A previous question asked about the radial and pseudoradial properties for the Tychonoff plank and its deleted variant. The current question does the same for the Dieudonné plank and its deleted variant.

The Dieudonné plank and deleted Dieudonné plank have the same underlying set as the Tychonoff plank and deleted Tychonoff plank, but with a finer topology.

Here the Dieudonné plank is the product space $$X=(\omega_1+1)\times(\omega+1)=[0,\omega_1]\times[0,\omega]$$ where the second factor $$[0,\omega]$$ is given its usual order topology and the first factor $$[0,\omega_1]$$ is given a topology finer than its order topology. Specifically, $$[0,\omega_1]$$ is given a Fortissimo topology where all $$\alpha<\omega_1$$ are isolated points and neighborhoods of $$\omega_1$$ are the cocountable sets containing $$\omega_1$$; that is, the topology on $$[0,\omega_1]$$ is the one-point Lindelöfication of $$[0,\omega_1)$$ with the discrete topology.

The deleted Dieudonné plank is the subspace of the Dieudonné plank obtained by removing the point $$\langle\omega_1,\omega\rangle$$.

Note: What Counterexamples in topology (example #89) calls "Dieudonné plank" is the "deleted Dieudonné plank" above.

The results are parallel to those for the Tychonoff planks:

• The Dieudonné plank is not radial.
• The Dieudonné plank is pseudoradial.
• The deleted Dieudonné plank is radial.

The proofs are very similar (parts of them even identical) to the proofs for the Tychonoff planks. I am posting them below, so pi-base can be updated to use that as reference.

First some preliminary results.

A topological space is called well-based if every point has a nbhd base well-ordered by reverse inclusion; or equivalently, if every point has a nbhd base totally ordered by inclusion. (See https://mathoverflow.net/questions/322162)

Clearly, a well-based space is radial.

Fact: The space $$[0,\omega_1]$$ with the Fortissimo topology is well-based.

(Reason: At the point $$p=\omega_1$$, the nbhds $$(\alpha,\omega_1]$$ for $$\alpha<\omega_1$$ form a totally ordered local base. And at each isolated point the corresponding singleton trivially forms a totally ordered local base.)

Lemma: Let $$Y$$ be a pseudoradial subspace of a space $$X$$. If $$A\subseteq X$$ is radially closed in $$X$$, then $$A\cap Y$$ is closed in $$Y$$.

This is not hard to check.

Proof that the Dieudonné plank is not radial:

Let $$A=\omega_1\times\omega$$. The point $$p=\langle\omega_1,\omega\rangle$$ is in the closure of $$A$$. Suppose by contradiction that there is a transfinite sequence $$(x_\alpha)_{\alpha<\lambda}$$ of points of $$A$$ converging to $$p$$. We can assume $$\lambda$$ is an infinite regular cardinal of smallest possible cardinality. As $$A$$ has cardinality $$\aleph_1$$, $$\lambda$$ is at most $$\omega_1$$. If $$\lambda=\omega$$, the sequence would be countable and disjoint from some neighborhood $$[\alpha,\omega_1]\times [0,\omega]$$ of $$p$$, which is not possible. So $$\lambda=\omega_1$$.

But, by considering suitable nbhds of $$p$$ we see that for each $$n\in\omega$$ the $$x_\alpha$$ are eventually in $$\omega_1\times[n+1,\omega)$$. So, if $$\pi_2$$ is the projection onto the second component, the cardinality of $$\{\alpha<\lambda:\pi_2(x_\alpha)\le n\}$$ is less than $$\lambda$$, hence countable. Since every $$\alpha$$ is in one of these sets, $$\lambda$$ can be expressed as a countable union of countable sets, which is impossible.

Proof that the deleted Dieudonné plank is well-based (hence radial):

Every point of the space has as open nbhd either a vertical slice $$\{\alpha\}\times[0,\omega]$$ or a horizontal slice $$[0,\omega_1]\times\{n\}$$ (with the Fortissimo topology). These are well-based. Hence the whole space is well-based.

Proof that the Dieudonné plank is pseudoradial:

Let $$X$$ be the Dieudonné plank. Let $$A\subseteq X$$ be radially closed and let $$p$$ be a point not in $$A$$. We have to exhibit a nbhd of $$p$$ disjoint from $$A$$.

If $$p$$ belongs to the deleted Dieudonné plank $$Y=X\setminus\{\langle\omega_1,\omega\rangle\}$$ (which is open in $$X$$), there is a nbhd of $$p$$ in $$Y$$ (hence in $$X$$) disjoint from $$A$$, since $$Y$$ is radial.

Now assume $$p=\langle\omega_1,\omega\rangle$$. The subspace $$\{\omega_1\}\times[0,\omega]$$ is radial, hence by the Lemma there is some $$m\in\omega$$ such that $$\{\omega_1\}\times[m,\omega]$$ is disjoint from $$A$$. For each $$k\in[m,\omega]$$ the point $$\langle\omega_1,k\rangle$$ is not in $$A$$, and the horizontal slice $$[0,\omega_1]\times\{k\}$$ is radial. So again by the Lemma there is some $$\alpha_k\in\omega_1$$ such that $$[\alpha_k,\omega_1]\times\{k\}$$ is disjoint from $$A$$. Let $$\alpha=\sup\{\alpha_k:k\in[m,\omega]\}$$. The set $$[\alpha,\omega_1]\times[m,\omega]$$ is a nbhd of $$p$$ disjoint from $$A$$.