Let $X = U \cup V$ be a topological space where $U$ and $V$ are open sets in $X$. Let $f : [0,1] \longrightarrow X$ be a loop in $X$. Then does there exist a subdivision $b_0,b_1,b_2, \cdots , b_m$ of $[0,1]$ such that $f(b_i) \in U \cap V$ and $f([b_{i-1},b_i])$ is contained either in $U$ or in $V$ for each $i=1,2, \cdots,m$?
Since $[0,1]$ is compact so by Lebesgue number lemma it is not hard to see that for the open cover $\{f^{-1} (U), f^{-1} (V) \}$ there exists $\delta > 0$ such that for any subset of $[0,1]$ having diameter less than $\delta$ is contained either in $f^{-1} (U)$ or in $f^{-1} (V)$. Now let us take a subdivision $b_0,b_1,b_2, \cdots , b_m$ of $[0,1]$ with norm or mesh less than $\delta$. Then for each $i$, $([b_{i-1},b_i])$ is contained either in $f^{-1} (U)$ or in $f^{-1} (V)$ i.e. for each $i$, $f([b_{i-1} , b_i])$ is contained either in $U$ or in $V$. Now if for each $i$, $f(b_i) \in U \cap V$ then we are through. If it is not the case then what will we do? Munkres in his book "Topology" suggests that if there exists some $i$ for which $f(b_i) \notin U \cap V$ then we can eliminate that $b_i$ from the subdivision to obtain a new subdivision $c_0,c_1,c_2, \cdots , c_{m-1}$ of $[0,1]$ which still satisfies the condition that $f([c_{i-1},c_i])$ is contained either in $U$ or in $V$ for each $i$. A finite number of repetitions of this process leads us to the desired subdivision.
Why is it so? If we omit $b_i$ from the subdivision then the length of the subinterval $[b_{i-1},b_{i+1}]$ may exceed $\delta$. Then how can we apply Lebesgue number lemma to show that $f([b_{i-1} , b_{i+1}])$ is contained either in $U$ or in $V$? I wonder! Please help me in this regard.
Thank you very much.
If $f(b_i)$ is not in $U\cap V$ then both sets $f([b_{i-1},b_i])$ and $f([b_i,b_{i+1}])$ lies in $U$ or both lies in $V$. If this is not the case, then WLOG we can assume that $f([b_{i-1},b_i])\subseteq U$ and $f([b_i,b_{i+1}])\subseteq V$ from where $f(b_i)\in U\cap V$, contrary to the assumption that $f(b_i)$ is not in $U\cap V$. Thus we can eliminate the point $b_i$ by considering the interval $[b_{i-1},b_{i+1}]$ which, by the above, has its image under $f$ in one of the sets $U$ or $V$.