The Tychonoff plank is the product $(\omega_1+1)\times(\omega+1)=[0,\omega_1]\times[0,\omega]$ where each factor has the order topology. The deleted Tychonoff plank is the Tychonoff plank with the point $(\omega_1,\omega)$ removed. I am wondering about the radial and pseudoradial properties for these spaces.
A space $X$ is called pseudoradial if every radially closed set is closed. Equivalently, for every non-closed set $A\subseteq X$ there is a point $p\in\overline A\setminus A$ and a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ with each $x_\alpha\in A$ and converging to $p$. (Here $\lambda$ is a limit ordinal and can always be taken to be a regular cardinal.)
A space $X$ is called radial if for every non-closed set $A\subseteq X$ and every point $p\in\overline A\setminus A$ there is a transfinite sequence $(x_\alpha)_{\alpha<\lambda}$ with each $x_\alpha\in A$ and converging to $p$.
For the Tychonoff plank, the results below are stated in the article by Horst Herrlich (1967) that introduced the pseudoradial and radial properties, but without justification.
- The Tychonoff plank is not radial. I give a proof below.
- The Tychonoff plank is pseudoradial.
- The deleted Tychonoff plank is radial. I give a proof below.
- The deleted Tychonoff plank is pseudoradial (because radial implies pseudoradial).
How to prove the Tychonoff plank is pseudoradial?
Would also appreciate if you could check the validity of my proofs below.
Proof that the Tychonoff plank is not radial:
Let $X=[0,\omega_1]\times[0,\omega]$ be the Tychonoff plank. Take $A=\omega_1\times\omega$ and $p=(\omega_1,\omega)$. The point $p$ 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 $p$. Taking a cofinal subsequence as necessary, we can assume $\lambda$ is a regular cardinal. Let $\pi_1$ and $\pi_2$ are the projections onto the first and second component. The sequence $(\pi_1(x_\alpha))_{\alpha<\lambda}$ is a sequence in $\omega_1$ that converges to $\omega_1$. So necessarily $\lambda\ge\operatorname{cof}(\omega_1)=\omega_1$; in particular $\lambda$ is uncountable.
On the other hand, by considering a suitable nbhd of $p$ we see that for each $n\in\omega$ the $x_\alpha$ are eventually in $\omega_1\times[n+1,\omega)$. So the cardinality of $\{\alpha<\lambda:\pi_2(x_\alpha)\le n\}$ is less than $\lambda$. But every $\alpha$ is in one of these sets, so $\lambda$ can be expressed as a countable union of sets, each of cardinality less than $\lambda$, which is not possible as $\lambda$ is a uncountable regular cardinal.
Observation:
Since the radial property is hereditary, it follows that the product of larger ordinals is not radial either. For example $(\omega_1+1)\times(\omega_1+1)$ is not radial. Is it pseudoradial?
Proof that the deleted Tychonoff plank is radial:
Let $X=([0,\omega_1]\times[0,\omega])\setminus\{(\omega_1,\omega)\}$ be the deleted Tychonoff plank. Suppose $A\subseteq X$ is not closed and $p=(p_1,p_2)\in\overline A\setminus A$.
If $p_1<\omega_1$, the nbhd $V=[0,p_1]\times[0,p_2]$ of p is compact Hausdorff and countable, hence metrizable (see here for example). As $p$ is in the closure of $V\cap A$, there is a sequence of points in $V\cap A$ converging to $p$.
If $p_1=\omega_1$, necessarily $p_2<\omega$ and $V=[0,\omega_1]\times\{p_2\}$ is a nbhd of $p$ homeomorphic to the radial space $\omega_1+1$. As $p$ is in the closure of $V\cap A$, there is a (transfinite) sequence of points in $V\cap A$ converging to $p$.
As an alternative to the the last proof, one can show the slightly stronger fact:
The deleted Tychonoff plank is well-based:
A 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.
The deleted Tychonoff plank is covered by the open sets $[0,\omega_1)\times[0,\omega]$ and $\{n\}\times[0,\omega_1]$ for $n\in\omega$. The first open set is first countable, hence well-based; the other open sets are homeomorphic to an ordinal space, hence also well-based. Since being well-based is a local property, the space is well-based.
Best Answer
Note that a equivalent definition for radial is this: for every set $A$ and limit point $p$ of $A$, there exists a transfinite sequence in $A$ converging to $p$.
Note that a equivalent definition for pseudoradial is this: for every radially closed set $A$ and limit point $p$ of $A$, there exists a transfinite sequence in $A$ converging to $p$.
Let $H$ be radially closed in the Tychonoff plank $X$. Let $h$ be a limit point of $H$. We will show that there exists a transfinite sequence in $H$ converging to $h$, and thus show $X$ is pseudoradial.
If $h$ belongs to the deleted Tychonoff plank $Y= X\setminus\{\langle\omega_1,\omega\rangle\}$, there is a transfinite sequence in $H$ converging to $h$ since $Y$ is radial.
Otherwise $h=\langle \omega_1,\omega\rangle$. Suppose we could not construct a transfinite sequence in $H$ converging to $h$. Then there exists $\{\omega_1\}\times(n,\omega]$ disjoint from $H$, and $(\alpha,\omega_1]\times\{\omega\}$ disjoint from $H$. Then for each $\langle\omega_1,N\rangle\in\{\omega_1\}\times(n,\omega]$ there is some $\beta_N$ with $(\beta_N,\omega_1]\times\{N\}$ disjoint from $H$, since $H$ is radially closed. Let $\beta=\sup(\{\alpha\}\cup\{\beta_N:N\in(n,\omega)\})$. Then $(\beta,\omega_1]\times(n,\omega]$ is a neighborhood of $h$ disjoint from $H$, a contradiction.
The theorem of http://sci-gems.math.bas.bg/jspui/handle/10525/551 alternatively implies that the product of any two successor ordinals is pseudoradial.