I'm trying to show that D = <mo fence="false" stretchy="false">{ ( x , x )

Aditya Erickson

Aditya Erickson

Answered question

2022-05-27

I'm trying to show that D = { ( x , x ) R 2 : x [ 0 , 1 ] } is a closed set, and conclude that D B ( R 2 ) (Borel subset).
And also I am a bit struggling to show that D is a closed set. Is it allowed to state D = [ 0 , 1 ] × [ 0 , 1 ] and [0,1] is closed in R , thus the cartesian product [ 0 , 1 ] × [ 0 , 1 ] is also closed?
And then I come to the next part. It is not really clear for me what a Borel sigma algebra means. I thought we have that D B ( R 2 ), because since D is open it is the complement of an open set. And since the Borel sigma algebra is the smallest sigma algebra containing all open sets, it must also contain all closed sets (by the properties of a sigma algebra.) Is this correct?

Answer & Explanation

pelankp

pelankp

Beginner2022-05-28Added 8 answers

Draw a picture. D is not [ 0 , 1 ] × [ 0 , 1 ]. D is closed because ( x n , x n ) ( u , v ) implies x n u and x n v, so u = v and ( u , v ) D.
Your argument for showing that D is Borel set is correct.
Cara Duke

Cara Duke

Beginner2022-05-29Added 2 answers

"Is it allowed to state D = [ 0 , 1 ] × [ 0 , 1 ]"
No, it is not allowed, because it is not true. The two sets are not identical. For example, (0,1) is an element of [ 0 , 1 ] × [ 0 , 1 ], but it is not an element of D.
"I thought we have that D B ( R 2 ), because since D is open it is the complement of an open set."
You are almost there. You probably mistyped, because D is not, in fact, open, it is closed. But yes, your reasoning is correct. R 2 D is an open set, and therefore in the sigma algebra. Because sigma algebras are closed under complements, D must also be in the sigma algebra.

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