Eα = [0, α) ∩ Q = {r ∈ Q : 0 ≤ r < α} for each α ∈ (0, 1). Compute U Eα and ∩Eα.
I tried to solve this problem with the sequence of α.
Guess the first one is [0,1) ∩ Q and the other {0}
Is there any hint to figure it out with concept of openness or interior points?
By definition,
$$\bigcup_{\alpha\in[0,1)}E_\alpha=\left\{\,r\in\Bbb Q\;|\;0\le r<\alpha\,,\,\text{for some}\;\;\alpha\in[0,1)\,\right\}=[0,1)\cap\Bbb Q$$
Or also
$$x\in[0,1)\cap\Bbb Q\implies0\le x<1,\,\text{and}\,x\in\Bbb Q\implies \exists\alpha\in[0,1) \left(\text{ for example}\,\alpha=x+\frac{1-x}2=\frac{x+1}2\right)$$
$$\text{s.t.}\;0\le x<\alpha\implies x\in E_\alpha\implies x\in\bigcup_{\alpha\in[0,1)}E_\alpha$$
About the intersection:
$$x\in\bigcap_{\alpha\in[0,1)}E_\alpha\implies\forall0\le\alpha<1\;,\;\;0\le x<1\;\text{and}\;x\in\Bbb Q\implies x=0$$
or also
$$0<x<1\;\text{and}\;x\in\Bbb Q\implies \exists \alpha\in (0,1)\;\left(\text{for example}\;\alpha'=\frac x2\right)\;\text{s.t.}$$
$$x\notin E_{\alpha'}\implies x\notin\bigcap_{\alpha\in[0,1)}E_\alpha$$
So you were right...