Show Borel Subsets of R2 for Horizontal & Vertical Sections

[sephiseraph]

Let E in R² be a Borel Set. Show that all horizontal and vertical sections

{ x : (x, y) in E }, { y : (x, y) in E }

of E are Borel subsets of R.

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I think I'm missing something out. My argument is that E is Borel, so E is formed of finitely many unions, intersections and complements of open sets in R², each of which has horizontal and vertical sections which are open in R (I have already shown this). Therefore the horizontal and vertical sections of E are formed of countably many unions, intersections and complements of open subsets of R, thus they are Borel.

Am I wrong?
I have been advised to approach this with a view to showing that the family of subsets of R² whose horizontal section is a Borel subset of R is a sigma-algebra on R² containing all the open sets. Which doesn't quite agree with my approach.

Any help would be appreciated.

 

[mmwave]

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Your approach is on the right track, but there are a few key points that you are missing. Let's break down the problem step by step:

1. First, we need to define what a Borel set is. A Borel set is any set that can be formed by starting with open sets in R2 and applying a finite number of operations such as union, intersection, and complement. This means that every open set in R2 is a Borel set, since it can be formed by taking the union of infinitely many open sets.

2. Now, let's consider the horizontal section of E, denoted by { x : (x, y) in E }. This set is formed by taking all the x-coordinates of points in E. Since E is a Borel set, it can be formed by taking unions, intersections, and complements of open sets in R2. This means that the horizontal section of E can also be formed by taking unions, intersections, and complements of open sets in R. Therefore, the horizontal section of E is a Borel subset of R.

3. Similarly, the vertical section of E, denoted by { y : (x, y) in E }, is formed by taking all the y-coordinates of points in E. By the same reasoning as above, the vertical section of E is also a Borel subset of R.

4. Finally, we need to show that the family of subsets of R2 whose horizontal section is a Borel subset of R is a sigma-algebra on R2 containing all the open sets. To do this, we need to show that this family satisfies the three properties of a sigma-algebra: it contains the empty set, it is closed under complement, and it is closed under countable unions.

- The empty set is a Borel subset of R since it can be formed by taking the union of zero open sets.
- If A is a subset of R2 whose horizontal section is a Borel subset of R, then the complement of A is also a subset of R2 whose horizontal section is a Borel subset of R. This is because the complement of A can be formed by taking the complement of each open set used to form A, and these complements will still be open sets in R.
- Finally, if A1, A2, ... are subsets of R2 whose horizontal sections