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 ?


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 $C \notin {\cal C}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.