1
$\begingroup$

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.

  • 0
    (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.2011-11-02

0 Answers 0