Understanding Induction as it Applies to Proving $A \cup (\cap_iB_i) = \cap_i(A \cup B_i)$ I am grappling with the concept of induction, and I'm trying to apply the concept to different proofs. I seem to be missing a piece of the conceptual puzzle though. Consider the following proof. (Question at the end)
Given $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$, then for an arbitrary index set $I$, $A \cup (\cap_iB_i) = \cap_i(A \cup B_i)$.
Proof:
Let $\mathcal{A}(n)$ be the assertion that $A \cup (\cap_nB_n) = \cap_n(A \cup B_n)$. Clearly $\mathcal{A}(1)$ is true, since $A \cup (B_1 \cap B_2) = (A \cup B_1) \cap (A \cup B_2)$.
Suppose now that $\mathcal{A}(n)$ is true. Then we may consider:
\begin{align*}
  (A \cup ( \cap_nB_n)) \cap B_{n+1} &= (A \cup B_{n+1}) \cup \big((\cap_nB_n) \cap B_{n+1}\big) \\
                                     &= (A \cap B_{n+1}) \cup (\cap_{n+1}B_{n+1}) \\
\implies x \in A \cap B_{n+1} \;\; \text{or} \;\; x \in \cap_{n+1}B_{n+1} &\implies x \in A \cup (\cap_{n+1}B_{n+1})
\end{align*}
Thus we have $\big(A \cup (\cap_nB_n)\big) \cap B_{n+1}= A \cup (\cap_{n+1}B_{n+1})$, and by induction, $ A \cup (\cap_{n+1}B_{n+1}) = \cap_{n+1}(A \cup B_{n+1})$.\
$\therefore A \cup (\cap_iB_i) = \cap_i(A \cup B_i)$.
$\blacksquare$
My Question: We used induction to show that the left hand side of the equality was true. Do we also need to perform the same manipulation to show the right hand side is true, or does the "suppose step" allow us to immediately make the deduction once we've shown that it is true for the left hand side? If so, why? Basically, am I correct in saying and by induction... in the above proof?
My fear is that we need to show the right hand side also works with our induction step to prove equality, but I can't really explain why I think that's true (or not true). It seems like the proof above is sufficient, since our "suppose step" should be immediately applicable.
I appreciate any guidance. Thanks!
 A: Just to make sure we are on the same page on everything. You proved that $$(A\cup(\cap_n B_i))\cap B_{n+1} = A\cup(\cap_{n+1} B_i)$$ and you used induction to prove that since $ (A\cup(\cap_n B_n)) = \cap_n (A\cup B_n)$
then $$(A\cup(\cap_n B_i))\cap B_{n+1} = (\cap_n (A\cup B_i))\cap B_{n+1} = \cap_{n+1} (A\cup B_i)$$
Resulting in what you wanted by joining that last equation with the first one (just a note to make sure you are using the induction hypothesis the right way, since at first I thought you were trying to apply it to the case of n+1).
An equality is symmetric so if you proved it for the LHS and it is equal to what you mean by the RHS then it is proved.
However, what induction tells you is that formula works for every FIXED integer, fixed being the keyword here. Therefore, you can't assume it works for the index set of the whole set of natural numbers, you would have to extend it, by saying for example suppose it doesn't work for the entire set of natural numbers, and try to reach a contradiction to your result over all but fixed integers.
A thing worth mentioning, I believe, is that you didn't prove it for an ARBITRARY index set but rather for subsets of natural numbers. This wouldn't work for $\cap_{x\in (0,1)} (A \cup B_x)$ for example, which is why a a direct proof would be more useful.
Hope this answers your question
