Show that if $A \subset B$ then $\sigma(A) \subset \sigma$ (B) Hello I am trying to show that for subsets $A,B \in \Omega$ with $A \subset B$ then $\sigma(A) \subset \sigma(B).$ The textbook I am studying is Probability and Measure Theory Second Edition by Robert B. Ash and the definition given in the book for a $\sigma$ field of a collection of subsets $A \in \Omega$ is such that $\sigma(A)$ is closed under complementation and countable union i.e.:
$\sigma(A)$ satisfies these properties:


*

*$\Omega \in \sigma (A).$

*If any subset $X \in \sigma(A)$ then $X^c \in \sigma(A).$

*If $X_1, X_2,  X_3, \ldots, \in \sigma (A)$ then $\left( \bigcup \limits_{i=1}^\infty X_i \right) \in \sigma(A).$
Based on this definition, to me it would seem rather obvious that A $\subset$ B would imply $\sigma(A) \subset \sigma(B)$ since the $\sigma$ field of a collection of sets that belong in the domain $\Omega$ expands the original collection of sets to include all possible combinations of the complements and unions of the original elements in the collection of sets so the assumption of $A \subset B$ would imply all the elements of $A$ are in $B$ so taking the $\sigma$ field of $B$ is really taking the $\sigma$ field of $A$ and then adding to it the $\sigma$ field of $(A - B)$ or $A \cap B^c$ i.e. $\sigma(B) = \sigma(A) + \sigma(A - B) = \sigma(A) \cup \sigma(A - B).$ To me this proof seems reasonable enough to lead to the result of $\sigma(A) \subset \sigma(B)$ but I just wanted to make sure if my argument is reasonable enough to lead to this conclusion or if there is anything else that can be added to strengthen it. 
 A: You've got the right idea, but you could be a bit more detailed by noting that $\sigma(A)$ is the smallest $\sigma$-algebra containing $A$ (i.e., it's the intersection of all $\sigma$-algebras containing $A$). 
If $A \subseteq B$, then any $\sigma$-algebra ${\cal F}$ containing $B$ will also contain $A$.   Thus, the intersection of all $\sigma$-algebras containing $A$ must be a subset of the intersection of all $\sigma$-algebras containing $B$.  If $B$ is much larger than $A$, the reverse inclusion need not hold.
A: 
"...expands the original collection of sets to include all possible combinations of the complements and unions of the original elements in the collection of sets..."

That is a weak point, since it will only be acceptable as mathematical proof if you define exactly what you mean with "all possible combinations of the complements and unions" and secondly prove that $\sigma(\mathcal A)$ falls together with it.
To prove it correctly it must be shown that an intersection of $\sigma$-fields is again a $\sigma$-field. If that is done (and it is not really difficult) then it can be concluded that $\sigma(\mathcal A)$ is the intersection of all $\sigma$-fields on $\Omega$ that contain $\mathcal A$. 
Then from $\mathcal A\subseteq\mathcal B\subseteq\sigma(\mathcal B)$ it follows that $\sigma(\mathcal B)$ is one of these $\sigma$-fields that contain $\mathcal A$ as subcollection, so that $\sigma(\mathcal B)$ is automatically a superset of that intersection of $\sigma$-fields.
