Lebesgue Covering Theorem : Suppose $\rho =\{G_n\}$ is a covering of a compact subset $K$ of $\mathbb R^p$. There exists a positive number $\lambda$ such that if $x,y \in K$ and $|x-y| < \lambda,$ then there is a set in $\rho$ containing both $x$ and $y$
Proof:
For each point $u \in K$, there is an open set $G_ {\alpha (u)} \in \rho$ containing $u$.
Let $\delta(u) >0$ be such that $|v-u|<2~ \delta(u),$ then $v \in G_ {\alpha (u)} ~~.....(1)$
Now, consider the open set $S(u)=\{v \in \mathbb R^p :|v-u|< \delta (u)\}$ and the collection $\gamma=\{S(u):u \in K\} $ of open sets.
Since, $S$ is a covering of the compact set $K$, then $K$ is contained in the union of a finite number of the sets in $S$, say in : $\{S(u_1),\cdots,S(u_n)\}~~.......(2)$.
We now define $\lambda$ to be the positive real number $:\lambda = \inf \{\delta(u_1),\cdots, \delta(u_n) \}$
If $x,y \in K$ and $|x-y|<\lambda$, then $x \in S(u_j) $ for some $j :1 \leq j \leq n$ so that $|x-u_j|<\delta(u_j)~~.........(3)$.
Since, $|x-y|<\lambda$, we have $|y-u_j| \leq |y-x|+|z-u_j| < 2 \delta(u_j) $.
According the definition of $\delta (u_j),$ we infer that both $x,y \in G_{\alpha(u_j)}$
Query:
EDIT:
Will an open set of radius $2 ~\delta(u)$ be always enclosed in the cover $\{G_n\} ?$
$(a)~~$As per my understanding, then open balls $\{B(u_1,\delta(u_1) ), ~~B(u_2,\delta(u_2)), \cdots,~~B(u_n,\delta(u_n))\}$ constitute an open finite cover of $K$.
Now, from $(3):$ If $x,y \in K$ and $|x-y|<\lambda$, then $x \in S(u_j) $
for some $j :1 \leq j \leq n$ so that $|x-u_j|<\delta(u_j)$ where $\lambda = \inf \{\delta(u_1),\cdots, \delta(u_n) \}$
But, this may not be always true as it maybe possible that if $|x-y|<\lambda$, then, $x$ is very near the boundary of one open set and $y$ is near the boundary of an another open set and the distance between them is $\lambda$.
$(b)~~$ I don't understand well, though, I do have a general idea, of the use of specifically defining $2~~\delta(u) >0$ be such that $|v-u|<2~ \delta(u),$ then $v \in G_ {\alpha (u)} $ . Please guide me on the same.
EDIT: $(c)~~$ We defined : Let $\delta(u) >0$ be such that $|v-u|<2~ \delta(u),$ then $v \in G_ {\alpha (u)}$ , where $G_{\alpha (u)}$ is an open set.
But, isn't it possible that $ |v-u|<2~\delta(u)$ but, $u,v$ do not belong to one common set, for example, when they are near the boundaries of two open sets : $G_ {\alpha (u_1)} $ and $G_ {\alpha (u_2)} $
Where could I be going wrong?
Thank you for your help
For (a), you can find such a $j$ simply because $\{ S(u_1), \dots ,S(u_n)\}$ is a covering of $K$. This has nothing to do with the fact that $\vert x-y\vert <\lambda$.
As for (b), you take $2\delta (u)$ to have a little more space. This ensures that if $x$ is in $S(u_j)$ and $y$ satisfies $\vert x-y\vert <\lambda$, hence $\vert x-y\vert<\delta(u_j)$, then $y$ will still be in $G_{\alpha(u_j)}$.