Is this Proposition Correct? For any Hausdorff Space $X$ we have $|X| \le 2^{d(X)s\Delta(X)} $ Is this proposition 4.1 right? It is from D. Basile, A. Bella, G J. Ridderbos, Weak extent, submetrizability and diagonal degrees. arXiv:1112.0883, 2011.

 A: I suspect (I do not have my copy of the Handbook of Set-theoretic Topology at hand) that the problem is that $\kappa\omega$ is not $T_3$, just $T_2$. Recall that it is H-closed but not compact, so it cannot be $T_3$. A $\sigma$-space is a $T_3$ (so including $T_1$) space with a $\sigma$-discrete network. We do know that $\kappa\omega$ is a Hausdorff space with a $\sigma$-discrete network (namely the singletons, as $\kappa\omega\setminus\omega$ is closed and discrete, and so are the singleton subsets of $\omega$). So the proof that $\sigma$-spaces have a $G^{\ast}_\delta$-diagonal (a.k.a $s\Delta(\kappa\omega) = \omega$) need not apply. 
So I'd check the definition (I think it assumes $T_3$ through-out the paper) and the proof for the $G^{\ast}_\delta$-diagonal as well, for (hidden) uses of regularity.   
Added later
I checked in the Gruenhage chapter of the named Handbook and indeed $T_3$ is the "culprit". On page 426 the author already states that all spaces are assumed to be regular and $T_1$ unless otherwise stated. And the theorem that $\sigma$-spaces have a $G^{\ast}_\delta$ diagonal uses theorem 2.11, which states that submetacompact spaces with a $G_\delta$ diagonal have a $G^{\ast}_\delta$ diagonal. We already know that $\kappa\omega$ does have indeed have a $G_\delta$ diagonal and it also subparacompact trivially (the singletons form $\sigma$-discrete closed refinement of any open cover, which is what is needed for this property) and so submetacompact (I think that this implication does not use regularity, going by Burke's chapter in the Handbook; otherwise we'd only have found another use of regularity). And indeed the proof of 2.11 (which is in lemma 2.12 in essence) makes have use of regularity explicitly (p. 433). 
So $\kappa\omega$ is a nice example of a subparacompact (possibly also submetacompact, see p. 370 in the diagram) Hausdorff space with a $G_\delta$ diagonal that does not have a $G^{\ast}_\delta$ diagonal, showing that indeed $T_3$ is essential in this statement.
