Difference between complete and closed set

Question

What is the difference between a complete metric space and a closed set?

Can a set be closed but not complete?

Answer 1

A metric space is complete if every Cauchy sequence converges (to a point already in the space). A subset $F$ of a metric space $X$ is closed if $F$ contains all of its limit points; this can be characterized by saying that if a sequence in $F$ converges to a point $x$ in $X$, then $x$ must be in $F$. It also makes sense to ask whether a subset of $X$ is complete, because every subset of a metric space is a metric space with the restricted metric.

It turns out that a complete subspace must be closed, which essentially results from the fact that convergent sequences are Cauchy sequences. However, closed subspaces need not be complete. For a trivial example, start with any incomplete metric space, like the rational numbers $\mathbb{Q}$ with the usual absolute value distance. Like every metric space, $\mathbb{Q}$ is closed in itself, so there you have a subset that is closed but not complete. If taking the whole space seems like cheating, just take the rationals in $[0,1]$, which will be closed in $\mathbb{Q}$ but not complete.

If $X$ is a complete metric space, then a subset of $X$ is closed if and only if it is complete.

Answer 2

In some sense, a complete metric space is "universally closed": A metric space $X$ is complete iff its image by any isometry $i : X \to Y$ is closed.

Indeed, if $X$ is complete, $i(X)$ is a complete subspace of $Y$ so $i(X)$ is closed in $Y$; moreover, if $X$ is closed in its completion then $X$ is complete itself.

Answer 3

An interesting thing about you question need to be noticed. A complete metric space $Y$ is a metric space $(Y,d_Y)$ such that every Cauchy sequence determined by the metric $d_Y$ is convergent for some point of $Y$. A closed subset $Y$ of a metric space $X$ is a set such that every convergent sequence of points in $Y$ converges to a point of $Y$ - with the metric $(X,d).$

This raise a funny observation. Let us prove that every complete subspace $Y$ is closed in $X$; that is, if $Y$ is endowed with the same metric as X and is complete relative to its metric, then $Y$ is closed in $X.$

Proof: Let $d_Y: Y \times Y \rightarrow \mathbb{R}$ be the induced metric of $d: X \times X \rightarrow \mathbb{R}.$ Then if $y \in X$ is limit of a convergent sequence of points in $Y$ with the metric $d$, say $\left(y^{k}\right)_{n\in\mathbb{N}}$, then for each $\epsilon>0$ there exist a $N_{\epsilon}$ such that $d(y_m,y_n)<\epsilon$ for all $n,m>N_{\epsilon}.$ Now, note that we can evaluate the function $d_Y$ at $(y_m,y_n),$ since each coordinate belongs to $Y$. Then for each $\epsilon>0$ there exist a $N_{\epsilon}$ such that $d_{Y}(y_m,y_n)<\epsilon$ for all $n,m>N_{\epsilon}.$ Hence $\left(y^{k}\right)_{n\in\mathbb{N}}$ is Cauchy and there exist a point $\bar{y} \in Y$ such that $d_Y(y^{k}, \bar{y}) \rightarrow 0$. Since $d$ is equal to $d_Y$ for every point of $Y$, then $d(\bar{y},y) \leq d(\bar{y},y^{k})+d(y^{k},y)\rightarrow 0$. Finally, we can conclude that $\bar{y}=y$ with $\bar{y} \in Y$ and $y$ belongs to $Y.$

This is a detailed proof. I found it quite funny to prove it without using a circular argument.