# Why "cylinder sets"?

If $I$ is any set of indexes, we define $E^I=\{(x_i)_{i\in I}:x_i\in E\,\,\forall i\in I\}$, $E$ being any set. Subsets of $E^I$ of the form $C_J=\{x_i\in B_i\,\,\forall i\in J\}$, where $B_i\in\mathcal{A}\,\,\forall i\in J$, $\mathcal{A}$ is a $\sigma$-algebra on $E$ and $J\subseteq I$ is finite, are called "cylinder sets", and form a basis of the product $\sigma$-algebra $\mathcal{A}^{\otimes I}$. Why are these sets called "cylinder sets"? What is the origin of this name?

• Hint: see what (visually) happens when we deal with $\mathbb{R}^2$. Feb 6, 2015 at 11:08
• The cylinder sets of $\mathbb{R}^2$ are products of borel sets of the real line. For example rectangles, i.e. products of intervals. In $\mathbb{R}^3$, the cylinder sets are parallelepipeds. So @Kolmin why not "rectangles", "generalized rectangles", "rectangle sets" or "parallelepiped sets"? Feb 6, 2015 at 11:37
• Don't shoot the messanger... I did not invent the terminology. :D Btw, behind jokes, the hint I gave you is exactly the one I found in a book to see why cylinder sets are called in such a way. Moreover, regarding your terminology, it sounds a bit problematic, because – take $\Re^2$ for example – cylinder sets are not bounded above and below, i.e. they are not actually rectangles. Indeed, they are cylinders. ;) Feb 6, 2015 at 18:03
• To me, a cylinder is a solid with circular basis extending in the third dimension for a possibly infinite height. This can be seen as a rotation solid, thus matching this definiton. Now obviously you can get cylinders as cylinder sets if you view $\mathbb{R}^3$ as the product of the plane and the line, as a circle is Borel in the plane and the third dimension is an interval. However, if $\mathbb{R}^3$ is the product of three lines, it is not straightforward to get a cylinder, since you have to first create the circle. Feb 6, 2015 at 19:01
• @MrArsGravis . In geometry a cylinder is $D\times \mathbb R$ where $D$ is a disc. In measure theory the disc can be anthing. That's it. In maths it is futile to get headaches about nomenclature. Jan 25, 2023 at 13:28

If $$|J|=1$$, i.e. $$J=\{i\}$$ for some $$i\in I$$, then the cylinder sets are simply the preimages of measurable subsets of $$E$$ w.r.t. the projection$$\pi_i:E^I\to E.$$Now why are the preimages of a projection called cylinder sets? Of course this is motivated by a very specific example:
Let $$E^3$$ be a $$3$$-dim. euclidean space and $$E^2$$ a $$2$$-dim. subspace. Now we can consider the obvious projection $$E^3\to E^2$$ (i.e. $$x\in E^3$$ is mapped to the intersection of $$E^2$$ and the line through $$x$$ perpendicular to $$E^2$$) and in this case it should be clear why the preimages are referred to as cylinders: At least the preimages of circles in $$E^2$$ are (infinite) cylinders in the usual sense.
In case that you complain that $$E^3$$ is not a product space: Pick any 1-dim. subspace $$E^1$$ orthogonal to $$E^2$$ and consider the obvious bijection $$E^1\times E^2\to E^3$$.