I was given a theorem in the section continuity of function in the book:
Theorem: Given a closed interval $[a,b]$ and an open cover $\phi$ of $[a,b]$, $\exists \lambda(\phi)$ such that $\forall A \subset [a,b]$ with $d(A):=sup\{|x_i-x_j||x_i,x_j \in A\} < \lambda$, then there is $I \in \phi$ such that $A \subset I$. First How to prove it? I was thinking $\lambda := inf\{|a_i-b_j|\}$ where $a_i,b_j$ are end points of some intervals in $\phi$.
Also, how is this theorem relate to continuity of functions?
I was thinking defining $f(x)=x$ on $[a,b]$ and then for all $\epsilon>0$, there is an open cover $I$ of $[a,b]$ such that $inf\{|a_i-b_i||(a_i,b_i)\in I\}<\epsilon$. Ok then there is $\lambda$ such that $\forall A \subset [a,b]$ with $d(A) < \lambda$, then there is $S \in I$ such that $max(A)-min(A) < \epsilon$.
