Brownian motion exit time

Question

I don't understand why $ \{a<B(s)<b, \forall s\in[0,1] \}\subset\{a < B(1)< b\}$. I'm almost certain that this must be a typo in my *book. But, I thought I would confirm it on the math stackexchange because a half-page proof follows which is based on this relation.

Obviously, $ \{a < B(1)< b\}\subset \{a<B(s)<b, \forall s\in[0,1] \}$ !?

*Introduction to Stochastic Calculus with Applications, third edition

Answer 1

Suppose $a< B(s) <b$ for all $s \in [0,1]$. Take $s=1$, so $a<B(1) < b$. The first requirement is more restrictive, and hence gives a smaller set.