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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. $\blacksquare$

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    $\begingroup$ 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. $\endgroup$ – Henno Brandsma Jun 24 at 6:23
  • $\begingroup$ the notion of being divisible (see here)also seems relevant. $\endgroup$ – Henno Brandsma Jun 24 at 6:26
  • $\begingroup$ @HennoBrandsma Thank you for the confirmation and for the information about uniformity. $\endgroup$ – caffeinemachine Jun 24 at 6:29
  • $\begingroup$ You're welcome. $\endgroup$ – Henno Brandsma Jun 24 at 6:36
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In this old answer (almost a duplicate but not quite IMO) you'll find an alternative shorter proof. I said in the comments: this fact about diagonal neighbourhoods is part of a larger theory on uniformities, in particular the fact that compact Hausdorff spaces have a unique uniformity, i.e. the set of all diagonal neighbourhoods.

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