbb{R}^\omega$? – Henno Brandsma Apr 4 '18 at 22:10
• continuously one-to-one in $R^{\omega}$ – Alexander Osipov Apr 6 '18 at 15:56

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.