Probability of a non-countable union of events of probability zero

Question

I can't understand a logical step from A First Course in Stochastic Processes by Karlin and Taylor. They define a random variable $U$ uniformly distributed on $[0,1]$ and a stochastic process $X_t = 1$ for $U = t$ and $X_t = 0$ otherwise, with a continuous index $t$. Then they note that ‘obviously’, $\operatorname{Pr}\{X_t \le \frac{1}{2}$ for all $0 \le t \le 1\} = 0$. How is this probability computed?

Paths {$X_t$} will consist strictly of zeros for all finite-dimensional sets of $t_i$ (and for countable infinite-dimensional as well, I guess), but they seem to include $1$ with probability $1$ when the whole interval is considered.

Answer 1

Whatever value $U$ takes, it is equal to some $t_0 \in [0,1]$. Therefore with probability $1$, $X_{t_0}=1$ for some value of $t_0$. This implies that not all of the $X_t$ can be simultaneously $ \leq \frac{1}{2}$.