Hatcher's proof of proposition (c) for covering spaces

Question

I am currently going through the proof of Hatcher's Algebraic Topology, and I am having some difficulty regarding one of his assertions in the section of the proof given below: I am unable, in particular, to understand his reasoning behind his last statement. Why is it that the compactness of $\{y_{0}\} \times I$ implies that we need only to consider a single neighbourhood $N$ such that $F(N \times [t_{i},t_{i+1}])$ is always contained in $U_{i}$?

Answer 1

I'd say he's taking $N$ as the intersection of all the $N_t$'s. It's a finite number of them, so it's still open.

Answer 2

Let $y_0\in Y$.

For every $t\in I$, there is an evenly covered (open) neighborhood of $F(y_0,t)$, let that be $U_t$.

$F$ is continuous, so there exists an open neighborhood around $(y_0,t)$ whose image is contained in $U_t$. This open neighborhood can be assumed to be of the form $N_t\times(a_t,b_t)$, for some open set $N_t$, since these form a basis for the product topology.

Since we're doing this for every $t\in I$, clearly we have $$\{y_0\}\times I\subset\bigcup_{t\in I}N_t\times(a_t,b_t)$$ i.e., this is an open cover of $\{y_0\}\times I$. But $\{y_0\}\times I$ is compact so, by definition of compactness, we can still cover it by just finitely many of these sets, i.e. $$\{y_0\}\times I\subset\bigcup_{t\in J}N_t\times(a_t,b_t)$$ for some finite $J\subset I$. In particular $$I\subset\bigcup_{t\in J}(a_t,b_t)$$ Now I'm not gonna prove this, but if you think about it, it's pretty intuitive that this means you have a finite partition $0=t_1<\ldots<t_m=1$ such that for every $i$, you have $[t_i,t_{i+1}]\subset(a_t,b_t)$ for some $t\in J$. Furthermore $$N:=\bigcap_{t\in J}N_t$$ is open since the intersection is finite and, for each $i$, $$N\times[t_i,t_{i+1}]\subset N_t\times(a_t,b_t)\implies F\left(N\times[t_i,t_{i+1}]\right)\subset F\left(N_t\times(a_t,b_t)\right)\subset U_t$$ for some $t\in J$. He just denoted $U_i:=U_t$ for this $t$.