Question 10.T.3: Let I0 = {(a, b) : a, b ∈ R}, I1 = {[a, b] : a, b ∈ R}, I2 =...

It suffices to prove that

\mathcal{B} ⊇ \mathcal{A(I}_{0}) ⊇ \mathcal{A(I}_{1}) ⊇ \mathcal{A(I}_{2}) ⊇ \mathcal{A(I}_{3}) ⊇ \mathcal{B}.

For the first inclusion, note that

(a, b) = \overset{∞}{\underset{n=1}{\cup}} [a +1/n, b) ,

which proves that \mathcal{B} ⊇ \mathcal{I}_{0}, and hence \mathcal{B} ⊇ \mathcal{A(I}_{0}). The second inclusion follows from the observation that

[a, b] = \overset{∞}{\underset{n=1}{\cap}} (a − 1/n, b + 1/n)

⇒ \mathcal{A(I}_{0}) ⊇ \mathcal{I}_{1}

⇒ \mathcal{A(I}_{0}) ⊇ \mathcal{A(I}_{1}).

Similarly, the equalities

(a, ∞) = \overset{∞}{\underset{n=1}{\cup}} [a + 1/n, n]

[a, ∞) = \overset{∞}{\underset{n=1}{\cap}} (a − 1/n, ∞)

[a, b) = [a, ∞)\setminus[b, ∞)

lead to the remaining inclusions.