Intersection of collection of sets when given one is a subset of the other... I am trying to prove a proposition in Topology, however, in order to do so...it appears I need the following to be true. Hopefully, someone can validate this along with my proof. Thanks!
Is the following true:
Let $\{I_{\alpha}\}_{\alpha \in A}$ and $\{E_{\beta}\}_{\beta \in B}$ be two arbitrary collection of sets. If $\{I_{\alpha}\}_{\alpha \in A} \subseteq \{E_{\beta}\}_{\beta \in B}$, then the following inclusion holds:
$$\bigcap_{\beta\in B} E_{\beta} \subseteq \bigcap_{\alpha \in A} I_{\alpha}.$$
Quite naturally, I can think of examples where this holds. For example, take the following two collection of sets:
$$I = \{\{a,b\}, \{a,b,c\}, \{a,b,c,d\}\} \implies \bigcap I = \{a,b\}.$$
$$E = \{\{a\},\{a,b\}, \{a,b,c\}, \{a,b,c,d\}, \} \implies \bigcap E = \{a\}.$$
And so we have that $I \subseteq E$ and $\bigcap E \subseteq \bigcap I$.
(My attempted) Proof:
Let $x \in \bigcap_{\beta\in B} E_{\beta}$, then for every $\beta \in B$ we have $x \in E_{\beta}$. This is where I get stuck. I know (probably) that at this step I need to use the fact that
$\{I_{\alpha}\}_{\alpha \in A} \subseteq \{E_{\beta}\}_{\beta \in B}$, however, I am not sure how to properly word this... maybe something like...for each $\alpha \in A$, there exists a $\beta \in B : 
I_{\alpha} = E_{\beta}$. Therefore, for each $\alpha \in A$ we have that $x \in I_{\alpha}$. This implies that $x \in \bigcap_{\alpha \in A} I_{\alpha}$. Hence, the following inclusion holds:
$$\bigcap_{\beta\in B} E_{\beta} \subseteq \bigcap_{\alpha \in A} I_{\alpha}.$$
Any help would be great, thanks!
 A: The indices actually make this harder than necessary. Suppose that $\mathscr{A}$ and $\mathscr{B}$ are collections of sets such that $\mathscr{A}\subseteq\mathscr{B}$; then $\bigcap\mathscr{B}\subseteq\bigcap\mathscr{A}$.
To prove this, suppose that $x\in\bigcap\mathscr{B}$; then $x\in B$ for each $B\in\mathscr{B}$. And $\mathscr{A}\subseteq\mathscr{B}$, so clearly $x\in A$ for each $A\in\mathscr{A}$, and therefore $x\in\bigcap\mathscr{A}$. Thus, $\bigcap\mathscr{B}\subseteq\mathscr{A}$.
Note that this is intuitively obvious when you think about it right: to get $\bigcap\mathscr{B}$ you’re intersecting all of the sets in $\mathscr{A}$ and possibly some more sets besides, and those extra sets, if then do anything at all, can only shrink the intersection even further.
A: Brian’s answer and intuition are great, and here is my pedagogically-inspired version of his index-free answer. I particularly wanted to give a proof that avoided “intuitively obvious” (or similar language, like “clearly.”)
Suppose that $\mathscr{A}\subseteq\mathscr{B}$. We need to show that $\bigcap\mathscr{B}\subseteq\bigcap\mathscr{A}$.
Let $b$ be a particular but arbitrary element of $\bigcap\mathscr{B}$. If we show that $b$ is in every element of $\mathscr{A}$, we can conclude that $b\in\bigcap\mathscr{A}$, and we are done.
Let $A\in\mathscr{A}$ be a particular but arbitrary element of $\mathscr{A}$. Because $\mathscr{A}\subseteq\mathscr{B}$, it follows that $A\in \mathscr{B}$. But $b$ is an element of $\bigcap\mathscr{B}$, so $b$ is an element of every set in $\mathscr{B}$, and in particular, $b$ is an element of $A$.
Since $A$ was arbitrary element of $\mathscr A$, $b$ is in every element of $\mathscr A$ and therefore in $\bigcap\mathscr A$.
A: One should think of the sets participating in an intersection as "adding more and more restrictions" to the resulting intersection.
The more restrictions you have, the smaller is the set you get in the end.
