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?


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.