Rudin RCA 5.6-7 Forms of Baire’s Theorem

Question

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

Answer 1

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.

Answer 2

In general (just from the topological definitions, no metric needed) we have that $X$ is not first category iff every countable intersection of open sense sets $U_n$ has non-empty (not dense) intersection. Completeness implies the latter easily enough.

But in a complete metric space, the completeness is "inside every open ball" (e.g. every closed ball $D(x,r)= \{y: d(x,y) \le r\}$ is also complete and the above fact holds within every open ball and so we get denseness of $\cap_n U_n$ (non-empty intersection in every open ball means dense in $X$).