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Let S be the set system consisting of the finite unions of bounded real intervals. It will now be checked if in $X=\mathbb{R}$ a) S is a ring, b) S is an algebra, c) S is a $\sigma$ ring

a) Let $A,B \in S$ with $A= \cup_{k}^{n}I_{k} ; B= \cup_{k}^{n}L_{k}$, where both I and L are sets and $k,n\in \mathbb{N}$. Then: $A\cup B = (\cup_{k}^{n} I_{k})\cup (\cup_{k}^{n}L_{k}) = \cup_{k}^{n}(I_{k}+L_{k}) \in S$ and $A-B = \cup_{k}^{n} I_{k} - \cup_{k}^{n}L_{k} = \cup_{k}^{n}(I_{k}-L_{k}) \in S$

and thus S is a ring in X.

b) since we showed that S is a ring, it is to show that $A\in S \Rightarrow A^{c}\in S$

$A^{c}=X\backslash(\cup_{k}^{n}I_{k})= \cup_{k}^{n}(X\backslash I_{k})\in S $

c) for this it is to show that for $(A_{n})_{n\ge 1} \in S$ also $\cup _{k=1}^{\infty}A_{n}$

Don't see a way to show this.

Does somebody see the right way. Please tell me.

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For b), are you sure that the complement of a bounded interval can be written as a finite union of bounded intervals? For c), put $A_k:=\left[2k,2k+1\right]$. –  Davide Giraudo Nov 2 '11 at 20:37
    
A couple of things: what is $I_k+L_k$? Is it just $I_k\cup L_k$? And is $A-B$ the set of elements of $A$ that aren't in $B$ (also denoted $A\backslash B$)? If so, it's not true that $\cup I_k \backslash \cup L_k$ is the same set as $\cup I_k\backslash L_k$: try $I_1=J_2=(0,2)$ and $I_2=J_1=(1,3)$. –  Shane O Rourke Nov 2 '11 at 20:45
    
(b) If $I$ and $J$ are disjoint, non-empty, bounded intervals in $\mathbb{R}$, $(\mathbb{R}\setminus I)\cup(\mathbb{R}\setminus J)=\mathbb{R}\ne\mathbb{R}\setminus(I\cup J)$. –  Brian M. Scott Nov 2 '11 at 21:08
    
(a) Presumably you allow the empty interval $(p,p)$ and the one point interval $[p,p]$. (b) A finite union of bounded intervals is bounded, so the complement of none of your sets is $S$; (c) A countable union of bounded intervals need not be bounded. –  André Nicolas Nov 2 '11 at 23:42

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