Is there any generalization of Lebesgue's number lemma? I have encountered a problem in learning algebra-topology these days. There is a basic and important result about the fundamental group: the fundamental group of $S^1$ is $\mathbb{Z}$. The proof of this conclusion uses the path lifting lemma and homotopy lifting lemma, and the proof of these two lemmas uses another lemma called Lebesgue's number lemma.
Lebesgue's number lemma:
If the metric space $(X,d)$ is compact and an open cover of $X$ is given, then there exists a number $\delta>0$ such that every subset of $X$ having diameter less than $\delta$ is contained in some member of the cover.
Such a number $\delta$ is called a Lebesgue number of this cover. The notion of a Lebesgue number itself is useful in other applications as well.
What I want to ask is: does this lemma have generalization form in more general space, such as compact non-metric space?
 A: Yes, but we need an additional structure to have a measure for the "diameter" of a set. A metric on $X$ does this very nicely. A well-known generalization of the concept of a metric space is that of uniform space. See https://en.wikipedia.org/wiki/Uniform_space. I shall not go into the details of the definition here, see the Wikipedia article or consult a textbook. The key concept is that of an entourage of the diagonal. A uniform structure $\mathcal U$ on a set $X$ is a collection of such entourages subject to suitable axioms. If $(x,y) \in V \in \mathcal U$, we symbolically write $\lvert x - y \rvert < V$. If $M \subset X$ has the property $M \times M \subset V$, we write $\text{diam} M < V$. This means that for all $x,y \in M$ we have $\lvert x - y \rvert < V$.
Here are some basic facts.


*

*Each uniform structure on a set $X$ induces a unique topology on $X$. A subset $G \subset X$ is open in this topology if each $x \in G$ admits an entourage $V$ such that the set $B(x,V) = \{ y \in X \mid \lvert x -y \rvert < V \} = \{ y \in X \mid (x,y) \in V \}$ of $x$ is contained in $G$. Note that $B(x,V)$ is not necessarily open in $X$, but it is easy to see that it contains an open neighborhood of $x$. Thus $B(x,V)$ can legitimately be called the $V$-neighborhood of $x$.

*Each metric $d$ on $X$ induces a natural uniform structure on $X$. A basis for this uniform structure is given by the sets $V_\varepsilon = \{ (x,y) \in X \times X \mid d(x,y) < \varepsilon \}$. Thus by definition $\lvert x - y \rvert < V_\varepsilon$ is equivalent to $d(x,y)  < \varepsilon$ and $\text{diam} M < V_\varepsilon$ means that $d(x,y)  < \varepsilon$ for all $x,y \in M$.

*Each compact Hausdorff space has a unique uniform structure inducing its topology. A basis for this uniform structure is given by the open neighborhoods of the diagonal in the space $X \times X$ with the product topology.
Let us now prove
Let $X$ be a compact uniform space. Then each open cover $\mathfrak U$ of $X$ admits an entourage $V$ such that each $M \subset X$ with $\text{diam} M < V$ is contained in a member of $\mathfrak U$.
To see this, note that for each $x \in X$ there exists an entourage $V_x$ such that the $V_x$-neighborhood $B(x,V_x)$ of $x$ is contained in some member of $\mathfrak U$. Choose an entourage $W_x$ such that $W_x \circ W_x = \{ (x,y) \in X \times X \mid \ \exists z \in X : (x,z),(z,y) \in W_x \} \subset V_x$. Since $X$ is compact, there exist finitely many $x_i$ such that $\bigcup_i B(x_i,W_{x_i}) = X$. Then $W = \bigcap W_{x_i}$ is an entourage. Now let $\text{diam} M < W$ and $\xi \in M$. We have $\xi \in B(x_i,W_{x_i})$ for some $i$, i.e. $(x_i,\xi) \in W_{x_i}$. For $x  \in M$ we have $(\xi,x) \in W \subset W_{x_i}$, thus $(x_i,x) \in W_{x_i} \circ W_{x_i} \subset V_{x_i}$ which means $x \in B(x_i,V_{x_i})$. Therefore $M \subset B(x_i,V_{x_i})$. The latter is contained in some member of $\mathfrak U$.
