The Deleted Tychonoff Plank

Ordinal Intervals and the Plank

The deleted Tychonoff plank is a useful example because its failure of normality is not visible from a small neighborhood around a single point. The obstruction comes from two entire edges: any open set containing one edge must intersect every open set containing the other. We first construct the space and verify the compact Hausdorff setting in which it sits.

Write \(\omega\) for the first infinite ordinal and \(\omega_1\) for the first uncountable ordinal. Give each ordinal interval \([0,\theta]\) its order topology: this is the topology generated by open intervals and open initial or final rays in the usual order, with neighborhoods at endpoints taken relative to \([0,\theta]\). In particular, a neighborhood of \(\omega\) in \([0,\omega]\) contains \((n,\omega]\) for some finite ordinal \(n\), and a neighborhood of \(\omega_1\) in \([0,\omega_1]\) contains \((\gamma,\omega_1]\) for some \(\gamma<\omega_1\).

For an open set in the product, a basic neighborhood has the form \(O\times V\), where \(O\) and \(V\) are open in their respective ordinal intervals. The product topology is being used here, not the box topology; for a product of two factors the basic neighborhoods are still rectangles.

The Ambient Plank Is Compact Hausdorff

Proof. Let \(\mathcal{U}\) be an open cover of \([0,\theta]\). We show that each initial segment \([0,\xi]\), for \(\xi\le\theta\), has a finite subcover. If not, let \(\xi\) be the least ordinal for which \([0,\xi]\) has no finite subcover. It cannot be \(0\), since a cover member containing \(0\) covers \([0,0]\). It cannot be a successor \(\eta+1\), since \([0,\eta]\) has a finite subcover by minimality, and one more cover member containing \(\eta+1\) covers \([0,\eta+1]\). If \(\xi\) is a nonzero limit ordinal, choose a cover member \(U\) containing \(\xi\). By the order-topology neighborhood basis at a limit ordinal, some \(\beta<\xi\) satisfies \((\beta,\xi]\subseteq U\). The set \([0,\beta]\) has a finite subcover by minimality; adding \(U\) gives a finite cover of \([0,\xi]\), again a contradiction. Thus \([0,\theta]\) is compact.

For Hausdorffness, take \(\alpha<\beta\) in \([0,\theta]\). If there is \(\gamma\) with \(\alpha<\gamma<\beta\), the open rays \([0,\gamma)\) and \((\gamma,\theta]\) are disjoint neighborhoods of \(\alpha\) and \(\beta\). If there is no such \(\gamma\), then \(\beta=\alpha+1\); the disjoint open rays \([0,\beta)\) and \((\alpha,\theta]\) contain \(\alpha\) and \(\beta\), respectively. Hence the ordinal interval is Hausdorff.

Finally, the product of two compact spaces is compact. To verify the finite-product fact here, let \(\mathcal{V}\) be an open cover of \(X\times Y\), with \(X\) and \(Y\) compact. For each \(x\in X\), compactness of \(Y\) gives finitely many members of \(\mathcal{V}\) covering \(\{x\}\times Y\). Around each \((x,y)\) choose a basic rectangle \(U_y\times V_y\) contained in its selected cover member. A finite number of the \(V_y\)'s cover \(Y\); intersect the corresponding \(U_y\)'s to obtain an open neighborhood \(U_x\) of \(x\). Then \(U_x\times Y\) is covered by the finitely many selected members. The sets \(U_x\) cover \(X\), so finitely many suffice; the associated finite collections cover \(X\times Y\). The product is Hausdorff because two distinct points differ in a coordinate, and disjoint neighborhoods in that Hausdorff factor give disjoint product neighborhoods. Applying these facts to the two ordinal intervals proves the claim about \(K\). \(\square\)

Regularity of the Deleted Plank

A space is regular if, whenever \(x\) is outside a closed set \(C\), there are disjoint open sets containing \(x\) and \(C\), respectively. A space is normal if every two disjoint closed sets have disjoint open neighborhoods. We use these definitions, so regularity here includes separation of a point from a closed set; the plank is also Hausdorff.

Proof. First, every compact Hausdorff space \(K\) is regular. Let \(C\) be closed in \(K\), and let \(x\notin C\). If \(C=\varnothing\), the claim follows by taking \(K\) and \(\varnothing\). Otherwise, for each \(y\in C\), Hausdorffness gives disjoint open neighborhoods \(U_y\) of \(x\) and \(V_y\) of \(y\). Compactness of \(C\) gives finitely many \(V_{y_1},\ldots,V_{y_r}\) covering \(C\). Then \(U=\bigcap_{i=1}^r U_{y_i}\) and \(V=\bigcup_{i=1}^r V_{y_i}\) are open, contain \(x\) and \(C\), respectively, and are disjoint: a point in \(U\cap V\) would belong to \(U_{y_i}\cap V_{y_i}\) for some \(i\). Thus \(K\) is regular.

Regularity passes to subspaces. Indeed, let \(Y\subseteq K\), let \(C\) be closed in \(Y\), and take \(y\in Y\setminus C\). By the definition of the subspace topology, \(C=Y\cap F_0\) for some closed \(F_0\subseteq K\). Since \(y\notin C\) and \(y\in Y\), we have \(y\notin F_0\). Regularity of \(K\) gives disjoint open neighborhoods of \(y\) and \(F_0\); intersecting both with \(Y\) gives disjoint open neighborhoods of \(y\) and \(C\) in \(Y\). Apply this to \(Y=P\subseteq K\). Hausdorffness also passes to subspaces: disjoint ambient neighborhoods of two distinct points restrict to disjoint neighborhoods in the subspace. Since \(K\) is compact Hausdorff, \(P\) is regular Hausdorff. \(\square\)

The Two Edges Cannot Be Separated

Proof. We show that the disjoint closed sets \(E\) and \(F\) have no disjoint open neighborhoods in \(P\). Suppose, to the contrary, that \(U\) and \(V\) are disjoint open subsets of \(P\) with \(E\subseteq U\) and \(F\subseteq V\).

For each \(\alpha<\omega_1\), the point \((\alpha,\omega)\) belongs to \(U\). Since \(U\) is open in the subspace \(P\), there is a product-open neighborhood in \(K\) whose intersection with \(P\) lies in \(U\). Its second-coordinate neighborhood contains a tail \((n_\alpha,\omega]\) for some finite ordinal \(n_\alpha\); by increasing the threshold if necessary, choose \(n_\alpha\in\mathbb{N}\). Thus there is an open neighborhood \(O_\alpha\) of \(\alpha\) in \([0,\omega_1]\) such that \[ O_\alpha\times(n_\alpha,\omega]\ \cap P\ \subseteq U. \] For each \(n\in\mathbb{N}\), let \(S_n=\{\alpha<\omega_1:n_\alpha=n\}\). At least one \(S_n\) is unbounded in \(\omega_1\). Otherwise, for each \(n\) choose a bound \(\delta_n<\omega_1\) for \(S_n\). The supremum of the countably many countable ordinals \(\delta_n\) is countable, hence is less than \(\omega_1\). It would bound their union, which is all of \(\omega_1\), a contradiction.

Fix \(n\) for which \(S_n\) is unbounded, and put \(m=n+1\). The point \((\omega_1,m)\) belongs to \(F\subseteq V\). Since \(m\) is isolated in \([0,\omega]\), and neighborhoods of \(\omega_1\) contain final tails, there is \(\gamma<\omega_1\) such that \[ (\gamma,\omega_1]\times\{m\}\subseteq V. \] Choose \(\alpha\in S_n\) with \(\alpha>\gamma\), possible because \(S_n\) is unbounded. Since \(\alpha\in O_\alpha\) and \(n<m\le\omega\), the point \((\alpha,m)\) lies in \(O_\alpha\times(n,\omega]\). It is not the deleted corner, so it belongs to \(P\), and therefore \((\alpha,m)\in U\). Also \(\alpha>\gamma\), so \((\alpha,m)\in(\gamma,\omega_1]\times\{m\}\subseteq V\). This contradicts \(U\cap V=\varnothing\). Thus \(E\) and \(F\) cannot be separated by disjoint open sets, and \(P\) is not normal. \(\square\)

Why the Example Matters

The deleted plank separates two properties that are easy to conflate. It is Hausdorff and regular, yet it is not normal. Its regularity follows from its position as a subspace of a compact Hausdorff space, but that argument does not establish normality: normality need not pass to arbitrary subspaces. Here the two closed edges make the failure explicit. Each finite horizontal level of a neighborhood of \(E\) can have its own threshold, but only countably many thresholds are available. Since \(\omega_1\) is uncountable, one threshold occurs on an unbounded set of first coordinates; a tail neighborhood of a suitable point of \(F\) then forces an intersection.

When testing separation in a product, verify both coordinates of every proposed point. A strict tail condition such as \(n<y\le\omega\) is different from a singleton condition such as \(y=2\); confusing them can invalidate an otherwise plausible neighborhood argument.