Suppose $X$ is uniform on $(0,1)$ and define the events $E_{a} = \{ X \neq a \}$. Show that $P(E_a) = 1$ but $P( \bigcap_a E_a ) = 0$.
Attempt
First, notice that
$$ P(E_a) = P(X \neq a) = P(X < a \; \text{or} \; X > a) = \int_0^a dx + \int_a^1 dx = a + (1-a) = 1 $$
Now, as for the second is where I get stuck. We have that
$$ \bigcap_a E_a = \bigcap_a \{ X <a \} \cup \{ X > a \} $$
but don't we get $1$ if we intersect all this events? Im having hard time visualizing this. If we proceed at say pick $E \in \bigcap_a E_a$, then
for all $a$, we have that $E= \{ X <a \} \cup \{X > a \} $. How do we obtain the empty set from here?
You didn't specify what values $a$ can take, but I assume that $a \in A \supseteq (0,1)$. Then, $\bigcap_a E_a$ is the event that $X$ does not take any values in $A$. In other words, $\omega \in \bigcap_A E_a$ implies that for all $a \in A$ $X(\omega) \neq a$. But $X$ is defined to be uniform on $(0,1)$. So $\bigcap_a E_a = \emptyset$.
You also asked: "Don't we get $1$ if we intersect this?" It seems like you're making a category error here. Each $E_a$ is a subset of an underlying sample space $(\Omega, \mathcal{F})$, not necessarily $(0,1)$. To say that $X$ is uniform on $(0,1)$ means that $X$ is a measurable function with domain $(\Omega, \mathcal{F})$ and codomain $(0,1)$ such that, for all subintervals $I$ of $(0,1)$, $P(X \in I)$ is equal to the length of $I$.