# $P(X)$ from $\sigma - \text{algebra}$ on $\{(1,2),(2,3)\}$

The title says the most.

I've been asked to show that, for a set $$X= \{1,2,3,4\}$$, the set of subsets $$G=\{(1,2),(2,3)\}$$ has a sigma algebra $$\sigma(G) = P(X)$$.

I understand that the sigma algebra of $$G$$ can be shown to be $$\sigma(G)=\{ (\emptyset,(2), (1,2),(2,3),(1,2,3),(1,2,3,4)) \}$$

But I can't figure out how to show that the remaining subsets are also part of the sigma algebra.

The standard notation should be $$G=\{\{1,2\},\{2,3\}\}\;.$$ It is not that $$G$$ "has a $$\sigma$$-algebra $$\sigma(G)$$" but that the $$\sigma$$-algebra generated by $$G$$. By definition, $$\sigma(G)$$ is the smallest sigma algebra containing $$G$$.

I understand that the sigma-algebra of $$G$$ can be shown to be...

This is incorrect. What you want to show is that $$\sigma(G)=P(X)$$, the power set of $$X$$, i.e., the set of all subsets of $$X$$.

To show $$\sigma(G)=P(X)$$, you first show that $$P(X)$$ is a $$\sigma$$-algebra by definition. This gives you $$P(X)\supset \sigma(G)$$.

Second, you need to show that $$\sigma(G)\supset P(X)$$.

By taking intersections, it is easy to show that $$\emptyset,\{k\}\in\sigma(G),\quad k=1,2,3,4.$$ Taking unions, you then show that any nonempty subset of $$X$$ is an element of $$\sigma(G)$$.

• SO for the second part, you essentially show that the unions of $\emptyset$ and K gives the subsets $\{1\}, \{2\}, \{3\}, \{4\}, \{1,2\}, \{2,3\}$ and beyond? Sep 14, 2020 at 16:41