Proof that in Сauchy metric space the union of a sequence of closed sets with no internal points has no internal points.
My attempt:
Let's proof the statement for two sets by the contradiction method: $$ \exists x_0 \in A_1 \cup A_2: \exists B(x_0,\epsilon) \subset A_1 \cup A_2 \implies x_0 \in A_1 \lor x_0 \in A_2 \lor x_0 \in A_1 \cap A_2. $$ $$ x_0 \in A_1 \implies B(x_0, \epsilon) \setminus \ \vartheta \subset A_2 \implies Int(B(x_0, \epsilon) \setminus \ \vartheta) \subset Int(A_2) = \emptyset \implies \operatorname{Int}(B(x_0, \epsilon) \setminus \ \vartheta) = \emptyset $$ $\operatorname{Int}(A)$ if a function that returns all inner points of the set.
$\vartheta \subset A_1$ is a set without inner points by definition of $A_1$ so we got the contradiction. If $x_0 \in A_2$ we got the same thing and if $x_0 \in A_1\cap A_2$ we'll pick another point. If $A_1 = A_2$ the statement is trivial.
By induction, we can use this statement for any finite collection of sets. Let the sequence $\{A_n|n\in\mathbb{N}\}$ satisfies the conditions of the statement. Let's define $\{B_n|n\in\mathbb{N}\}$ such that $\bigcup_{n \in \mathbb{N}}{B_n} = \bigcup_{n \in \mathbb{N}}{A_n}$ and $(B_i \cap B_j \neq \emptyset) \iff (i=j)$.
Then as I understand we need to define the sequence $\{x_n \mid n \in \mathbb{N}\}, x_n \to x_0, n \to \infty$ somehow to show that from the Cauchy condition we got the result, but I didn't realize how yet.
Thank you for your time.
I suppose by Cauchy metric sapce you mean complete metric sapce. I don't think a direct proof of this is possible. The result is an easy consequence of BCT (Baire Category Theorem). If $C_n$ is a sequence of closed sets and $C=\cup C_n$ contains an interior point then there is an open ball $B(x,r) \subset C$. Now the closed ball $\overset {-} B(x,r/2)$ is contained in $C$ and it is a complete metric space, hence of second category. By BRT at least one of the C$_n$'s must have an interior point.