# Exercise on refinement of a cover

For any cover $$\cal{U}$$ of a topological space $$X$$ there is a finest partition $$P(\cal{U})$$ refined by $$\cal{U}$$. Each element of $$\cal{U}$$ is contained in a unique element of $$P(\cal{U})$$. Willard, General topology, Ex. 36A. I ask for a proof please. Also if we take the cover $$[0,1/3]\cup[1/4,2/3]\cup[1/2,1]$$ of $$[0,1]$$ which is the finest partition of the proposition?

For your specific part: suppose that $$\mathcal{P}$$ is a partition of $$[0,1]$$ that refines your cover $$\{[0,\frac13], [\frac14,\frac23],[\frac12,1]\}$$.

Then some $$P_1 \in \mathcal{P}$$ satisfies $$[0,\frac13] \subseteq P_1$$ and also some $$P_2 \in \mathcal{P}$$ exists such that $$[\frac14,\frac23]\subseteq P_2$$. But $$\frac14$$ lies in both and so this forces $$P_1=P_2$$ (two elements of a partition are disjoint or equal). Similarly, a $$P_3$$ containing $$[\frac12,1]$$ must equal $$P_2$$ (and thus $$P_1$$!) as $$\frac12$$ is in both. It follows that $$P_1=P_2=P_3$$ and the partition $$\mathcal{P}$$ equals the quite boring $$\{[0,1]\}$$.

This gives an idea for a partition (which is really the set of classes of an equivalence relation) for refining any $$\mathcal{U}$$:

Define $$x \sim y$$ iff there is a "path in $$\mathcal{U}$$ from $$x$$ to $$y$$":

$$x \sim y \iff \exists n\ge 1, \exists U_1, \ldots, U_n \in \mathcal{U}: (x \in U_1) \land (y \in U_n) \land (\forall 1 \le i \le n-1: U_i \cap U_{i+1} \neq \emptyset\tag{1}$$

and let $$\mathcal{P}$$ be the set of equivalence classes. It's indeed easy to check that this defines an equivalence relation on $$X$$ and that the classes refine the members of $$\mathcal{U}$$ (if $$x,y \in U \in \mathcal{U}$$ then $$x \sim y$$) and is as required.

Bonus theorem: $$X$$ is connected iff for every open cover $$\mathcal{U}$$ of $$X$$ and every $$x,y \in X$$ there is a path in $$\mathcal{U}$$ from $$x$$ to $$y$$. This is quite handy in some proofs involving local compactness, or local (path-)connectedness etc.