Here is the version of the theorem being used in my class. It is from the textbook Introduction to Topology by Gamelin and Greene.
Beginning of proof
Let $\{U_n \}_{n=1}^{\infty}$ be a sequence of dense open subsets of a complete metric space $X$. Then $\cap_{n=1}^{\infty} E_n$ is also dense in $X$.
We went to show that $\cap_{n=1}^{\infty} U_n$ dense $\iff \overline{\cap_{n=1}^{\infty} U_n}=X$
$\iff \forall x\in X, x\in \overline{\cap_{n=1}^{\infty} U_n}$ ($x$ is adherent to $\cap_{n=1}^{\infty} U_n$)
$\iff \forall x\in X, \forall \epsilon >0$,$\exists y\in B_\epsilon(x)\cap(\cap_{n=1}^{\infty} U_n)$
Proof:
Let $x\in X$, $\epsilon >0$
Since $U_1$ is dense, $x\in X=\overline U_1 \iff \text{x is adherent to}\ U_1$. So we have that $\exists y_1\in B_\epsilon(x)\cap U_1$.
Since $U_1$ is open, $\exists r_1>0$ such that $B_{r_1}(y_1)\subset U_1$
Since $U_2$ is dense, $\exists y_2\in B_{r_1}(y_1) \cap U_2$ ($y_1$ adherent to $U_2$)
Since $U_2$ is open, $\exists r_2$ such that $B_{r_2}(y_2)\subset U_2$
By shrinking $r_1$, we may arrange that $r_1<1$ and $\overline{B_{r_1}(y_1)}\subset U_1\cap B_\epsilon(x)$.
By shrinking $r_2$, we may arrange that $r_2<\frac{1}{2}$ and $\overline{B_{r_2}(y_2)}\subset U_2\cap B_{r_1}(y_1)$.
Continuing in this manner, we get that $\{y_n\}\subset X$ and a sequence $\{r_n\}$ such that $0<r_n<\frac{1}{n}$.
$\overline{B_{r_n}(y_n)}\subset U_n\cap B_{r_{n-1}}(y_{n-1})$.
It follows that $\overline{B_{r_n}(y_n)}\subset B_{r_{n-1}}(y_{n-1})\subset B_{r_{n-2}}(y_{n-2})\subset … \subset B_{r_1}(y_1)\subset B_\epsilon(x)$
If $m>n$, $y_m\in B_{r_n}(y_n) \iff d(y_n,y_m)<r_n \to 0$ $\implies \{y_n\}$ is Cauchy.
Since $X$ is complete, $y_n\to y\in X$
Claim: $y\in B_\epsilon (x)\cap(\cap_{n=1}^{\infty} U_n)$ ($\iff y\in \overline{\cap_{n=1}^{\infty}U_n}$)
Proof of claim: If $m>n, y_m\in B_{r_n}(y_n)$ If we remove the first $n$ terms, this sequence is a subset of $B_{r_n}(y_n)$
So $\{y_{n+1},y_{n+2},…\}\to y$
$\implies y\in \overline{B_{r_n}(x_n)}\subset U_n$ (by definition of closure)
$\implies y\in B_\epsilon (x)$
$y\in \cap_{n=1}^{\infty} U_n$ ($y$ is in each $U_n$).
End of proof
Now, I am confused by the parts
"By shrinking $r_1$, we may arrange that $r_1<1$ and $\overline{B_{r_1}(y_1)}\subset U_1\cap B_\epsilon(x)$.
By shrinking $r_2$, we may arrange that $r_2<\frac{1}{2}$ and $\overline{B_{r_2}(y_2)}\subset U_2\cap B_{r_1}(y_1)$."
Is this because the intersection $U_1\cap B_\epsilon(x)$ is open so we can find $r_1$ small enough to stay in the set? Why can we conclude further that the closure of this open ball is contained in the intersection?
Best Answer
You are right for the first part. For your second question, we may find an $r<1$ such that$B_r(y_2)\subset U_2\cap B_{r_1}(y_1)$. Then we can take $r_2=\frac r2$, since in a metric space, by the triangle inequality, we have $\overline{B_r(x)} \subset B_{2r}(x)$.