(Criticize the proof): Closed subset of metric space is complete.

Question

Suppose $X$ is a closed subset, then if $\{x_n\}$ is a Cauchy sequence in $X$ which converges to $x \in Y$ in some bigger space $Y$, then I show that $x$ is a limit point of $X$ by the following argument, every ball around $x$ will contain some $x_n$ and hence every ball intersects $X$. Hence $x \in X$. Therefore every Cauchy sequence is convergent. This proves that every closed set is complete, which we know is wrong. So where did I made a mistake?

Answer 1

I think you have proven that every closed subset $X$ of a complete metric space $Y$ is itself complete.

If $Y$ is not complete, even if $X$ is closed in it, you cannot invoke some bigger complete space $Y'$, because then it's not guaranteed anymore that $X$ is closed in $Y'$.

Try with concrete examples: take $X=[0,1]\cap\mathbb Q$. $X$ is closed in $\mathbb Q$, but $\mathbb Q$ itself is not complete. On the other hand, you can embed $\mathbb Q\subset\mathbb R$, but then $X$ is not closed in $\mathbb R$ anymore.

Answer 2

Your argument fails for $X\subset Y$ dense. If we take $(x_n)_n$ the sequence of decimal approximations (by defect) of $\sqrt 2$, then $(x_n)_n$ is a Cauchy sequence in $\mathbb Q$ but does not converge in $\mathbb Q$. However, for any $\epsilon>0$ we have $(-\epsilon+\sqrt 2,\epsilon+\sqrt 2)\cap \mathbb Q \neq \emptyset$, since $\mathbb Q$ is dense in $\mathbb R$.

Answer 3

If $(Y,d)$ is a complete metric space then for $X \subseteq Y$, the space $(X,d)$ (same metric, restricted to $X$) is complete iff $X$ is closed in $Y$.

If $X$ is closed: take a Cauchy sequence $(x_n)_n$ in $(X,d)$. This is also a Cauchy sequence in $(Y,d)$ (this only depends on the mutual distances of points of the form $x_n$, and these are measured with $d$ in both cases). As $Y$ is complete under $d$, there is some $y \in Y$ such that $x_n \to y$. As all $x_n$ are in $X$, $y$ is in the closure of $X$, hence $y \in X$ (as $X$ is closed). So $y$ is also the limit of $(x_n)_n$ in $(X,d)$.

If $(X,d)$ is complete: suppose $y$ is in the closure of $X$. Then there is a sequence $(x_n)_n$ from $X$ that converges (in $(Y,d)$) to $y$. As a convergent sequence is Cauchy (this holds in all metric spaces) $(x_n)_n$ is a Cauchy sequence in $(X,d)$ as well. As $(X,d)$ is by assumption complete, there is some $x \in Y$ such that $x_n \to x$ under $d$. As limits of sequences are unique, $y = x \in X$ and so $y \in X$, and $X$ is indeed closed, as it contains all points of its closure.