Let $(X,d)$ be a metric space: If $Y\subset X $ is a complete subspace then $Y$ is closed
From the lecture class I get the solution, however I have a question concerning one of the main step:
Solution: Assume $Y$ is not closed: Let $y\in \bar{Y}\backslash Y$ so $B_{1/n}(y)\cap Y\neq \emptyset$ $\forall n\in \mathbb{N}$.Pick for each $n\in\mathbb{N}$ an element $x_n\in B_{1/n}(y)\cap Y$. For $\varepsilon >0$ let $N^{-1}<\frac{\varepsilon}{2}$ for $n\in\mathbb{N}$. Then $\forall n,m \geq N$ \begin{equation} d(x_m,x_n)\leq d(x_m,y)+d(y,x_n)\leq \varepsilon. \end{equation} By construction $x_n \to y$ as $n\to \infty$. So $(x_n)_{n\geq 1}$ is a Cauchy-sequence, which limit is not in $Y$. So not closed implies not complete!
Question: Why is $B_{1/n}(y)\cap Y\neq \emptyset$ ?