Let $X$ be a real separable Banach space and $X^*$ be its topological dual. We call sets of the form $$C=\{x\in X~|~ (f_1(x),\dots,f_n(x))\in C_0\in B(\mathbb{R}^n)\}$$

a Cylinder set with Base $C_0$ and cut points $f_1\dots f_n$. I want to show the collection of cylinder sets constitute an algebra but I don't see how to define unions. Denote $\mathscr{C}$ the collection of cylinder sets, then choosing $C_0=\mathbb{R}$ and $f\neq 0$, we have $C=X\in \mathscr{C}$.

For $C_{C_0}=C=\{x\in X~|~ (f_1(x),\dots,f_n(x))\in C_0\in B(\mathbb{R}^n)\}\in \mathscr{C}$, we have $$C_{C_0^c}=\{x\in X~|~ (f_1(x),\dots,f_n(x))\in C_0^c\in B(\mathbb{R}^n)\}\in \mathscr{C}$$ Now for unions, I don't see how to define unions of sets with different cut off points, namely let for example $$A=\{x| (f_1(x),f_2(x))\in A_0\in B(\mathbb{R}^2)\},\quad B=\{x| (g_1(x),g_2(x),g_3(x))\in B_0\in B(\mathbb{R}^3)\}$$ with $f_1\neq g_1$ and $f_2\neq g_2$. How is the union $A\cup B$ represented ?


It's much easier to try to prove that the collection of cylinder sets is closed under finite intersections. Since you already know that it is closed under complements, this will be sufficient to see that it is an algebra.

If $$A = \{x \in X: (f_1(x), \dots, f_n(x)) \in A_0\} \qquad B = \{x \in X: (g_1(x), \dots, g_m(x)) \in B_0\}$$ then we have that $$A \cap B = \{x \in X: (f_1(x), \dots f_n(x), g_1(x), \dots, g_m(x)) \in A_0 \times B_0\}$$ and so $A \cap B$ is also a cylinder set.

It is also possible to just write down a representation of $A \cup B$ as a cylinder set directly. We have $$A \cup B = \{ x \in X: (f_1(x), \dots, f_n(x), g_1(x), \dots g_m(x)) \in (\mathbb{R}^n \times B_0) \cup (A_0 \times \mathbb{R}^m) \}.$$

To come up with this, you can either just see it is the right expression directly or deduce it from the simpler representations you already know for intersections and complements.

  • $\begingroup$ Where does it go wrong for countable intersection/unions? $\endgroup$ – badatmath Jul 20 at 13:38
  • $\begingroup$ If you had countable intersections/unions then the collection of cut points I use to write down the representation would no longer be finite. $\endgroup$ – Rhys Steele Jul 20 at 13:39

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