I am to prove $A_0\cap(\bigcup_{i=1}^n A_i) = \bigcup_{i=1}^n (A_0\cap A_i), n\ge 2$ by induction.
I started out like this:
Step 1: Prove that $A_0\cap(\bigcup_{i=1}^n A_i) = \bigcup_{i=1}^n (A_0\cap A_i)$ holds for $n = 2$
$$A_0\cap(A_1\cup A_2) = (A_0\cap A_1)\cup (A_0\cap A_2) $$
Step 2: Assume it holds true for $n = k$
$$A_0\cap(\bigcup_{i=1}^k A_i) = \bigcup_{i=1}^k (A_0\cap A_i)$$
Step 3: $n = k+1$
\begin{align*} \bigcup_{i=1}^{k+1} (A_0\cap A_i) &=\bigcup_{i=1}^k (A_0\cap A_i)\cup(A_0\cap A_{k+1}) \\ &= (A_0\cap A_1)\cup (A_0\cap A_2)...\cup(A_0\cap A_k)\cup(A_0\cap A_{k+1}) \\ &= \bigcup_{i=1}^{k+1}(A_0\cap A_{k+1}) \end{align*}
Am I right in this, or if not, where did I go wrong?
Assuming
$$A_0\cap(\bigcup_{i=1}^k A_i) = \bigcup_{i=1}^k (A_0\cap A_i)$$
Then it follows that
$$[A_0\cap(\bigcup_{i=1}^k A_i)] \cup [A_0\cap A_{k+1}] = [\bigcup_{i=1}^k (A_0\cap A_i)] \cup [A_0\cap A_{k+1}] $$
Hence (By Applying the "factoring" Property to the Left hand side)
$$A_0\cap[(\bigcup_{i=1}^k A_i) \cup A_{k+1}] = [\bigcup_{i=1}^k (A_0\cap A_i)] \cup [A_0\cap A_{k+1}] $$
Thus (re-indexing)
$$A_0\cap(\bigcup_{i=1}^{k+1} A_i) = \bigcup_{i=1}^{k+1} (A_0\cap A_i) $$ As is required