Difference between complete and closed set

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.

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.

Completeness asks for a space: does every Cauchy sequence converge to a limit in that space? This is the case for R.

A subset of a space is closed if it contains its limit points. It should be intuitive that if you are a subset of R, then any sequence in your subset that converges must converge in R. Now the question is: will that point still be in my set? If so, it is closed. That's why, by definition, a closed interval in a complete space must be complete.

Steps towards showing that any closed interval in R is complete.