I have seen similar question to mine, but have not been able to find an approach such as this approach to this question. So the problem is to prove that any complete metric space without isolated points, is uncountable. My attempt so far has been the following: From the definition of isolated point we know that all such elements belong to the set $ X\setminus X' $. Since there is no such element, we conclude that $ X=X' $. Because $ X\neq\emptyset$, and $ X=X' $, we deduce that $ X $ is infinite. We maintain that $ X $ is uncountable. Suppose on the contrary that it is countable. Then we assume it to be $ X=\lbrace x_{1},x_{2},\cdots\rbrace $. Let $ x_{1}\neq y_{1}\in X $ and $ r_{1}<1 $ be respectively a point and a positive real number, such that $ x_{1}\notin N_{r_{1}}[y_{1}] $.(The notation $ N_{r_{1}}[y_{1}] $ simply means a close ball with center $ y_{1} $ and radius $ r_{1} $.) Then, we choose point $ x_{2}\neq y_{2}\in X $ and $ r_{2}<\frac{1}{2} $ to be a point and a positive real number such that$ N_{r_{2}}[y_{2}]\subset N_{r_{1}}(y_{1}) $(Should I state the usage of "Axiom of Choice" here?If yes,how?). Because $ X $ is an infinite set, we are allowed to repeat this process infinitely(Is this true?). Therefore we attain a chain of closed balls: \begin{align*} N_{r_{1}}[y_{1}] \supset N_{r_{2}}[y_{2}] \supset\cdots\qquad;r_{n}<\frac{1}{n} \end{align*} Let $ b_{n}=\lbrace N_{r_{n}}[y_{n}]\rbrace $ be a sequence in $ X $. We claim that $ \lbrace b_{n}\rbrace $ is Cauchy.(I have no idea how to proceed further.) Could you please assist me with the questions embedded in the text and suggest a way to complete the proof?
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It is a good idea to go via contradiction. Now take a look at the set $A_i:=X\setminus x_i$. What is the intersection $\cap_{i\in\mathbb{N}} A_i$? What property of $A_i$ can be deduced by the fact that $x_i$ is no isolated point and what follows for the intersection $\cap_{i\in\mathbb{N}} A_i$?
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