User:IssaRice/Lebesgue number lemma and Tao's proof that sequential compactness implies covering compactness

One typical statement of the Lebesgue number lemma is the following: Let $(X,d)$ be a metric space, and let $Y$ be a sequantially compact subset of $X$. Let $(V_{\alpha }{)}_{\alpha \in A}$ be a collection of open sets which covers $Y$. Then there exists a number $\lambda >0$ such that for each $y\in Y$ there is some $\alpha \in A$ such that $B(y,\lambda )\subseteq V_{\alpha }$. Here $\lambda$ is called a Lebesgue number (any smaller positive number is also a Lebesgue number, so this number is not unique). Thus the Lebesgue number lemma asserts the existence of a Lebesgue number.

The idea is that we can find some fixed radius $\lambda$, and consider balls $B(y,\lambda )$ as we move around in $Y$. We might think that at some of these places, our ball $B(y,\lambda )$ will be "broken up" by not being entirely within some single $V_{\alpha }$. But the Lebesgue number lemma says this is not the case, that we can move around freely within $Y$ and always have our balls entirely living inside some $V_{\alpha }$ (of course, the specific index $\alpha$ may change as we move around). The balls $B(y,\lambda )$ are "basically points", but with the helpful property that they have positive diameter.

In Analysis II, Terence Tao shows something similar when he proves that sequential compactness implies covering compactness for metric spaces. Specifically, he defines for each $y\in Y$ the number $r(y):=sup\{r>0:B(y,r)\subseteq V_{\alpha }\text{ for some }\alpha \in A\}$ (the set here is non-empty because the collection covers $Y$, so $r(y)$ is positive). Then he takes the infimum of all these $r(y)$s, $r_0:=inf\{r(y):y\in Y\}$. He then shows that $r_0>0$ (this is the content of Case 1 of his proof).

It looks like $r_0$ is playing the role of $\lambda$ in the usual statement of the Lebesgue number lemma, but this is not exactly right (rather, any number less than $r_0$ is a Lebesgue number).

But in fact, the statement that $r_0>0$ is equivalent to the Lebesgue number lemma, and this is not hard to show.

Suppose $r_0>0$. Let $y\in Y$ be arbitrary. Then $r(y)\in \{r(y):y\in Y\}$ so $r(y)\ge r_0$. Now consider any number $\lambda >0$ such that $\lambda <r_0$. We have $r(y)=sup\{r>0:B(y,r)\subseteq V_{\alpha }\text{ for some }\alpha \in A\}\ge r_0>\lambda >0$. The important thing is that $\lambda$ cannot be an upper bound for the set $\{r>0:B(y,r)\subseteq V_{\alpha }\text{ for some }\alpha \in A\}$. Thus there exists some $r$ in this set such that $r>\lambda$, which means that $B(y,\lambda )\subseteq B(y,r)\subseteq V_{\alpha }$ for some $\alpha \in A$. We have just shown the Lebesgue number lemma.

Now suppose the Lebesgue number lemma is true. Then we are given some $\lambda >0$. Thus if $y\in Y$ then there is some $\alpha \in A$ such that $B(y,\lambda )\subseteq V_{\alpha }$. Since $r(y)$ is the supremum of such radii, we have $r(y)\ge \lambda$. Now taking the infimum over $y$, we have $r_0\ge \lambda >0$.

Actually I think maybe $r_0=sup\{\lambda :\lambda \text{ is a Lebesgue number}\}$.