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 (this latter property is what allows us to extract a finite subcover of $(V_{\alpha })_{\alpha \in A}$).

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 $r(y)\geq r_{0}$. Now consider any number $\lambda >0$ such that $\lambda <r_{0}$. We have $r(y)=\sup\{r>0:B(y,r)\subseteq V_{\alpha }{\text{ for some }}\alpha \in A\}\geq r_{0}>\lambda >0$. 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 $r(y)\geq \lambda$. Now taking the infimum over $y$, we have $r_{0}\geq \lambda >0$.

Actually I think maybe $r_{0}=\sup\{\lambda :\lambda {\text{ is a Lebesgue number}}\}$.