# A halving neighborhood theorem for compact Hausdorff spaces.


# Question

Let $$X$$ be a compact Hausdorff space and $$\Delta_2(X)$$ denote the set $$\set{(x, x):\ x\in X}$$ in $$X\times X=:X^2$$. I want to confirm that the following is true (a proof if supplied below).

Theorem 1. Let $$X$$ be a compact Hausdorff space and $$O$$ be a neighborhood of $$\Delta_2(X)$$ in $$X^2$$. Then there is a neighborhood $$Q$$ of $$\Delta_2(X)$$ such that whenever $$(x, y)$$ and $$(y, z)$$ are in $$Q$$ for some $$x, y, z\in X$$, we have $$(x, z)\in O$$.

The motivation for this is to generalize, to compact Hausdorff spaces, the following fact about metric spaces that if "$$d(x, y), d(y, z)< \varepsilon/2$$ then $$d(x, z)< \varepsilon$$." My larger gaol was to have a device which allows mimicking proofs in topological dynamics for compact metric spaces to arbitrary compact Hausdorff spaces.

The purpose of this post is two-fold. One is to verify my proof below, and the other is to get a shorter proof of the theorem above. (If you do not want to read my proof and supply your own proof then please go ahead and share!) I am somewhat apprehensive about my proof since it is longer that what seems necessary and also that it took me many iterations to get the details right, for I had incorrectly proven it multiple times in the process.

# Purported Proof

Lemma 2. Let $$X$$ be a compact Hausdorff space and $$A$$ be a closed set in $$X$$. Let $$U$$ be neighborhood of $$A$$. Then there is a neighborhood $$O$$ of $$A$$ in $$X$$ such that $$\bar O\subseteq U$$.

Proof. Restatement of the fact that compact Hausdorff spaces are normal.

Lemma 3. Let $$X$$ be a compact Hausdorff space and $$O$$ be a neighborhood of $$\Delta_2(X)$$ in $$X^2$$. Then there is an open cover $$\mc V$$ of $$X$$ such that $$(V\cup V')\times (V\cup V') \subseteq O$$ whenever $$V, V'\in \mc V$$ are such that $$V\cap V'\neq \emptyset$$.

Proof. We say that an open cover $$\mc U$$ of $$X$$ if good if $$\overline{\bigcup_{U\in \mc U} U\times U}$$ is contained in $$O$$. It is clear from Lemma 2 and from compactness of $$X$$ that finite good open covers of $$X$$ exist. Also, given an open cover $$\mc U$$ of $$X$$, we say that $$G$$ in $$\mc U$$ is well-behaved if whenever $$G\cap U\neq \emptyset$$ for some $$U$$ in $$\mc U$$, we have $$(G\cup U)\times (G\cup U)$$ is contained in $$O$$.

Let $$\mc U=\set{U_1, \ldots, U_m, G_1, \ldots, G_n}$$ be a be an arbitrary finite good open cover of $$X$$, where each $$G_i$$ is well-behaved and each $$U_i$$ is not well-behaved. If $$m=0$$ then we are done. So assume that $$m\geq 1$$. It automatically follows that then $$m\geq 2$$. We will construct a finite good open cover of $$X$$ which has fewer ill-behaved elements. By Lemma 2 we know that there is a neighborhood $$Q$$ of $$\Delta_2(X)$$ which contained $$\overline{\bigcup_{U\in \mc U} U\times U}$$ such that $$\bar Q\subseteq O$$.

Let $$K$$ be the boundary of $$U_1\cup \cdots \cup U_{m-1}$$. For each $$p$$ in $$K$$, let $$W_p$$ be a neighborhood of $$p$$ in $$X$$ such that $$W_p$$ is contained in $$G_i$$ whenever $$W_p\cap G_i\neq \emptyset$$, and $$(W_p\cup U_i)\times (W_p\cup U_i)\subseteq Q$$ whenever $$W_p\cap U_i$$ is not empty. The existence of $$W_p$$ can be established by a compactness argument. Since $$K$$ is compact, there is a finite set $$F$$ of $$K$$ such that $$\set{W_p:\ p\in F}$$ covers $$K$$. Define $$U_m'=U_m\setminus \overline{U_1\cup \cdots \cup U_{m-1}}$$ and $$\mc U' = \set{U_1, \ldots, U_{m-1}, U_m', G_1, \ldots, G_n} \cup \set{W_p:\ p\in F}$$ It is easy to check that $$U_m'$$ as well as each $$G_i$$ is well-behaved in $$\mc U'$$. Also, $$\overline{\bigcup_{U'\in \mc U'} U'\times U'}$$ is contained in $$Q$$, and hence $$\mc U'$$ is a good open cover. This finishes the proof. $$\blacksquare$$

Theorem 4. Let $$X$$ be a compact Hausdorff space and $$O$$ be a neighborhood of $$\Delta_2(X)$$ in $$X^2$$. Then there is a neighborhood $$Q$$ of $$\Delta_2(X)$$ such that whenever $$(x, y)$$ and $$(y, z)$$ are in $$Q$$ for some $$x, y, z\in X$$, we have $$(x, z)\in O$$.

Proof. Let $$\mc U$$ be an open cover of $$X$$ such that whenever $$U$$ and $$U'$$ in $$\mc U$$ are such that $$U\cap U' \neq \emptyset$$, we have $$(U\cup U')\times (U\cup U')$$ is contained in $$O$$. Such an open cover is furnished by Lemma 3. Define $$Q=\bigcup_{U\in \mc U} U\times U$$. Now let $$(x, y)$$ and $$(y, z)$$ be in $$Q$$ for some $$x, y, z\in X$$. Then there are $$U$$ and $$U'$$ in $$\mc U$$ such that $$(x, y)\in U\times U$$ and $$(y, z)\in U'\times U'$$. Thus $$U\cap U'$$ is non-empty, and thus $$(U\times U')\times (U\times U')$$ is contained in $$O$$. But since $$(x, z)$$ is in $$(U\cup U')\times (U\cup U')$$, we see that $$(x, z)\in O$$, and we are done. $$\blacksquare$$

• Look up the theorem that the set of neighbourhoods of $\Delta_2(X)$ forms a uniformity for compact Hausdorff $X$. It's in fact a step in the proof of the theorem that a compact Hausdorff space has a unique compatible uniformity. A uniformity is the notion you're looking for. The "halving" fact is one of its axioms. – Henno Brandsma Jun 24 at 6:23
• the notion of being divisible (see here)also seems relevant. – Henno Brandsma Jun 24 at 6:26
• @HennoBrandsma Thank you for the confirmation and for the information about uniformity. – caffeinemachine Jun 24 at 6:29
• You're welcome. – Henno Brandsma Jun 24 at 6:36