# Covering Lemma 1

Internet was a bit weird lately, and I didn’t want to lose a post half-way through, so I decided to wait in writing this. I seem to have a weakness when it comes to figuring out how metric properties and measure properties interplay. It is almost inevitable that you will need to call upon some “covering lemma” to get the job done. These are used extensively in differentiation, but also less-known in generally defining measures that interact nicely if your space already has a metric.

Lebesgue Number: Given a compact metric space $(X, d)$, then for any open cover, there exists $\delta>0$ such that if a set has diameter less than $\delta$ it is contained in one of the members of the cover.

Proof: Really you just do all the stadard tricks, and then tie them together. So let $\{U_i\}_{i=1}^n$ be a finite subcover of our cover. (We ignored the whole space being a member for trivial reasons). Now define $C_i=X\setminus U_i$ and a function $f: X\to \mathbb{R}$ that basically averages the distances to each C_i. This gives us $f(x)=\frac{1}{n}\sum_{i=1}^nd(x,C_i)$. Now for any x, we know that $x\in U_i$ for some i, and this set is open so there is an $\epsilon>0$ small enough so that $B_\epsilon (x)\subset U_i$, thus $d(x, C_i)\geq\epsilon$ yielding $f(x)\geq \frac{\epsilon}{n}$.

Since f is continuous and X compact, f achieves its min. We call this $\delta$. Now if E is a set with diameter less than $\delta$, then we can pick any $x_0\in E$ and get that $B_\delta (x_0)\supset E$. Now $\delta\leq f(x_0)\leq d(x_0, C_k)$ where $C_k$ is chosen as the maximal distance to any of the C’s. But then we have it since $U_k=X\setminus C_k\supset E$. Thus any set of diameter less than delta is contained in a member of the covering.

There are some interesting other ways of proving this. This is I think considered the “standard” since Munkres does it this way.

OK. So I guess I’ll do a series of these. I was going to do two in this post, but I don’t feel like doing the other now that I just typed that out. Also, if I do a series I can do the three that I feel are most important instead of just the two easier ones.