Ytukyg

Textbook Text:

"In $mathbb{R}^{2}$, let $mathfrak{R}$ be the collection of all rectangles of the type $(a_1, b_1] times (a_2, b_2]$ where $-infty leq a_i < b_i leq infty$ for $i = 1, 2$ with the understanding that $(a_i, infty] = (a_i, infty)$.

Let $mathfrak{A}$ be the collection of all finite unions of members of $mathfrak{R}$. We have $mathfrak{R} subset mathfrak{A}$ since every $A in mathfrak{R}$ is the union of finitely many, actually one, members of $mathfrak{R}$ so that $A in mathfrak{A}$.­ We regard $emptyset$ as the union of 0 members of $mathfrak{R}$ so that $emptyset in mathfrak{A}$.

It is easily verified that $mathfrak{A}$ is an algebra of subsets of $mathbb{R}^2$."

Showing that $mathfrak{A}$ is an algebra of subsets of $mathbb{R}^2$:

1) A = $(-infty, infty] times (-infty, infty] = mathbb{R}^{2} in mathfrak{R}$, so $mathbb{R}^{2} in mathfrak{A}$.

2) By definition, $mathfrak{A}$ is the collection of finite unions of members of $mathfrak{R}$. So, let $C = cup_{i=1}^{N_1} E_{i}, D = cup_{i=1}^{N_2} F_{i}$ where $E, F$ are arbitrary collections of members of $mathfrak{R}$. Then $C, D in mathfrak{A}$ and $ C cup D = (cup_{i=1}^{N} E_{i}) cup (cup_{i=1}^{N} F_{i}) in mathfrak{A}$.

3) Let $A in mathfrak{A}$. From 2), if $A, A^{c} in mathfrak{A}$ then $A cup A^{c} = mathbb{R}^{2} in mathfrak{A}$. So $A in mathfrak{A} implies A^{c} in mathfrak{A}$.

Is this a correct proof to show that $mathfrak{A}$ is an algebra of subsets of $mathbb{R}^2$?

Edit:

2nd Attempt on proving $A in mathfrak{A} implies A^{c} in mathfrak{A}$

3) Let $A = cup_{i=1}^{N} (a_{1i}, b_{2i}] times (a_{2i}, b_{2i}]$

Then
begin{align*}
A^{c} &= cap_{i=1}^{N} Big[(a_{1i}, b_{1i}] times (a_{2i}, b_{2i}]Big]^{c}\
&= cap_{i=1}^{N} (-infty, a_{1i}] cup (b_{1i}, infty] times (-infty, a_{2i}] cup (b_{2i}, infty]\
&= (-infty, max_{iin N}(a_{1i})] cup ( min_{iin N}(b_{1i}), infty] times (-infty, max_{iin N}(a_{2i})] cup ( min_{iin N}(b_{2i}), infty]\
&= (-infty, max_{iin N}(a_{1i})] times (-infty, max_{iin N}(a_{2i})] cup (-infty, max_{iin N}(a_{1i})] times (-infty, max_{iin N}(b_{2i})] cup (-infty, max_{iin N}(b_{1i})] times (-infty, max_{iin N}(a_{2i})] cup (-infty, max_{iin N}(b_{1i})] times (-infty, max_{iin N}(b_{2i})]
end{align*}

Therefore as $A^{c}$ is a finite union of rectangles, $A in mathfrak{A} implies A^{c} in mathfrak{A}$.

An algebra is closed by finite intersections.

The intersection of two rectangles is a rectangle.

Thus the intersection of two finite unions of rectangles is a finite union of intersections of two rectangles; hence a finite union of rectangles.

The complement of (a,b]×(c,d] is

(oo,a]×R $cup$ (b,oo)×R $cup$ R×(oo,c] $cup$ R×(d,oo).

Thus the complement of a finite union of rectangles is the finite intersection of complements, which from above is a finite union of rectangles.

(-oo,oo]×(-oo,oo] = R$^2$ is incorrect.

The rectangles need to be defined to avoid that.