Finite union of open/closed set is algebra.

Let $\displaystyle \xi = \{ (a,b] : a<b,a,b \in \mathbb {R} \}$

Let A be the finite union of elements from $\displaystyle \xi$

Prove that A is an algebra.

Proof so far.

Let $\displaystyle E \in A$, then $\displaystyle E = \bigcup _{k=1}^n (a_k,b_k]$

I need to show that:
1. $\displaystyle E^c \in A$
2. $\displaystyle \bigcup ^m _{n=1} E_n \in A$

Claim 2 I think should be pretty straight forward, $\displaystyle \bigcup ^m _{n=1} E_n \in A = \bigcup ^m _{n=1} \bigcup _{k=1}^n (a_k,b_k] \in A$ since the union of unions of those sets should still be in the same form.

Claim 1 I'm in trouble, I have the following:

$\displaystyle E^c= \{ \bigcup _{k=1}^n (a_k,b_k] \} ^c = \bigcap _{k=1}^n(a_k,b_k]^c= \bigcap _{k=1}^n((- \infty ,a_k] \cup (b_k , \infty ))$

$\displaystyle = \bigcap _{k=1}^n( - \infty ,a_k] \cup \bigcap _{k=1} ^n (b_k, \infty )$

Now, will this be equal to $\displaystyle \{ \bigcup _{r>0} \bigcap _{k=1}^n (a_k-r,a_k] \} \cup \{ \bigcap _{r>-} \bigcap _{k=1}^n (b_k,b_k+ \frac {r}{n} ] \}$? In which case I can conclude it is indeed in A?

Thank you.