My problem:

Suppose $(X,d)$ is a metric space and $A=\bigcup_{i \in I}B(x_i,r_i) \subset X$ is totally bounded where $B(x,r)=\{y \in X: d(x,y)<r \}$.

I know that if $A$ is totally bounded, $A \subset \bigcup_{j \in \mathbb{N}}B(y_j,3q_j)$ such that the balls $B(y_j,q_j)$ are mutually disjointed.

But in my notes I wrote that if $A=\bigcup_{i \in I}B(x_i,r_i)$ where $I$ is arbitrary, then we can conclude that $A=\bigcup_{i \in I}B(x_i,r_i) \subset \bigcup_{n \in \mathbb{N}} B(x_{i_n},5r_{i_n})$ such that $B(x_{i_n},r_{i_n})$ are mutually disjointed.

In other words we can use the same balls of $A$ to obtain the covering property.

  • $\begingroup$ check Vitali's Theorem in the book Measure Theory and Fine Properties of Functions, Evans-Gariepy $\endgroup$ – alphaomega Jul 16 at 12:27

Your Answer

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

Browse other questions tagged or ask your own question.