Can the Sets A and B be Considered Elements in a Sigma-Algebra?

[P3X-018]

If we define two sets as

A = \left\{ (x,y)\in \mathbb{R} \vert x+y<a \right\}
B = \left\{ (x,y)\in \mathbb{R} \vert x-y<a \right\}

and then define the system of sets,

\mathbb{D} = \left\{A\vert a\in \mathbb{R}\right\}\cup\left\{B\vert a\in \mathbb{R}\right\}

and then let \sigma(\mathbb{D}) be the \sigma-algebra generated by the system of set \mathbb{D}. To show that the sets

C = \left\{ (x,y)\in \mathbb{R} \vert x+y\leq a \right\}
D = \left\{ (x,y)\in \mathbb{R} \vert x-y\leq a \right\}

are elements in \sigma(\mathbb{D}) for every a\in \mathbb{R}.
Some questions arises. Can the sets A and B be considered as elements in \mathbb{D} and hence in \sigma(\mathbb{D}) too, or are they just subsets of \mathbb{D}?
Now I want to express C as a union of A and the set S = \left\{ (x,y)\in \mathbb{R} \vert x+y=a \right\}, and similar for D. But then I have to express the new set in some way to show that it is contained in \sigma(\mathbb{D}). (Maybe there are easier approaches?). I was considering to write the new set S as

S = \bigcap_{n=1}^{\infty}\left\{ (x,y)\in \mathbb{R} \vert x+y<a+\frac{1}{n} \right\}\cap \left\{ (x,y)\in \mathbb{R} \vert x+y\geq a \right\}

Where the last set is the complement of A. But will this set even converge towards S, or will it just be empty?

 

[EnumaElish]

I think A and B are not subsets. {A} and {B} are subsets. But, isn't \sigma defined over subsets of \mathbb{D}?

I think S is empty, but \bar S, defined with an < (instead of an <) is non-empty. See http://en.wikipedia.org/wiki/Nested_sequence_of_intervals

 

[P3X-018]

So A and B are elements in \sigma(\mathbb{D}), right? And yes \sigma(\mathbb{D}) is defined as the smallest \sigma-algebra containing \mathbb{D}.

Hmm... Now you have changed your answer of weather S is empty or not. I thought that maybe it would be non-empty because of the way we (in class) defined an interval of the type (a,b] as

(a,b] = \bigcap_{n=1}^{\infty}\left(a,\, b+\frac{1}{n}\right)

as a intersection of open intervals.
But hmm... then again maybe this wasn't even a good inspiration, since it seems different now that I type it...

But how would you suggest to write S from the 'building blocks' of \sigma(\mathbb{D})?

EDIT:
The question really is just, will the infinity intersection be

\bigcap_{n=1}^{\infty}\left\{ (x,y)\in \mathbb{R} \,\vert\, x+y&lt;a+\frac{1}{n}\right\} = \left\{ (x,y)\in \mathbb{R} \,\vert\, x+y\leq a\right\}

Because that is the set I'm interested in. And that's why I considered that if (a,b] can be writting in that way, why can't the above equation be true?
By this expression S becomes completely useless, since I can just set C equal to that expression, and C will then consist of an infinite intersection of A sets. But then we're back to my first question, are A and B elements in the \sigma-algebra even though it is generated by the union of A and B, so that C is can be considered as an element in the algebra?

 

[P3X-018]

Ok instead of making yet another edit, I'll make a new reply, getting confusing wit so edits now.
I think I was confused about \mathbb{D}, since I saw it a union of 2 sets, but it's actually a union of 2 system of sets. So the \mathbb{D} is actually,

\mathbb{D} = \left\{ A \subset \mathbb{R}^2 \,\vert\, A\in \{A_a\,\vert\, a\in\mathbb{R}\}\quad\mathrm{or}\quad A\in \{B_a\,\vert\, a\in\mathbb{R}\} \right\}

and I think that it is clear from this, that A and B are contained in \mathbb{D} and hence also \sigma(\mathbb{D}).

 

[EnumaElish]

I think I agree with
\bigcap_{n=1}^{\infty}\left\{ (x,y)\in \mathbb{R}^2 \,\vert\, x+y&lt;a+1/n\right\} = \left\{ (x,y)\in\mathbb{R}^2\vert x+y\le a\right\}

I think I was right the first time, then confused myself with nested intervals.