# Is D a borel subset?

I need to show that $$D=\{(x,x)\in \mathbb{R}^2 :x \in [0,1]\}$$ is a closed set, and conclude that $$D \in B(\mathbb{R}^2)$$ (Borel subset).

I am a bit struggling to show that D is a closed set. Is it allowed to state $$D=[0,1]\times[0,1]$$ and $$[0,1]$$ is closed in $$\mathbb{R}$$, thus the cartesian product $$[0,1]\times[0,1]$$ 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 $$D \in B(\mathbb{R}^2)$$, 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?

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– 5xum
Nov 29 '21 at 9:49

Draw a picture. $$D$$ is not $$[0,1]\times [0,1]$$. $$D$$ is closed because $$(x_n,x_n) \to (u,v)$$ implies $$x_n \to u$$ and $$x_n \to v$$, so $$u=v$$ and $$(u,v) \in D$$.

Your argument for showing that $$D$$ is Borel set is correct.

Is it allowed to state $$D=[0,1]\times[0,1]$$

No, it is not allowed, because it is not true. The two sets are not identical. For example, $$(0,1)$$ is an element of $$[0,1]\times [0,1]$$, but it is not an element of $$D$$.

I thought we have that $$D∈B(\Bbb R^2)$$, 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. $$\mathbb R^2\setminus D$$ 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.

• Ah yes, thank you I understand it now. I indeed mistyped. Nov 26 '21 at 12:54