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Consider flipping a coin $n$ times. Define the sample space as $$ \Omega = \{(r_1,r_2,r_3,\dots); r_i = 0 \text{ or }1\} $$ Define subsets of the sample space as $$ A_{a_1a_2\dots a_n} = \{(r_1,r_2,\dots )\in \Omega; r_i =a_i \text{ for } 1\leq i \leq n\} $$ where $r_i$ is $0$ if the $i$th coin flip is tails and $1$ if it is heads.

Define a set $\mathcal{J}$ by $$ \mathcal{J} = \{ A_{a_1a_2\dots a_n}; n\in \mathbb{N}, a_1,a_2,\dots ,a_n \in \{ 0,1\}\} \cup \{\emptyset , \Omega\} $$

I want to show that $\mathcal{J}$ is a semi-algebra, and am currently thinking about whether it is closed under finite intersections.

My question is: If $B = A_{a_1a_2\dots a_m}$, $C = A_{a_1a_2\dots a_n}, m<n$, then $B\cap C$ = C, correct? Because I believe $C\subset B$ (since $C$ restricts more values of $r_i$ than $B$ does).

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This is indeed true (if we have the same $a_i$ values on both). But these sets cannot both be in $\mathcal{J}$, as in your definition, we only have sets with exactly $n$ conditions. Note that you define everything for $\Omega$ with sequences of length $n$ only, which is weird considering the question title.

And these are all pairwise disjoint (if only one $a_i$ differs, the conditions are incompatible), which makes showing them to be a semi-algebra pretty trivial: if $A,B \in \mathcal{J}$ and $A \neq B$ (and both unequal to $\emptyset, \Omega$, as these are trivial cases), then $A \cap B = \emptyset$ and $A \setminus B = A$.

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  • $\begingroup$ Thank you very much for your help. Also, regarding the weird definition of $\Omega$, I just realized I have (had, since I edited it out) a typo in the post. Sorry. $\endgroup$ – majmun Jan 4 '16 at 23:25
  • $\begingroup$ Wait, sorry by I fail to see how my definition implies sets with exactly $n$ conditions? Shouldn't it apply for all $n$ such that $n\in \mathbb{N}$? Ah, perhaps that was a consequence of the typo? If so, sorry but then it changes your answer. At least, it changes the latter part of your answer. I am still marking your answer as accepted because your first sentence does indeed answer what I was asking. $\endgroup$ – majmun Jan 4 '16 at 23:57

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