Prove $\bigcap_\alpha P(A_\alpha)=P(\bigcap_\alpha A_\alpha)$
Where $P$ denotes the power set.
Definition: $\bigcap_\alpha A_\alpha=\{\forall\alpha\in A:x\in\ A_\alpha\}$
My work:
$x \in \bigcap_\alpha P(A_\alpha)\iff \forall\alpha\in P(A):x\in\ P(A_\alpha)\iff\forall\alpha\in A:x\in\ A_\alpha\iff \forall\alpha\in A:x\in\bigcap_\alpha A_\alpha $
Here i'm stuck. Can someone help me?
On
It's easier if you prove the two inclusions separately.
First a lemma: If $X\subseteq Y$, then $P(X)\subseteq P(Y)$.
Indeed, a subset of $X$ is also a subset of $Y$.
With the lemma at hand, and setting $A=\bigcap_\alpha A_\alpha$, we can conclude from $A\subseteq A_\alpha$ that $P(A)\subseteq P(A_\alpha)$. Since this holds for every $\alpha$, we have $$ P(A)\subseteq \bigcap_{\alpha}P(A_\alpha). $$
Conversely, suppose $C\in\bigcap_{\alpha}P(A_\alpha)$. Then $C\in P(A_\alpha)$, for every $\alpha$. Thus, by definition, $C\subseteq A_\alpha$, for every $\alpha$, and therefore $C\subseteq A=\bigcap_{\alpha}A_\alpha$, proving that $C\in P(A)$.
On
We show $$\bigcap_\alpha P(A_\alpha)=P(\bigcap_\alpha A_\alpha)$$
Note that an element of $\bigcap_\alpha P(A_\alpha)$ is a set which is a subset of every $A_\alpha$.
Thus it is a subset of the intersection $\bigcap_\alpha A_\alpha$
Therefore it is an element of the power set $ P(\bigcap_\alpha A_\alpha)$
On the other hand every element of $ P(\bigcap_\alpha A_\alpha)$ is a subset of $\bigcap_\alpha A_\alpha$
Thus it is a subset of every $A_\alpha$
Therefore it is an element of every $P(A_\alpha)$.
Thus it belongs to $\bigcap_\alpha P(A_\alpha)$
You should be more careful in your use of quantifiers and in the definition of arbitrary intersections. My proof: $$x \in \bigcap_{\alpha}\mathcal{P}(A_{\alpha}) \Leftrightarrow \forall \alpha(x \in \mathcal{P}(A_{\alpha})) \Leftrightarrow \forall \alpha(x \subseteq A_{\alpha}) \Leftrightarrow x \subseteq \bigcap_{\alpha}A_{\alpha} \Leftrightarrow x \in \mathcal{P}(\bigcap_{\alpha} A_{\alpha}) $$
The third logical equivalence is the only one that demands proof (although it is not hard to see, just note that conjunctions are distributive over the universal quantifier).