Let $A_n$ be open and dense subsets of $\mathbb R^n$ for each $n\in \mathbb N$. We need to prove that the intersection of all the $A_n$ is dense. (Note: we have not studied Baire Category Theorem)
I know that we can prove that a set $S$ is dense in some set $A$ if for every nonempty open set $U\subset A$, $U\cap S\neq\emptyset$. So my logic is that take some $U\subset A$, and prove $U\cap(A_1\cap(A_2\cap(…A_n)…)\neq\emptyset$ by using the fact that each of those sets is open and dense. Is this logic correct? If so how would I get this $U$?
Best Answer
Let $W$ be nonempty open set.
Obviously we can define sets $V_n$ such $V_n \subset (\cap_{1 \le i \le n} A_i$) $\cap W$, $\overline{V_n}$ $\subset V_{n - 1}$ and $diam\ V_n \to 0$.
If $v_i \in V_i$ then $v_i$ is сauchy sequence.
So $\cap_{1 \le i \le \infty} V_i$ = $\{\lim v_i\}$ and $\cap_{1 \le i \le \infty} V_i$ $\subset (\cap_{1 \le i \le \infty} A_i$) $\cap W \ne \emptyset$.