# The Tychonoff Plank

The Tychonoff plank is the product space $[0,\omega_1] \times [0,\omega]$, that is, the Tychonoff Plank is the product of the successor of the first uncountable ordinal and the successor of the first infinite ordinal. The deleted Tychonoff plank is $X=[0,\omega_1] \times [0,\omega]-\lbrace{(\omega_1,\omega)}\rbrace$, i.e., the Tychonoff plank minus the corner point $(\omega_1,\omega)$. This is another famous example of topological spaces that often appear in first year topology courses. My goal here is to present two facts. The deleted Tychonoff plank is not normal, thus showing that the Tychonoff plank $X=[0,\omega_1] \times [0,\omega]$ not hereditarily normal. The second point is that for the deleted Tychonoff plank, the one-point compactification and the Stone-Cech compactification conincide. This is due to the fact that any continuous real-valued function defined on the deleted Tychonoff plank can be extended to a continuous function on the product space $X=[0,\omega_1] \times [0,\omega]$ (see Theorem 19.12 in [Willard]).

First we show the deleted Tychonoff plank $X$ is not normal. Let $H=\lbrace{(\alpha,\omega):\alpha<\omega_1}\rbrace$ and $K=\lbrace{(\omega_1,n):n<\omega}\rbrace$. These are two disjoint closed sets in $X$. Let $H \subset U$ and $K \subset V$ where $U$ and $V$ are open in $X$. I will show $U \cap V \neq \phi$.

First define a pressing down function on $H$. For each limit ordinal $\alpha<\omega_1$, there is an integer $m(\alpha)$ and there is a countable ordinal $g(\alpha)<\alpha$ such that $[g(\alpha),\alpha] \times [m(\alpha),\omega] \subset U$. By the pressing down lemma, there is $\delta<\omega_1$ and there is a stationary set $S$ such that all points in $S$ are mapped to $\delta$ by $g$. Choose an uncountable $S_0 \subset S$ such that for each $\alpha \in S_0$, $m(\alpha)=j$ for some $j$. Consider $K_0=\lbrace{(\omega_1,n):j \leq n<\omega}\rbrace \subset K$.

Claim. $K_0 \subset Cl(U)$ where $Cl$ is the closure operator. Fix $x=(\omega_1,n)$. Fix an open set $O$ containing $x$, say $O=[\gamma,\omega_1] \times \lbrace{n}\rbrace$ where $\delta<\gamma<\omega_1$. Choose $\alpha \in S_0$ such that $\delta<\gamma<\alpha$. Based on how $g$ is obtained, $[\delta,\alpha] \times [j,\omega] \subset U$. It follows that $[\gamma,\alpha] \times \lbrace{n}\rbrace \subset U$ and $[\gamma,\alpha] \subset [\gamma,\omega_1] \times \lbrace{n}\rbrace=O$. So any open set $O$ containing $x$ contains points of $U$. Thus the claim is established.

Since $K_0 \subset K \subset V$, $V$ would contain points of $U$. Thus the deleted Tychonoff plank $X$ is not normal.

To established the second point, let $f:X\rightarrow\mathbb{R}$ a continuous function. Then $f$ can be  extended to include the corner point. Consider the restriction of $f$ to the following horizontal segments of the deleted Tychonoff plank.
$A=\lbrace{(\alpha,\omega):\alpha<\omega_1}\rbrace$
$B_n=\lbrace{(\alpha,n):\alpha \leq \omega_1}\rbrace$ for each $n<\omega$.

Any continuous function is eventually constant on these horizontal segments. For the proof, see this post. So there is $\alpha_0<\omega_1$ such that $f((\alpha,\omega))=x_0$ for all $(\alpha,\omega) \in [\alpha_0,\omega_1) \times \lbrace{\omega}\rbrace$. For each $n<\omega$, there is $\beta_n<\omega_1$ such that $f((\alpha,n))=y_n$ for all $(\alpha,n) \in [\beta_n,\omega_1] \times \lbrace{n}\rbrace$. Choose some $\alpha>\alpha_0,\beta_n$ for all $n$.

It is clear that the sequence $\lbrace{y_n}\rbrace$ converges to $x_0$. So by defining $f((\omega_1,\omega))=x_0$, $f$ is still a continuous function.

Reference
[Willard] Willard, S., [1970] General Topology, Addison-Wesley Publishing Company, Inc.