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I am trying to prove that collection of all finite dimensional cylinder sets is an algebra but not $\sigma$-algebra. Cylinder sets are defined as:

$\mathcal{B}_n$ is defined as the smallest $\sigma-$ algebra containing the rectangles $\{(x_1,x_2,\ldots,x_n): x_1 \in I_1,\ldots,x_n \in I_n \}$ where $I_1,I_2,\ldots,I_n$ are intervals in $\mathbb{R}$.

An $n$- dimensional cylinder set in $\mathbb{R}^{\infty} := \mathbb{R}\times \mathbb{R}\times...$ is defined as the set $\{\textbf{x}: (x_{1},x_2,\ldots,x_n) \in B\},\ B \in \mathcal{B}_n$.

I am unable to visualize how to prove that finite unions belong in the collection, and also why some infinite unions won't belong to the collection.

Can anyone help me with this ?

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I presume that by a cylinder set you mean a set of the form $B \times \mathbb{R} \times \mathbb{R} \times \cdots$, with $B \in {\cal B_n}$ for some $n$. Let ${\cal C}_n$ denote the $n$ dimensional cylinder sets and ${\cal C} = \cup_n {\cal C}_n$.

Note that if $C \in {\cal C}_n$ then $C \in {\cal C}_m$ for all $m \ge n$.

If $C_1,...,C_k$ are cylinder sets, then all of the sets are in some ${\cal C}_p$ for some $p$, hence $C_1 \cup \cdots \cup C_k \in {\cal C}_p$, hence ${\cal C}$ is closed under finite unions.

It might be easier to show that ${\cal C}$ is not closed under infinite intersections.

Let $C_k = \{0\}\times \cdots\times\{0\} \times \mathbb{R} \times \cdots$ ($k$ zeros), and note that $C=\cap_k C_k = \{(0,0,0,\cdots) \}$. We see that $C \notin {\cal C}$.

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  • $\begingroup$ Thanks a lot. You mean infinite intersections, right ? $\endgroup$ – pikachuchameleon May 13 '15 at 8:09
  • $\begingroup$ @AshokVardhan: Yes indeed, I fixed the typo! $\endgroup$ – copper.hat May 13 '15 at 14:47
  • $\begingroup$ Can we also say that $ \mathcal{C} $ cannot contain any singleton sets using the same logic that $\{(0,0,...\}$ cannot belong ? $\endgroup$ – pikachuchameleon May 14 '15 at 13:12
  • $\begingroup$ Every element of ${\cal C}$ ends in $\times \mathbb{R} \times \mathbb{R} \times \cdots$, so it cannot contain any singleton sets. $\endgroup$ – copper.hat May 14 '15 at 14:25
  • $\begingroup$ You say "it might be easier" - is it even possible to prove this through unions? $\endgroup$ – Dahn Oct 30 '15 at 10:34

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