Help with a sigma-algebra problem with random variables (show $\sigma(X_S)\subseteq \sigma(X_T)$ if $S\subseteq T$) My problem is as follows:

Let $X_S$ and $X_T$ be two stochastic processes where $S,T$ are index sets. Let $\sigma(X_S)$ and $\sigma(X_T)$ denote the sigma-algebra generated by $X_S$ and $X_T$. Show that $\sigma(X_S)\subseteq \sigma(X_T)$ if $S\subseteq T$.

Here, sigma-algebra generated by a random variable is defined as the following.

Suppose $X:(\Omega,\cal{F}) \to$ $(E,\cal E)$, then the sigma-algebra generated by $X$ is $\sigma(X)=\sigma\{X^{-1}(A):A\in \cal E \}$, i.e. the sigma-algebra generated by $\{X^{-1}(A):A\in \cal E \}$.

This claim is actually very intuitive, but I need help to write a formal proof. I can give an intuitive example of this claim for the case of finite index sets.

Suppose $S=\{1\}$ and $T=\{1,2\}$ and so $X_S=\{X_1\}$ and $X_T=\{X_1,X_2\}$. Let $\Omega$ be a set of students, and treat $X_1,X_2$ as some properties for each $\omega\in \Omega$. Now let $X_1(\Omega)=\{male, female\}$, $X_2(\Omega)=\{undergrad, grad\}$.
Clearly $\sigma(X_S)=\{\emptyset, \Omega, m, f\}$ where "$m$" denotes the set of all male students for convenience and so is "$f$" for the set of all female students.
Likewise, $\sigma(X_T)=\{\emptyset, \Omega,m\&u, f\&u,m\&g,f\&g,m,f,u,g\}$ where "$u$" denotes the set of all undergraduate students and "$g$" denotes the set of all graduate students. I use things like $m\&g$ to represent students carrying both properties.
Thus we can see $\sigma(X_S)\subseteq \sigma(X_T)$.

To write out a formal proof, we need to show for any $A\in \sigma(X_S)$ we have $A\in \sigma(X_T)$. I can intuitively explain this is true but anyone can help with a formal proof? I am new to such stuff so I really need help. Thank you!
 A: Let $(\Omega,\mathcal{A})$ and $(E,\mathcal{F})$ be measurable spaces and $X: (\Omega,\mathcal{A}) \to (E,\mathcal{F})$ be a mapping. Then $X$ is measurable if, and only if,
$$X^{-1}(F) \in \mathcal{A}$$
for any $F \in \mathcal{F}$. Since, by definition, $\sigma(X)$ is the smallest $\sigma$-algebra containing all these sets, this means that $\sigma(X)$ is the smallest $\sigma$-algebra $\mathcal{A}'$ such that $X: (\Omega,\mathcal{A}') \to (E,\mathcal{F})$ is measurable. If we want to show the relation $\sigma(X) \subseteq \mathcal{A}'$ for some sigma-algebra $\mathcal{A}'$, it therefore suffices to check that $X: (\Omega,\mathcal{A'}) \to (E,\mathcal{F})$ is measurable.

So, suppose that $S \subseteq T$. Then, by definition of $\mathcal{A}' := \sigma(X_T)$, we know that $X_t: (\Omega,\mathcal{A}') \to (E,\mathcal{F})$ is measurable for any $t \in T$. In particular, for $s \in S \subseteq T$, we get that $$X_s: (\Omega,\mathcal{A}') \to (E,\mathcal{F})$$ is measurable. Hence, by the above considerations, $\sigma(X_s) \subseteq \sigma(X_T)$. Since this holds for any $s \in S$, this proves $\sigma(X_S) \subseteq \sigma(X_T)$.
