# Paracompactness properties of the deleted Tychonoff plank

I am looking for a justification of paracompactness and related properties for the deleted Tychonoff plank.

The deleted Tychonoff plank is the space $$X=((\omega_1+1)\times(\omega+1))\setminus\{\langle\omega_1,\omega\rangle\}=([0,\omega_1]\times[0,\omega])\setminus\{\langle\omega_1,\omega\rangle\}$$ where each factor has the order topology.

Counterexamples in topology (space #87) lists the following properties without justification:

1. $$X$$ is not paracompact.
2. $$X$$ is not metacompact.
3. $$X$$ is not countably paracompact.
4. $$X$$ is countably metacompact.

I give a proof of the first three below, and would appreciate if you could check it. For the fourth one:

How to prove the deleted Tychonoff plank is countably metacompact?

Not paracompact follows immediately from not metacompact.

Proof that $$X$$ is not metacompact:

The metacompact property is hereditary with respect to closed sets. And the space contains as a closed subspace a copy of the ordinal space $$\omega_1$$, which is not metacompact.

Proof that $$X$$ is not countably paracompact:

Let $$\mathscr U$$ be the countable open cover of $$X$$ consisting the various horizontal slices $$[0,\omega_1]\times\{n\}$$ for $$n\in\omega$$, together with $$[0,\omega_1)\times[0,\omega]$$. Suppose $$\mathscr V$$ is an open refinement of $$\mathscr U$$ and let's show that it cannot be locally finite.

Each point $$\langle\omega_1,n\rangle$$ is in some $$V_n\in\mathscr V$$. Necessarily $$V_n\subseteq[0,\omega_1]\times\{n\}$$. So there is some $$\alpha_n<\omega_1$$ such that $$V_n$$ contains $$[\alpha_n,\omega_1]\times\{n\}$$. Take $$\alpha=\sup\{\alpha_n:n\in\omega\}$$. Then $$\alpha<\omega_1$$ and if $$W$$ is any nbhd of $$\langle\alpha,\omega\rangle$$, $$W$$ must contain some vertical interval $$\{\alpha\}\times[m,\omega]$$ for some $$m\in\omega$$ and hence meet all the $$V_n$$ with $$n\ge m$$. This shows that the open cover $$\mathscr V$$ is not locally finite.

• Your proofs look sound to me. Dec 3, 2023 at 9:32

Let $$\mathcal U$$ be a countable open cover of $$X$$.
$$\omega_1\times\{\omega\}$$ is countably compact so take a finite subcover $$\mathcal U_\omega$$ of $$\mathcal U$$ covering it.
Each $$(\omega_1+1)\times\{n\}$$ is compact and open, so take a finite subcover of $$\mathcal U$$ and refine it to miss all other points of the plank. Call this $$\mathcal U_n$$.
It follows that $$\bigcup_{n\leq\omega}\mathcal U_n$$ is a point finite open refinement of $$\mathcal U$$ covering the space.