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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 $B_{1/n}(y)\cap Y\neq \emptyset$ ?

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  • $\begingroup$ Because y is in the closure of Y so y is a limit point of Y. $\endgroup$ – Zack Ni Aug 2 '16 at 7:47
  • $\begingroup$ That is the exact meaning of the fact that $y \in \bar{Y}/Y$ : the fact that $y \in \bar{Y}$ means that we can choose a sequence tending to $y$ whose elements are in $Y$, which are the $x_n$. However, $y$ itself is not in $Y$. $\endgroup$ – астон вілла олоф мэллбэрг Aug 2 '16 at 7:48
  • $\begingroup$ Ok thank you @ZackNi so maybe I have trouble with the definition of the closure: could you recall it to me? $\endgroup$ – MorganeMaPh Aug 2 '16 at 7:53
  • $\begingroup$ @sure a closure of set W is the set that includes all of the limit point of W. $\endgroup$ – Zack Ni Aug 2 '16 at 7:55
  • $\begingroup$ Ok but just to be sure: if $y$ is in $\bar{Y}$ then $B_{1/n}(y)$ is not completely IN $\bar{Y}$, right? $\endgroup$ – MorganeMaPh Aug 2 '16 at 7:57

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