# is the set F = {A|A or A^c is finite} over R a sigma algebra?

• Apr 5th 2011, 08:02 AM
rosh3000
is the set F = {A|A or A^c is finite} over R a sigma algebra?
define $F = \{ A | A \mbox{ or } A^{c} \mbox{is finite} \} (\Omega = \mathbb{R}). \mbox{ Is F a } \sigma \mbox{-algebra over } \mathbb{R} ?$

My anwser: Yes as
a) if empty set = $A^{c}$ then A = $\mathbb{R} = \Omega$ which belongs to F
b) WLOG if A is finite let B = $A^{c}$ is uncoutnable but $B^{c}$ is fintire thus B belongs in F.
c) a union of finite sets is finite thus belong in F
• Apr 5th 2011, 08:11 AM
Plato
Quote:

Originally Posted by rosh3000
define $F = \{ A | A \mbox{ or } A^{c} \mbox{is finite} \} (\Omega = \mathbb{R}). \mbox{ Is F a } \sigma \mbox{-algebra over } \mathbb{R} ?$

My anwser: Yes as
a) if empty set = $A^{c}$ then A = $\mathbb{R} = \Omega$ which belongs to F
b) WLOG if A is finite let B = $A^{c}$ is uncoutnable but $B^{c}$ is fintire thus B belongs in F.
c) a union of finite sets is finite thus belong in F

For each $n\in\mathbb{Z}^+$ is it true that $\{n\}\in\mathcal{F}~?$

Is it true that $\bigcup\limits_n {\left\{ n \right\}}\in\mathcal{F}~?$
• Apr 5th 2011, 08:36 AM
rosh3000
Yes clearly finite,
No union is infinte and complement is uncountable infinte thus not in f

Thank you