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.


1 Answer 1


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 $\sigma(G)$.

  • $\begingroup$ 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? $\endgroup$ Sep 14, 2020 at 16:41

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