In Rudin's Real and Complex Analysis, he mentions two (equivalent) forms of Baire's theorem,
$(1)$: for a complete metric space $X$, the intersection of every countable collection of dense open subsets of $X$ is dense in $X$, and
$(2)$: no complete metric space is of the first category
It's not hard to see that $(1) \implies (2)$
For $(2) \implies (1)$, I'm wondering if the following is correct:
For contrapositive, assume that $\{U_i\}_{i=1}^\infty$ is a collection of dense open sets in $X$ s.t. $A = \cap_{i=1}^\infty U_i$ is not dense in $X$. Then there is $x \in X$ s.t. x is not a limit point of $A$, and so there is a ball $B(x; \delta)$, $\delta > 0$, not in $A$. So $B(x; \delta) \subset A^c = \cup_{i=1}^\infty (U_i)^c$.
But then the complete metric space formed by $Y = B(x; \delta)$ is the union of the nowhere dense sets $\{(U_i)^c \cap Y\}_{i=1}^\infty$, contradicting $(2)$.
Initially I tried to show that if $(1)$ fails to hold for a complete metric space $X$, then $X$ is of the first category. But I take it this is not necessarily possible to show (and not necessary for the proof)?
Background: I'm just trying to verify that I understood the contents of this post: The equivalence of different forms of Baire's category theorem in Rudin's Real and Complex Analysis
Best Answer
Your proof is correct provided you already know that there is an equivalent metric on $B(x,\delta)$ which makes it complete. With the original metric this is not complete (except in trivial cases).
If $V$ is an open set in a complete metric sapec $(X,d)$ then $D(x,y)=d(x,y)+|\frac 1 {d(x,X\setminus V)}-\frac 1 {d(y,X\setminus V)}|$ defines an equivalent metric which makes $V$ complete.