# a question on Tychonoff Plank

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?

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

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 $$\operatorname{st}(K,\mathscr{U})=T$$.