Recently, I meet an example called Tychonoff Plank. Let $\omega^*=[0,\omega]$ and $\omega_1^*=[0,\omega_1]$ are linear ordering topological space. Let $T=\omega^*\times \omega_1^*\setminus \{\langle \omega^*,\omega_1^* \rangle\}$. Then $T$ is called Tychonoff Plank. Is this space is star compact?

  • $\begingroup$ What is $\; \operatorname{St} \;$ in the definition of star compact? $\;\;\;$ $\endgroup$
    – user57159
    Commented Jan 26, 2012 at 1:02
  • $\begingroup$ $St(K, \mathscr{U})=\cup\{u\in \mathscr{U}: u \cap K \neq \emptyset\}$ $\endgroup$
    – Paul
    Commented Jan 26, 2012 at 1:09

1 Answer 1


Yes, it is.

Suppose that $\mathscr{U}$ is an open cover of $T$. First we’ll take care of most of $\{\omega\}\times\omega_1$. For each $\alpha\in\omega_1$ there is some $U_\alpha\in\mathscr{U}$ containing $\langle\omega,\alpha\rangle$, which must contain a basic open nbhd of $\langle\omega,\alpha\rangle$ of the form $(n_\alpha,\omega]\times(\beta_\alpha,\alpha]$ for some $n_\alpha\in\omega$ and $\beta_\alpha<\alpha$. By the pressing-down lemma there is a $\beta\in\omega_1$ such that $\{\alpha\in\omega_1:\beta_\alpha=\beta\}$ is stationary and therefore cofinal in $\omega_1$. For $k\in\omega$ let $A_k=\{\alpha\in\omega_1:\beta_\alpha=\beta\land n_\alpha=k\}$; clearly there is an $m\in\omega$ such that $|A_m|=\omega_1$. Let $p=\langle\omega,\beta+1\rangle$; then $\operatorname{st}(p,\mathscr{U})\supseteq (m,\omega]\times(\beta,\omega_1)$. Note that we may replace $\beta+1$ here by any countable ordinal greater than $\beta$.

Next we take care of $\omega\times\{\omega_1\}$. For each $n\in\omega$ there must be some $U_n\in\mathscr{U}$ such that $\langle n,\omega_1\rangle\in U_n$, so we can choose a strictly increasing sequence $\langle\xi_n:n\in\omega\rangle$ in $\omega_1$ such that $\xi_0>\beta$ and $\{n\}\times[\xi_n,\omega_1]\subseteq U_n$ for each $n\in\omega$. Let $\eta=\sup_n\,\xi_n >\beta$, and let $K_0=\omega^*\times\{\eta\}$; clearly $K_0$ is compact, and $T\setminus\operatorname{st}(K_0,\mathscr{U})\subseteq\omega^*\times[0,\eta]$. But $\omega^*\times[0,\eta]$ is compact, so $K=K_0\cup\big(\omega^*\times[0,\eta]\big)$ is compact, and clearly $\operatorname{st}(K,\mathscr{U})=T$.


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