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.

  • 1
    $\begingroup$ Your proofs look sound to me. $\endgroup$ Dec 3, 2023 at 9:32

1 Answer 1


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.


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